Answer: B, Companies passed on production and transportation costs to consumers
Explanation:
A higher oil price occurred when companies passed on production and transportation costs to consumers.
Cause of high price of oilThe oil producing companies spend so much money in producing crude oil from the reservoirs to the surface. They also spend money in processing and transporting the crude oil to the end users or consumers.
The final price of the oil depends on the total amount spent by these companies in producing the hydrocarbons.
Thus, a higher oil price occurred when companies passed on production and transportation costs to consumers.
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A mechanic applies a force of 60N at a distance of 80 cm from the pivot on a wheel wrench. What is the size of the moment?
Answer:
48 Nm
Explanation:
Moment, or torque, is the cross product of radius and force vectors.
τ = r × F
τ = (0.80 m) (60 N)
τ = 48 Nm
A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring
Answer:
a) The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex], b) The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].
Explanation:
This question is incomplete and complete version will be presented herein:
A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring (b) Find the angular frequency of the block 's vibrations.
a) Since spring is hung from the ceiling and is stretched by action of gravity on 0.573 kilogram block. According to the Hooke's Law, force experimented by the spring is directly proportional to elongation. An expression describing the phenomenon is presented and described below: (System at equilibrium - Newton's Second Law)
[tex]m\cdot g = k\cdot \Delta x[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]\Delta x[/tex] - Spring linear deformation, measured in meters.
Now, the spring constant is cleared in this equation and outcome is computed: ([tex]m = 0.573\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta x = 0.198\,m[/tex])
[tex]k = \frac{m\cdot g}{\Delta x}[/tex]
[tex]k = \frac{(0.573\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.198\,m}[/tex]
[tex]k = 28.381\,\frac{N}{m}[/tex]
The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex].
b) Let suppose that mass-spring system is experimenting a simple harmonic motion, so that angular frequency is equal to:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
Given that [tex]k = 28.381\,\frac{N}{m}[/tex] and [tex]m = 0.573\,kg[/tex], the angular frequency, measured in radians per second, of the block is:
[tex]\omega = \sqrt{\frac{28.381\,\frac{N}{m} }{0.573\,kg} }[/tex]
[tex]\omega = 7.038\,\frac{rad}{s}[/tex]
The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].
A freight car moves along a frictionless level railroad track at constant speed. The freight car is open on top. A large load of coal is suddenly dumped into the car. What happens to the speed of the freight car
Answer:
The speed of the freight car decreases.
Explanation:
According to the law of conservation of momentum indicates that for colliding in an isolated system, the total momentum pre and post collision is same for the two objects this is done because the momentum that one item has lost is same for the momentum that the other received
In the given situation, the freight car travels at constant speed along a frictionless railroad line. The top floor freight car is open. Then a huge load of coal is dumped inside the car.
Therefore the speed of the freight car decreased by applying the law of conservation of momentum i
Unit conversion
The choices are in units
A,GA,MA,uA,kA,mA,nA,pA. Pick one the units
Answer:
1.234567 kA
Explanation:
The prefix k stands for kilo-, or 10³. The prefix m stands for milli-, or 10⁻³. The sum shown is ...
1.234 kA + 0.000567 kA = 1.234567 kA
An object, with mass 70 kg and speed 21 m/s relative to an observer, explodes into two pieces, one 4 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame
Answer:
K = 3.9 kJ
Explanation:
The kinetic energy ([tex]K_{T}[/tex]) added is given by the difference between the final kinetic energy and the initial kinetic energy:
[tex] K_{T} = K_{f} - K_{i} [/tex]
The initial kinetic energy is:
[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} [/tex]
Where m₁ is the mass of the object before the explosion and v₁ is its velocity
[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} = \frac{1}{2}70 kg*(21 m/s)^{2} = 1.54 \cdot 10^{4} J [/tex]
Now, the final kinetic energy is:
[tex] K_{f} = \frac{1}{2}m_{2}v_{2}^{2} + \frac{1}{2}m_{3}v_{3}^{2} [/tex]
Where m₂ and m₃ are the masses of the 2 pieces produced by the explosion and v₁ and v₂ are the speeds of these pieces
Since m₂ is 4 times as massive as m₃ and v₃ = 0, we have:
[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}v_{2}^{2} + \frac{1}{2}*\frac{1}{5}m_{1}*0 [/tex] (1)
By conservation of momentum we have:
[tex] p_{i} = p_{f} [/tex]
[tex] m_{1}v_{1} = m_{2}v_{2} + m_{3}v_{3} [/tex]
[tex] m_{1}v_{1} = \frac{4}{5}m_{1}v_{2} + \frac{1}{5}m_{1}*0 [/tex]
[tex] v_{2} = \frac{5}{4}v_{1} [/tex] (2)
By entering (2) into (1) we have:
[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}(\frac{5}{4}v_{1})^{2} = \frac{1}{2}*\frac{4}{5}70 kg(\frac{5}{4}*21 m/s)^{2} = 1.93 \cdot 10^{4} J [/tex]
Hence, the kinetic energy added is:
[tex] K_{T} = K_{f} - K_{i} = 1.93 \cdot 10^{4} J - 1.54 \cdot 10^{4} J = 3.9 \cdot 10^{3} J [/tex]
Therefore, the kinetic energy added to the system during the explosion is 3.9 kJ.
I hope it helps you!
a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B suspended by a cord of length l =2.2m. The block then swings through a maximum angle of theta = 60. Determine (a) the initial speed of the bullet vo, (b) the impulse imparted by the bullet on the block, (c) the force on the cord immediately after the impact
Answer:
(a) v-bullet = 399.04 m/s
(b) I = 2.38 kg m/s
(c) T = 2.59 N
Explanation:
(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.
The potential energy is given by:
[tex]U=(M+m)gh[/tex] (1)
U: potential energy
M: mass of the wood block = 1 kg
m: mass of the bullet = 6g = 6.0*10^-3 kg
g: gravitational constant = 9.8m/s^2
h: distance to the ground
The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:
[tex]h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m[/tex]
The potential energy is:
[tex]U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J[/tex]
Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:
[tex]U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}[/tex]
Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:
[tex]Mv_1+mv_2=(M+m)v[/tex] (2)
v1: initial velocity of the wood block = 0m/s
v2: initial speed of the bullet
v: speed of bullet and block = 2.38m/s
You solve the equation (2) for v2:
[tex]M(0)+mv_2=(M+m)v[/tex]
[tex]v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}[/tex]
The speed of the bullet before the impact with the wood block is 399.04 m/s
(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:
[tex]I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}[/tex]
The impulse is 2.38 kgm/s
(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:
[tex]T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N[/tex]
The force on the cord after the impact is 2.59N
Answer:
The initial speed of the bullet [tex]V_o = 777.97m/s[/tex]The force on the cord immediately after the impact = [tex]19.71N[/tex]Explanation:
Apply the law of conversion of energy
[tex]V_f = \sqrt{2gh}[/tex]
where,
h = height of which the bullet and block rise after impact
[tex]h = L - Lcos\theta\\\\h = 2.2 - (2.2*cos60)\\\\h = 1.1m[/tex]
Therefore,
[tex]V_f = \sqrt{2gh}\\\\V_f = \sqrt{2*9.8*1.1}\\\\V_f = 4.64m/s[/tex]
From conservation of momentum principle, [tex]m_Bv_B = 0[/tex]
[tex]m_ov_o + m_Bv_B = (m_b+m_B)V_f\\\\0.006V_o = (0.006+1)*4.64\\\\V_o = 777.97m/s[/tex]
C) The force in the cable is due to the centrfugal force of the system, which is due to the motion of the system is a curved path and weight of the system
[tex]F = \frac{m_b+m_B}{L}V_f^2 + (m_b+m_B)g\\\\F = \frac{0.006+1}{2.2}*4.64^2 + (0.006+1)9.81\\\\F = 19.71N[/tex]
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You are watching an object that is moving in SHM. When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left. Part A How much farther from this point will the object move before it stops momentarily and then starts to move back to the left
Answer:
2.95m
Explanation:
The farthest distance the object can move is the radius of the circle of which the Simple harmonic motion is assumed to be a part
But V = w× r; where V is velocity,
w is angular velocity and r is radius.
Also,
a= w2r; where a is linear acceleration
but a = v× r ; by comparing both equations
Hence r = a/v =8.6/2.45 =3.51m
But the horizontal distance of the motion is given by:
X = rcosx ; where x is the angle
X is the distance covered.
We know that the maximum value of cos x is 1 which is 0°
When the object moves in a fashion directly parallel to an horizontal distance, maximum distance would be reached and hence:
X = r=3.51m
Meaning the object needs to travel 3.51-0.56=2.95m further.
Note: the acceleration of the motion is constant whether it is swinging towards the left or right.
When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left and the amplitude of motion A = 0.732 m.
What is Amplitude of motion?
The distance between the central and extreme points for a moving particle is known as the amplitude of motion.
The given data to find the amplitude of motion,
Object displaced = 0.560 m
Velocity = 2.45 m/s
Acceleration = 8.60 m/s²
Starting with sine:
x(t) = Asin(ωt)
so that t = 0, x = 0
x(t) = 0.56 m = Asin(ωt)
v(t) = x(t)'= 2.45 m/s = Aωcos(ωt)
a(t) = v(t)'= -8.60 m/s² = -Aω²sin(ωt)
x(t) / a(t) = Asin(ωt) / -Aω²sin(ωt)
0.56m / -8.60 m/s² = -1 / ω²
ω² = 15.3571 rad^2/s^2
ω = 3.91881 rad/s
x(t) / v(t) = Asin(ωt) / Aωcos(ωt)
0.560m / 2.45m/s = tan(3.91t) / 3.91rad/s
0.8937= tan(3.91t)
t = 0.176 s
x(0.176) = Asin(3.59×0.176)
0.65 m= Asin(0.631)
A = 0.732 m is the amplitude of motion.
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A worker with spikes on his shoes pulls on rope that is attached to a box that is resting on a flat, frictionless frozen lake. The box has mass m, and the worker pulls with a constant tension T at an angle θ = 40 ∘ above the horizontal. There is a strong headwind on the lake, which produces a horizontal force Fw that is pointed in the opposite direction than the box is being pulled. Draw a free-body diagram for this system. Assume that the worker pulls the box to the right. If the wind force has a magnitude of 30 N, with what tension must the worker pull in order to move the box at a constant velocity?
Answer:
a
The free body diagram is shown on the first uploaded image
b
The tension on the rope is [tex]T=39.16 \ N[/tex]
Explanation:
From the question we are told that
The mass of the box is m
The tension on the box is T
The angle at which it is pulled is [tex]\theta = 40^o[/tex]
The force produced by the strong head wind is [tex]Fw = 30 \ N[/tex]
At equilibrium the net force acting on the block along the horizontal axis is zero i.e
[tex]Tcos \theta -F_w = 0[/tex]
substituting values
[tex]Tcos (40) -30 = 0[/tex]
[tex]Tcos (40) = 30[/tex]
[tex]T(0.76604)) = 30[/tex]
[tex]T=39.16 \ N[/tex]
At a time when mining asteroids has become feasible, astronauts have connected a line between their 3220-kg space tug and a 6240-kg asteroid. They pull on the asteroid with a force of 362 N. Initially the tug and the asteroid are at rest, 311 m apart. How much time does it take for the ship and the asteroid to meet
-- F = m a ... ==> a = F/m
-- The tension in the rope is 362 N. That same force acts on the asteroid and on the tug, pulling them together.
-- The asteroid's acceleration is 362N / 6240 kg = 0.058 m/s², headed for a point on the rope somewhere between the asteroid and the tug.
-- The tug's acceleration is 362 N / 3220 kg = 0.112 m/s², also headed for a point on the rope somewhere between the tug and the asteroid.
-- So now we have a gap between them, initially 311 m long, closing with a speed that starts at zero and accelerates at 0.170 m/s² .
-- D = (1/2) a T²
311 m = (1/2) (0.170 m/s²) (T²)
T² = 311 m / 0.085 m/s²
T = √(311/0.085) seconds
T = 60.41 seconds
The answer I get is so durn near 60 seconds (1 minute) that it suggests two things to me: ==> That's where the weird numbers of 362N and 311m came from, and ==> there's a good chance that my answer is correct.
Note: It's important to me that you know that 5 points for this one is really cheap and chintzy, and the reason I decided to try it was only to see whether I could.
A roller coaster car is going over the top of a 15-m-radius circular rise. The passenger in the roller coaster has a true weight of 600 N (therefore a mass of 61.2 kg). At the top of the hill, the passengers "feel light," with an apparent weight of only 360 N. How fast is the coaster moving
Answer:
v = 7.67 m/s
Explanation:
The equation for apparent weight in the situation of weightlessness is given as:
Apparent Weight = m(g - a)
where,
Apparent Weight = 360 N
m = mass passenger = 61.2 kg
a = acceleration of roller coaster
g = acceleration due to gravity = 9.8 m/s²
Therefore,
360 N = (61.2 kg)(9.8 m/s² - a)
9.8 m/s² - a = 360 N/61.2 kg
a = 9.8 m/s² - 5.88 m/s²
a = 3.92 m/s²
Since, this acceleration is due to the change in direction of velocity on a circular path. Therefore, it can b represented by centripetal acceleration and its formula is given as:
a = v²/r
where,
a = centripetal acceleration = 3.92 m/s²
v = speed of roller coaster = ?
r = radius of circular rise = 15 m
Therefore,
3.92 m/s² = v²/15 m
v² = (3.92 m.s²)(15 m)
v = √(58.8 m²/s²)
v = 7.67 m/s
A woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.80 rev/s about an axis through its center. The disk has mass 110 kg and radius 4.0 m. Calculate the magnitude of the total angular momentum of the woman–disk system. (Assume that you can treat the woman as a point.)
Answer:
The angular momentum is [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
Explanation:
From the question we are told that
The mass of the woman is [tex]m = 50 \ kg[/tex]
The angular speed of the rim is [tex]w = 0.80 \ rev/s = 0.8 * [\frac{2 \pi}{1} ] = 5.024 \ rad \cdot s^{-1}[/tex]
The mass of the disk is [tex]m_d = 110 \ kg[/tex]
The radius of the disk is [tex]r_d = 4.0 \ m[/tex]
The moment of inertia of the disk is mathematically represented as
[tex]I_D = \frac{1}{2} m_d r^2_d[/tex]
substituting values
[tex]I_D = \frac{1}{2} * 110 * 4^2[/tex]
[tex]I_D = 880 \ kg \cdot m^2[/tex]
The moment of inertia of the woman is
[tex]I_w = m * r_d^2[/tex]
substituting values
[tex]I_w = 50 * 4^2[/tex]
[tex]I_w =800\ kg[/tex]
The moment of inertia of the system (the woman + the large disk ) is
[tex]I_t = I_w + I_D[/tex]
substituting values
[tex]I_t = 880 +800[/tex]
[tex]I_t =1680 \ kg \cdot m^2[/tex]
The angular momentum of the system is
[tex]L = I_t w[/tex]
substituting values
[tex]L = 1680 * 5.024[/tex]
[tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy of the surrounding electrons). The following masses are given:
5727Co: 56.936296u
5726Fe: 56.935399u
Express your answer in millions of electron volts (1u=931.5MeV/c2) to three significant figures.
A negligible amount of this energy goes to the resulting 5726Fe atom as kinetic energy. About 90 percent of the time, after the electron-capture process, the 5726Fe nucleus emits two successive gamma-ray photons of energies 0.140MeV and 1.70 102MeV in decaying to its ground state. The electron-capture process itself emits a massless neutrino, which also carries off kinetic energy. What is the energy of the neutrino emitted in this case?
Express your answer in millions of electron volts.
Answer:
Explanation:
⁵⁷Co₂₇ + e⁻¹ = ²⁷Fe₂₆
mass defect = 56.936296 + .00055 - 56.935399
= .001447 u
equivalent energy
= 931.5 x .001447 MeV
= 1.3479 MeV .
= 1.35 MeV
energy of gamma ray photons = .14 + .017
= .157 MeV .
Rest of the energy goes to neutrino .
energy going to neutrino .
= 1.35 - .157
= 1.193 MeV.
Scenario 2: Use the following information to answer questions 3 and 4:
Your client, Jim, is interested in weight control. He weighs 75kg.
3. If Jim walks 3.3 mph (0% grade), how long must he walk to expend 300 kcal total?
A. 52 min
B. 42 min
C. 65 min
D. 99 min
4. If Jim exercises at an intensity of 6 kcal/min, what is the leg ergometer work rate?
A. 47 watts
B. 90 watts
C. 61 watts
D. 71 watts
Answer:
A. 52 min
.A. 47 watts
Explanation:
Given that;
jim weighs 75 kg
and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.
Using the following relation to determine the amount of calories burned per minute while walking; we have:
[tex]\dfrac{MET*weight (kg)*3.5}{200}[/tex]
here;
MET = energy cost of a physical activity for a period of time
Obtaining the data for walking with a speed of 3.3 mph From the standard chart for MET, At 3.3 mph; we have our desired value to be 4.3
However;
the calories burned in a minute = [tex]\dfrac{4.3*75 (kg)*3.5}{200}[/tex]
= 5.644
Therefore, for walking for 52 mins; Jim burns approximately 293.475 kcal which is nearest to 300 kcal.
4.
Given that:
mass m = 75 kg
intensity = 6 kcal/min
The eg ergometer work rate = ??
Applying the formula:
[tex]V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7[/tex]
where ;
[tex]V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}[/tex]
[tex]V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}[/tex]
[tex]V_O_2 ( intensity ) = 0.0012[/tex]
∴[tex]0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min[/tex]
Converting to watts;
Since; 6.118kg-m/min is = 1 watt
Then 291.66 kgm /min will be equal to 47.67 watts
≅ 47 watts
An object with a mass of 1500 g (grams) accelerates 10.0 m/s2 when an
unknown force is applied to it. What is the amount of the force
Answer:
15N
Explanation:
F=ma
m=1500g = 1.5kg
a=10m/s2
1.5×10=15 N
Answer:15000gms^-2
Explanation:
F=m×a
m=1500g, a=10ms^-2
F=(1500×10)gms^-2
F=15000gms^-2
g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber, and from the bottom hangs a 10 gram lead weight. The density of lead is 11.3 g/cm3. What fraction of the bobber's volume is submerged, as a percent of the total volume
Answer:
Explanation:
total weight acting downwards
= 3g + 10g
13 g
volume of lead = 10 / 11.3 = .885 cm³
Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g
Buoyant force on lead = .885 x 1 x g
total buoyant force = vg + .885 g
For floating
vg + .885 g = 13 g
v = 12.115 cm³
total volume of bobber
= 4/3 x 3.14 x 2³
= 33.5 cm³
fraction of volume submerged
= 12.115 / 33.5
= .36
= 36 %
The fraction of the bobber's volume submerged as a percent of the total volume is 36.2 %.
The given parameters;
diameter of the bobber, d = 4 cmmass of the bobber, m = 3 gmass of the lead, m = 10 gdensity of the lead, ρ = 11.3 g/cm³The volume of the bobber is calculated as follows;
[tex]V = \frac{4}{3} \pi \times r^3\\\\V = \frac{4}{3} \pi \times (2)^3\\\\V = 33.52 \ cm^3[/tex]
The buoyant force experienced by the bobber due to the volume submerged is calculated as follows;
[tex]F _b= \rho Vg\\\\F_b = 1 \times V \times g\\\\F_b = Vg[/tex]
The volume of the lead is calculated as follows;
[tex]V = \frac{mass}{density} \\\\V = \frac{10}{11.3} \\\\V = 0.885 \ cm^3[/tex]
The buoyant force experienced by the lead due to the volume submerged is calculated as follows
[tex]F_b = \rho Vg\\\\F_b = 0.885 g[/tex]
The total buoyant force is calculated as;
[tex]Vg + 0.885g = (3+ 10)g\\\\g(V + 0.885) = 13g\\\\V+ 0.885 = 13\\\\V = 13 -0.885\\\\V = 12.12 \ cm^3[/tex]
The fraction of the bobber's volume submerged as a percent of the total volume is calculated as follows;
[tex]= \frac{12.12}{33.52} \times 100\%\\\\= 36.2 \ \%[/tex]
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An unstrained horizontal spring has a length of 0.36 m and a spring constant of 320 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.033 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.
Answer:
1.been both -ve charged or both +be charged particles
2. 3.52mC
Explanation:
For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.
Now the Force sustain by the extended string is
F = Ke;
Where K is the force constant of the string, 320 N/m
e is the extension,0.033 m
F = 320 × 0.033 =10.56N
2.But according to columns law of charge;
F = kQ1 Q2
But Q1=Q2{ since the charge are of the same magnitude}.
Hence F = KQ^2
Where K is columns constant =9×10^9F/m
Hence Q=√F/K
Q= √10.56/9×10^9
=3.52×10^-3C
= 3.52mC
The brakes of a car are applied to give it an acceleration of -2m/s^2. The car comes to a stop in 3s. What was its speed when the brakes were applied?
Answer:
So if its acceleration is -2m/s^2 that means every second the initial velocity would be subtracted by 2. So since it took 3 seconds 2*3=6. The initial velocity was 6 m/s
When solving vector addition problems you can use either the graphical
method or the
Answer :the resultant of two vectors can be found using either the parallelogram method or the triangle method. don't know if this was helpful ?
Explanation:
Answer:
Analytical method.
If you secure a refrigerator magnet about 2mmfrom the metallic surface of a refrigerator door and then move the magnet sideways, you can feel a resistive force, indicating the presence of eddy currents in the surface.
A)Estimate the magnetic field strength Bof the magnet to be 5 mTand assume the magnet is rectangular with dimensions 4 cmwide by 2 cmhigh, so its area A is 8 cm2. Now estimate the magnetic flux ΦB into the refrigerator door behind the magnet.
Express your answer with the appropriate units.
B)If you move the magnet sideways at a speed of 2 cm/s, what is a corresponding estimate of the time rate at which the magnetic flux through an area A fixed on the refrigerator is changing as the magnet passes over? Use this estimate to estimate the emf induced under the rectangle on the door's surface.
Express your answer with the appropriate units.
Answer:
(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v
Explanation:
Solution
Given that:
A refrigerator magnet about = 2 mm
The estimated magnetic field strength of the magnet is = 5 m T
The Area = 8 cm²
Now,
(A) The magnetic flux ΦB = BA
Thus,
ΦB = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²
So,
ΦB = 4* 6 ^ ⁻6 T m²
(B)By applying Faraday's Law we have the following formula given below:
Ε = Bℓυ
Here,
ℓ = 2 cm the same as 2 * 10 ^⁻2 m
B = 5 m T = 5 * 10 ^ ⁻3 T
υ = 2 cm/s = 2 * 10 ^ ⁻2 m/s
Thus,
Ε = (5 * 10 ^ ⁻3 T) * (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v
E =2 * 10 ^ ⁻6 v
A) The magnetic flux ΦB into the refrigerator door behind the magnet :
4 * 6⁻⁶ Tm²B) The estimated emf induced under the rectangle on the door's surface ;
2 * 10⁻⁶ vGiven data :
magnetic field strength of magnet ( B ) = 5 mT
size of refrigerator magnet = 2 mm
Area of magnet ( A ) = 4 * 2 = 8 cm²
A) Determine the magnetic flux ΦBwhere ; ΦB = BA
ΦB = ( 5 * 10⁻³ ) * ( 4 * 10⁻² ) * ( 2 * 10⁻² ) Tm²
= 4 * 6⁻⁶ Tm²
B) Determine estimated emf inducedTo determine the estimated emf we will apply Faraday's law
Ε = Bℓυ ---- ( 2 )
where : B = 5 * 10⁻³ T, ℓ = 2 * 10⁻² m, υ = 2 * 10⁻² m/s
insert values into equation 2
E = ( 5 * 10⁻³ ) * ( 2 * 10⁻² ) * ( 2 * 10⁻² )
= 2 * 10⁻⁶ v
Hence we can conclude that The magnetic flux ΦB is 4 * 6⁻⁶ Tm² and The estimated emf induced is 2 * 10⁻⁶ v
Learn more about magnet flux : https://brainly.com/question/4721624
A CD is spinning on a CD player. In 220 radians, the cd has reached an angular speed of 92 r a d s by accelerating with a constant acceleration of 14 r a d s 2 . What was the initial angular speed of the CD
Answer:
The initial angular speed is [tex]w_i = 48 \ rad/s[/tex]
Explanation:
From the question we are told that
The angular displacement is [tex]\theta = 220 \ rad[/tex]
The angular speed is [tex]w_f = 92 \ rad/s[/tex]
The acceleration is [tex]\alpha = 14 \ rad/s^2[/tex]
Generally the initial angular speed can be evaluated as
[tex]w_f ^2 = w_i ^2 + 2 * \alpha * \theta[/tex]
=> [tex]w_i ^2 = w_f ^2 - 2 * \alpha * \theta[/tex]
substituting values
=> [tex]w_i ^2 = 92 ^2 - 2 * 14 * 220[/tex]
=> [tex]w_i ^2 = 2304[/tex]
=> [tex]w_i = 48 \ rad/s[/tex]
assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves the uzzel of the cannon at a highet of 200 m.( the cannon is at the edge of the cliff) A: find the horizontal distance the cannon travles. B: when does the cannon ball reach the ground? C: find the maximum highet the cannon ball reaches.
Answer:
A. xmax = 131.49 m
B. t = 8.74 s
C. ymax = 220.33 m
Explanation:
A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:
[tex]y=y_o+v_osin\theta-\frac{1}{2}gt^2[/tex] (1)
yo: height from the projectile is fired = 200m
vo: initial velocity of the projectile = 25m/s
g: gravitational acceleration = 9.8 m/s^2
θ: angle between the direction of the initial motion of the ball and the horizontal = 53°
t: time
You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.
When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:
[tex]0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2[/tex] (2)
You use the quadratic formula to obtain the value of t:
[tex]t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s[/tex]
You use the positive value because it has physical meaning.
Now, you can calculate the horizontal range of the projectile by using the following formula:
[tex]x_{max}=v_ocos\theta t[/tex]
[tex]x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m[/tex]
The cannon ball travels a horizontal distance of 131.49 m
B. The cannon ball reaches the canon for t = 8.74s
C. The maximum height is obtained by using the following formula:
[tex]y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}[/tex] (3)
By replacing in the equation (3) the values of all parameters you obtain:
[tex]y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m[/tex]
The maximum height reached by the cannon ball is 220.33m
A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance . For the next loading, the spring is compressed a distance . How much faster does the second dart leave the gun compared with the first
Complete question is;
A dart is inserted into a spring - loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is compressed a distance 2x. How much faster does the second dart leave the gun compared to the first?
Answer:
The second dart leaves the gun two times faster than the first one.
Explanation:
If we assume there was no energy loss during the spring - dart energy transfer, we can easily apply the principle of conservation of energy. So;
Potential energy = kinetic energy
Thus;
½kx² = ½mv²
Making velocity "v" the subject, we have;
v = √(kx²/m)
Since the initial distance is "x", thus initial launching velocity is;
v1 = √(kx²/m)
Since next distance is 2x, thus, second launch velocity is;
v2 = √(k(2x)²/m)
Expanding, we have;
v2 = √(4kx²/m)
v2 = 2√(kx²/m)
Comparing this to the one gotten for v1 earlier, we can see that it is double v1.
So, v2 = 2v1
Hence, The second dart leaves the gun two times faster than the first one.
The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic field used to align the spins, with proportionality constant 5 hz/T. If the strength of the applied field is known to be 20 T plus or minus 3 T, which of the following correctly describes the uncertainty in the INVERSE frequency (1/frequency)?
A. 3/2000s
B. 3/5s
C. 1/15s
D. 1/4
Complete Question
The complete question is shown on the first uploaded image
Answer:
The uncertainty in inverse frequency is [tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]
Explanation:
From the question we are told that
The value of the proportionality constant is [tex]k = 5 \frac{Hz }{T}[/tex]
The strength of the magnetic field is [tex]B = 20 \ T[/tex]
The change in this strength of magnetic field is [tex]\Delta B = 3 \ T[/tex]
The magnetic field is given as
[tex]B = \frac{k}{\frac{1}{w} }[/tex]
Where [tex]w[/tex] is frequency
The uncertainty or error of the field is given as
[tex]\Delta B = \frac{k }{[\frac{1}{w}^]^2 } \Delta [\frac{1}{w} ][/tex]
The uncertainty in inverse frequency is given as
[tex]\Delta [\frac{1}{w} ] = \frac{\Delta B}{k [\frac{1}{w^2} ]}[/tex]
[tex]\Delta [\frac{1}{w} ]= \frac{\Delta B}{k (B)^2 }[/tex]
substituting values
[tex]\Delta [\frac{1}{w} ]= \frac{3}{5 (20)^2 }[/tex]
[tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]
what is the speed of light in quartz
Answer:
1.95 x 10^8 m/s.
Explanation:
Answer:
the answer is 1.95 x 10^8 m/s
Explanation:
Why do some nucleus emit electrons?
Answer:
In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus. In beta-plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.
Explanation:
Hope this helps!
A single loop of wire with an area of 0.0820 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.220 T/s .
Required:
a. What emf is induced in this loop?
b. If the loop has a resistance of 0.600Ω, find the current induced in the loop.
Answer:
a) emf = 0.01804 V
b) I = 0.03 A
Explanation:
a) The emf is calculated by using the following formula:
[tex]|emf|=|\frac{d\Phi_B}{dt}|=|\frac{d(A\cdot B)}{dt}|[/tex] [tex]=A|\frac{dB}{dt}|[/tex]
A: area of the loop = 0.0820m^2
B: magnitude of the magnetic field
dB/dt: change of the magnetic field, in time: 0.220 T/s
Where ФB is the magnetic flux, the surface vector and magnetic vector are perpendicular between them, and the area A is constant.
You replace the values of A and dB/dt in the equation (1):
[tex]|emf|=(0.082m^2)(0.220T/s)=0.01804V[/tex]
b) The current in the loop is:
[tex]I=\frac{emf}{R}[/tex]
R: resistance of the loop = 0.600Ω
[tex]I=\frac{0.01804V}{0.600\Omega}=0.03A=30mA[/tex]
a. The emf induced in this loop is 18.04mV.
b. The current induced in the loop is 30.06mA.
a. We know that,
[tex]flux(\phi)=B*A[/tex]
Where B is magnetic field and A is the area.
[tex]emf=\frac{d\phi}{dt}=A*\frac{dB}{dt}[/tex]
Given that, Area , [tex]A=0.0820m^{2},B=3.80T,\frac{dB}{dt}=0.220T/s[/tex]
Substituting all values in above equation.
[tex]emf=0.0820*0.220=0.01804V=18.04mV[/tex]
b. Resistance, [tex]R=0.600ohm[/tex]
Current induced in the loop is,
[tex]I=\frac{emf}{R}=18.04/0.6=30.06mA[/tex]
Hence, the emf induced in this loop is 18.04mV.
The current induced in the loop is 30.06mA.
Learn more:
https://brainly.com/question/15477464
A uniform thin spherical shell of mass M=2kg and radius R=0.23m is given an initial angular speed w=18.3rad/s when it is at the bottom of an inclined plane of height h=3.5m, as shown in the figure. The spherical shell rolls without slipping. Find wif the shell comes to rest at the top of the inclined plane. (Take g-9.81 m/s2, Ispherical shell = 2/3 MR2 ).Express your answer using one decimal place.
Answer:
47.8rad/s
Explanation:
For energy to be conserved.
The potential energy sustain by the object would be equal to K.E
P.E = m× g× h = 2 × 9.81× 3.5= 68.67J
Now K.E = 1/2 × I × (w1^2 - w0^2)
I = 2/3 × M × R2
= 2/3 × 2 × (0.23)^2= 0.0705
Hence
W1 = final angular velocity
Wo = initial angular velocity
From P.E = K.E we have;
68.67J = 1/2 × 0.0705 × (w1^2 - w0^2)
(w1^2 - w0^2) = 1948.09
W1^2 = 1948.09 + (18.3^2)
W1^2=2282.98
W1 = √2282.98
=47.78rad/s
= 47.8rad/s to 1 decimal place.
On April 13, 2029 (Friday the 13th!), the asteroid 99942 mi Apophis will pass within 18600 mi of the earth-about 1/13 the distance to the moon! It has a density of 2600 kg/m^3, can be modeled as a sphere 320 m in diameter, and will be traveling at 12.6 km/s.
1)If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver?
2)The largest nuclear bomb ever tested by the United States was the "Castle/Bravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases 4.184x10^15 J of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?
Answer:
Explanation:
Volume of asteroid = 4/3 x π x 160³
= 17.15 x 10⁶
mass = volume x density
= 17.15 x 10⁶ x 2600
= 445.9 x 10⁸ kg
kinetic energy
= 1/2 x 445.9 x 10⁸ x( 12.6 )² x 10⁶
= 35.4 x 10¹⁷ J .
2 )
energy of 15 megaton
= 4.184 x 10¹⁵ x 15 J
= 62.76 x 10¹⁵ J
No of bombs required
= 35.4 x 10¹⁷ / 62.76 x 10¹⁵
= 56.4 Bombs .
A trap-jaw ant snaps its mandibles shut at very high speed, good for catching small prey. But these ants also slam their mandibles into the ground; the resulting force can launch the ant into the air for a quick escape. If a 12 mg ant hits the ground with an average force of 47 mN for a time of 0.13 ms, at what speed will it leave the ground?
Answer:
FInal speed (v) = 0.509 m/s (Approx)
Explanation:
Given:
Mass of ant (m) = 12 mg
Force (f) = 47 N
Time taken (t) = 0.13 ms
Find:
FInal speed (v) = ?
Computation:
Initial velocity (u) = 0
Impulse = change in momentum
Force × TIme = change in momentum
47 × 0.13 = mv - mu
6.11 = 12 (V)
FInal speed (v) = 0.509 m/s (Approx)
An arrow is shot from a height of 1.55 m toward a cliff of height H. It is shot with a velocity of 26 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.99 s later.
(a) Draw a sketch of the given example. Include the x-y coordinate system.
(b) What is the height of the cliff?
(c) What is the maximum height reached by the arrow along its trajectory?
(d) What is the arrow's impact speed just before hitting the cliff?
Answer:
Explanation:
vertical component of the velocity of arrow
= 26 sin 60 = 22.516 m
height reached by it after 3.99 s
h = ut - 1/2 g t²
= 22.516 x 3.99 - .5 x 9.8 x 3.99²
= 89.83 - 78
11.83 m
Total height of cliff = 1.55 + 11.83
= 13.38 m
c ) maximum height covered s
v² = u² - 2gs
0 = u² - 2gs
s = u² / 2g
= 22.516² / 2 x 9.8
= 25.86
maximum height reached
= 25.86 + 1.55
= 27.41 m
d )
vertical speed after 3.99 s
v = u - gt
= 22.516 - 9.8 x 3.99
= -16.586
Horizontal component will remain unchanged
Horizontal component = 26 cos 60
= 13 m /s
Resultant of two velocities
= √ 13²+ 16.568²
= 21 m /s