Answer:
by using it's buoyant or floating effect by Archimedes.
the buoyant force act on the astronauts body and make he/ she feels like in low gravity.
the buoyant force equation is
F = Density of liquid x earth gravitational field x volume of astronauts body and suit.
the Weight of astronauts in the pools will be less than in the land or air.
Weight in water = weight in air/land - buoyant force
so the astronauts will feel like in the outer space with low gravity.
A car travels around an oval racetrack at constant speed. The car is accelerating:________.
A) at all points except B and D.
B) at all points except A, B, C, and D.
C) everywhere, including points A, B, C, and D.
D) nowhere, because it is traveling at constant speed.
2) A moving object on which no forces are acting will continue to move with constant:_________
A) Acceleration
B) speed
C) both of theseD) none of these
Answer:
1A,2D,3B
Explanation:
hope this helps
commune time to work ( physics) i need help pls :(
How much work must be done on a 10 kg snowboard to increase its speed from 4 m/s to 6 m/s?
Answer:
100 J
Explanation:
Work = change in energy
W = ΔKE
W = ½ mv² − ½ mv₀²
W = ½ m (v² − v₀²)
W = ½ (10 kg) ((6 m/s)² − (4 m/s)²)
W = 100 J
A type of friction that occurs when air pushes against a moving object causing it to negatively accelerate
a) surface area
b) air resistance
c) descent velocity
d) gravity
Answer:
Air resistance
Answer B is correct
Explanation:
The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.
hope this helps
brainliest appreciated
good luck! have a nice day!
An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (constant) acceleration of the vehicle during this time? Group of answer choices
Answer:
Dear Kaleb
Answer to your query is provided below
Acceleration of the vehicle is 12m/s^2
Explanation:
Explanation for the same is attached in image
A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 45.0kg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.700 kg, traveling perpendicular to the door at 12.0m/s just before impact
A) Find the final angular speed of the door.
answer in rad/s
B) Does the mud make a significant contribution to the moment of inertia?
Yes or No
Answer:
0.19rad/s and Yes
Explanation:
From the principle of conservation of momentum it means momentum before and after collision is the same.
Momentum before collision is 0.700 kg×12 = 8.4Ns
Momentum of the door = mass of door × velocity of door
8.4Ns = mass of door × velocity of door
Velocity of door = 8.4Ns/45 =0.19m/s
But velocity V= w×r ;
w-angular velocity
r- raduis = width
w= 0.19/1m = 0.19rad/s
2. Yes it did because it resisted The moment of inertia and ensued the locking of the door.
Davina accelerates a box across a smooth frictionless horizontal surface over a displacement of 18.0 m with a constant 25.0 N force angled at 23.0° below the horizontal. How much work does she do on the box? A. 176 J B. 414 J C. 450 J D. 511 J Group of answer choices
Answer:
W = 414 J, correct is B
Explanation:
Work is defined by
W = ∫ F .dx
where F is the force, x is the displacement and the point represents the dot product
this expression can also be written with the explicit scalar product
W = ∫ F dx cos θ
where is the angle between force and displacement
for this case as the force is constant
W = F x cos θ
calculate
W = 25.0 18.0 cos (-23)
W = 414 J
the correct answer is B
The battery on your car has a rating stated in ampere minute which permit you to estimate the length of time a fully charged battery could deliver any particular current before discharge. Approximately how much energy is stored by a 50 ampere minute 12 volt battery
Answer:
Thus, the energy stored by a 50 Ampere minute battery is found to be 36 KJ.
Explanation:
The power delivered by a battery is given by the formula:
P = VI
where,
P = Power Delivered by battery in 1 second
V = Voltage of battery = 12 volt
I = Current stored in battery
But, if we multiply both sides of equation by time (t), then:
Pt = VIt
where,
Pt = Power x Time = E = Energy Stored = ?
It = Rating of Battery = (50 A.min)(60 sec/min) = 3000 A.sec
Therefore,
E = (12 volt)(3000 A.sec)
E = 36000 J = 36 KJ
Thus, the energy stored by a 50 Ampere minute battery is found to be 36 KJ.
A low C (f=65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and
its mass density is 5.0 g/m2, determine the tension of the wire.
Answer:
T = 676 N
Explanation:
Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g[tex]/m^{2}[/tex] = 0.005 kg
A stationary wave that is set up in the string has a frequency of;
f = [tex]\frac{1}{2L}[/tex][tex]\sqrt{\frac{T}{M} }[/tex]
⇒ T = 4[tex]L^{2}[/tex][tex]f^{2}[/tex]M
Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.
But M = L × ρ = (2 × 0.005) = 0.01 kg/m
T = 4 × [tex]2^{2}[/tex] ×[tex]65^{2}[/tex] × 0.01
= 4 × 4 ×4225 × 0.01
= 676 N
Tension of the wire is 676 N.
Two students are pushing their stalled car down the street. If the net force exerted on
the car by the students is 1000 N at an angle of 20° below horizontal, the horizontal
component of the force is:
(a) greater than 1000 N.
(b) less than 1000 N.
(c) sum of the pushing force and the weight of the students.
(d) (a) and (b)
(e) (a) and (c)
Answer:
B
Explanation:
Because the force has 2 components (horizontal and vertical), the horizontal component must be smaller than the total force. The Pythagorean theorem only adds positive values (because they're squared) so it makes sense. Using trigonometry, 100*cos(-20) yields a horizontal force of around 939.7N, which is less than 1000N.
An object will sink in a liquid if the density of the object is greater than that of the liquid. The mass of a sphere is 0.723 g. If the volume of this sphere is less than ________ cm3, then the sphere will sink in liquid mercury (density
Answer:
= 0.0532 cm^3
Explanation:
The computation of volume of the sphere is shown below:-
[tex]Density = \frac{Mass}{Volume}[/tex]
Where,
Density = 13.6 g/cm^3
Mass of sphere = 0.723 g
now we will put the values into the above formula to reach volume of the sphere which is here below:-
[tex]Volume = \frac{0.723}{13.6}[/tex]
= 0.0532 cm^3
Therefore for computing the volume of the sphere we simply applied the above formula.
How many ohms of resistance are in a 120–volt hair dryer that draws 7.6 amps of current?
From Ohm's law . . . Resistance = (voltage) / (current)
Resistance = (120 volts) / (7.6 Amperes)
Resistance = 15.8 Ω
A tank circuit consists of an inductor and a capacitor. Give a simple explanation for why the magnetic field in the induc- tor is strongest at the moment that the separated charge in the capacitor reaches zero.
Answer:
If you pull a permanent magnet rapidly away from a tank circuit, what is likely to happen in that circuit?
Charge will oscillate in the tank's capacitor and inductor.
Explanation:
Mark Watney (Matt Damon in the Martian movie) and Marvin the Martian (Looney Tunes cartoon character) are having an argument on the surface of Mars (negligible air resistance). They are testing out their new potato launcher that fires projectiles at a constant speed. Mark launches his potato at an angle of 60◦ and Marvin launches his identical potato at an angle of 30◦ . Without any calculations try to answer the following questions, and justify each answer.
(A) Which potato lands farther away from the launcher (potatoes are launched from ground level)?
(B) Which potato spends more time in the air before hitting the ground
(C) Which potato has a greater speed just before it hits the ground?
Answer:
A) The two potatoes cover the same horizontal distance from the launcher.
B) Mark's potato spends more time in the air than Marvin's potato before hitting the ground.
C) Marvin's potato hits the ground with a greater speed than Mark's potato
Explanation:
A) For projectile motion, the final horizontal distance of the projectile from where it was initially launched (its range) is given as
R = (u² sin 2θ)/g
where
u = initial velocity of the projectile
θ = angle above the horizontal at which the projectile was launched = 30°, 60°
g = acceleration due to gravity on Mars
Since, u and g are the same for Mark and Marvin, sin 2θ would determine which range is higher.
Sin (2×60°) = sin 120°
Sin (2×30°) = sin 60°
Sin 120° = Sin 60°
Hence, the two potatoes cover the same horizontal distance from the launcher.
B) Time spent in the air for a projectile is given as
T = (2u sin θ)/g
Again, since u and g are the same for Mark and Marvin on Mars, sin θ will give the required idea of whose potato spends more time in the air.
Sin 60° = 0.866
Sin 30° = 0.50
Sin 60° > Sin 30°
Hence, Mark's potato spends more time in the air than Marvin's potato.
C) The horizontal velocity for projectile motion is constant all through the motion and is equal to u cos θ
u cos 60° < u cos 30°
And the initial vertical velocity is u sin θ
Final vertical velocity
= (initial vertical velocity) - gt
g = acceleration due to gravity on Mars
T = time of flight
For Mark,
initial vertical velocity = u sin 60°, greater than Marvin's u sin 30°
And Mark's potato's time of flight is greater as established in (B) above.
But for Marvin
initial vertical velocity = u sin 30°, less than Mark's u sin 60°
And Marvin's potato's time of flight is lesser as established in (B) above
So, at the end of the day, the final vertical velocity is almost the same for both Mark's and Marvin's potatoes.
Hence, the horizontal component of the final velocity edges the final speed of the potatoes just before hitting the ground in Marvin's favour.
Hope this Helps!!!
Block 1, of mass m1 = 2.50 kg , moves along a frictionless air track with speed v1 = 27.0 m/s. It collides with block 2, of mass m2 = 33.0 kg , which was initially at rest. The blocks stick together after the collision.A. Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.B. Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically.C. what is the change deltaK= Kfinal- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.
Answer:
a
The total initial momentum of the two-block system is [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
b
The magnitude of the final velocity of the two-block system [tex]v_f = 1.9014 \ m/s[/tex]
c
the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is
[tex]\Delta KE =- 847.08 \ J[/tex]
Explanation:
From the question we are told that
The mass of first block is [tex]m_1 = 2.50 \ kg[/tex]
The initial velocity of first block is [tex]u_1 = 27.0 \ m/s[/tex]
The mass of second block is [tex]m_2 = 33.0\ kg[/tex]
initial velocity of second block is [tex]u_2 = 0 \ m/s[/tex]
The magnitude of the of the total initial momentum of the two-block system is mathematically repented as
[tex]p_i = (m_1 * u_1 ) + (m_2 * u_2)[/tex]
substituting values
[tex]p_i = (2.50* 27 ) + (33 * 0)[/tex]
[tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
According to the law of linear momentum conservation
[tex]p_i = p_f[/tex]
Where [tex]p_f[/tex] is the total final momentum of the system which is mathematically represented as
[tex]p_f = (m_+m_2) * v_f[/tex]
Where [tex]v_f[/tex] is the final velocity of the system
[tex]p_i = (m_1 +m_2 ) v_f[/tex]
substituting values
[tex]67.5 = (2.50+33 ) v_f[/tex]
[tex]v_f = 1.9014 \ m/s[/tex]
The change in kinetic energy is mathematically represented as
[tex]\Delta KE = KE_f -KE_i[/tex]
Where [tex]KE_f[/tex] is the final kinetic energy of the two-body system which is mathematically represented as
[tex]KE_f = \frac{1}{2} (m_1 +m_2) * v_f^2[/tex]
substituting values
[tex]KE_f = \frac{1}{2} (2.50 +33) * (1.9014)^2[/tex]
[tex]KE_f =64.17 J[/tex]
While [tex]KE_i[/tex] is the initial kinetic energy of the two-body system
[tex]KE_i = \frac{1}{2} * m_1 * u_1^2[/tex]
substituting values
[tex]KE_i = \frac{1}{2} * 2.5 * 27^2[/tex]
[tex]KE_i = 911.25 \ J[/tex]
So
[tex]\Delta KE = 64.17 -911.25[/tex]
[tex]\Delta KE =- 847.08 \ J[/tex]
A standing wave on a string that is fixed at both ends has frequency 80.0 Hz. The distance between adjacent antinodes of the standing wave is 12.0 cm. What is the speed of the waves on the string, in m/s
Answer:
v = 19.2 m/s
Explanation:
In order to find the speed of the string you use the following formula:
[tex]f=\frac{v}{2L}[/tex] (1)
f: frequency of the string = 80.0Hz
v: speed of the wave = ?
L: length of the string = 12.0cm = 0.12m
The length of the string coincides with the wavelength of the wave for the fundamental mode.
Then, you solve for v in the equation (1), and replace the values of the other parameters:
[tex]v=2Lf=2(0.12m)(80.0Hz)=19.2\frac{m}{s}[/tex]
The speed of the wave is 19.2 m/s
Assuming 100% efficient energy conversion, how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery?
Complete question is;
Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes.
Answer:
Amount of water required to charge the battery = 7.35 m³
Explanation:
The formula for Potential energy of the water at that height = mgh
Where;
m = mass of the water
g = acceleration due to gravity = 9.8 m/s²
h = height of water = 50 cm = 0.5 m
We know that in density, m = ρV
Where;
ρ = density of water = 1000 kg/m³
V = volume of water
So, potential energy is now given as;
Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J
Now, formula for energy of the battery is given as;
E = qV
We are given;
q = 50 A.min = 50 × 60 = 3,000 C
V = 12 V
Thus;
qV = 3,000 × 12 = 36,000 J
E = 36,000 J
At a 100% conversion rate, the energy of the water totally powers the battery.
Thus;
(4900V) = (36,000)
4900V = 36,000
V = 36,000/4900
V = 7.35 m³
When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing is 273 mph relative to the airplane. (a) What suction pressure is developed on the wing at that point? (b) What is the pressure at the leading edge (a stagnation point) of the wing?
Answer:
P1 = 0 gage
P2 = 87.9 lb/ft³
Explanation:
Given data
Airplane flying = 200 mph = 293.33 ft/s
altitude height = 5000-ft
air velocity relative to the airplane = 273 mph = 400.4 ft/s
Solution
we know density at height 5000-ft is 2.04 × [tex]10^{-3}[/tex] slug/ft³
so here P1 + [tex]\frac{\rho v1^2}{2}[/tex] = P2 + [tex]\frac{\rho v2^2}{2}[/tex]
and here
P1 = 0 gage
because P1 = atmospheric pressure
and so here put here value and we get
P1 + [tex]\frac{\rho v1^2}{2}[/tex] = P2 + [tex]\frac{\rho v2^2}{2}[/tex]
0 + [tex]\frac{2.048 \times 10^{-3} \times 293.33^2}{2}[/tex] [tex]= P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}[/tex]
solve it we get
P2 = 87.9 lb/ft³
A mass m at the end of a spring vibrates with a frequency of 0.72 Hz . When an additional 700 g mass is added to m, the frequency is 0.64 Hz . Part A What is the value of m? Express your answer using two significant figures.
Answer:
The value of m is 2635.294 grams.
Explanation:
Let suppose that mass-spring system has a simple harmonic motion, to this respect the formula for frequency is:
[tex]f = \frac{\omega}{2\pi}[/tex]
Where [tex]\omega[/tex] is the angular frequency, measured in radians per second.
For a mass-spring system under simple harmonic motion, the angular frequency is:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
Where:
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m[/tex] - Mass, measured in kilograms.
The following equation is obtained after replacing angular frequency in frequency formula:
[tex]f = \frac{1}{2\pi}\cdot \sqrt{\frac{k}{m} }[/tex]
As this shows, frequency is inversely proportional to the square root of mass. Hence, the following relationship is deducted:
[tex]f_{1}\cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{2}}[/tex]
If [tex]m_{2} = m_{1} + 700\,g[/tex], [tex]f_{1} = 0.72\,hz[/tex] and [tex]f_{2} = 0.64\,hz[/tex], the resulting expression is simplified and then initial mass is found after clearing it:
[tex]f_{1} \cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{1}+700\,g}[/tex]
[tex]f_{1}^{2} \cdot m_{1} = f_{2}^{2}\cdot (m_{1} + 700\,g)[/tex]
[tex]\left(\frac{f_{1}}{f_{2}} \right)^{2}\cdot m_{1} = m_{1} + 700\,g[/tex]
[tex]\left[\left(\frac{f_{1}}{f_{2}}\right)^{2} - 1\right]\cdot m_{1} = 700\,g[/tex]
[tex]m_{1} = \frac{700\,g}{\left(\frac{f_{1}}{f_{2}} \right)^{2}-1}[/tex]
[tex]m_{1} = \frac{700\,g}{\left(\frac{0.72\,hz}{0.64\,hz} \right)^{2}-1}[/tex]
[tex]m_{1} = 2635.294\,g[/tex]
The value of m is 2635.294 grams.
A particle with a charge of 5.1 μC is 3.02 cm from a particle with a charge of 2.51 μC . The potential energy of this two-particle system, relative to the potential energy at infinite separation, is
Answer:
U = 3.806 J
Explanation:
The potential energy between the two charges q1 and q2, is given by the following formula:
[tex]U=k\frac{q_1q_2}{r}[/tex] (1)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
q1 = 5.1*10^-6 C
q2 = 2.51*10^-6 C
r: distance of separation between particles = 3.02cm = 3.02*10^-2 m
You replace the values of all parameters in the equation (1):
[tex]U=(8.98*10^9Nm^2/C^2)\frac{(5.1*10^{-6}C)(2.51*10^{-6}C)}{3.02*10^{-2}m}\\\\U=3.806J[/tex]
The potential energy of the two particle system is 3.806 J
EXPLANATION ⛔
A 20 gram mass is suspended from meter rod at 20cm. The meter rod is balanced on 40cm mark. Weight of meter rod is
A. 0.4N
B. 0.6N
C. 6N
D. 60N
Answer:b
Explanation:I’m just trynna get more money dude
(a) What is the cost of heating a hot tub containing 1440 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take heat loss to surroundings into account? The cost of electricity is 9.00¢/(kW · h) and the specific heat for water is 4184 J/(kg · °C). $ 67 Incorrect: Your answer is incorrect. How much heat is needed to raise the temperature of m kg of a substance? How many joules are in 1 kWh? (b) What current was used by the 220 V AC electric heater, if this took 3.45 h? 88.2 Correct: Your answer is correct. A
Answer:
a) [tex]E = 6.024\,USD[/tex], For m kilograms, it is 4184m J., 3600000 joules, b) [tex]i = 88.200\,A[/tex]
Explanation:
a) The amount of heat needed to warm water is given by the following expression:
[tex]Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})[/tex]
Where:
[tex]m_{w}[/tex] - Mass of water, measured in kilograms.
[tex]c_{w}[/tex] - Specific heat of water, measured in [tex]\frac{J}{kg\cdot ^{\circ}C}[/tex].
[tex]T_{f}[/tex], [tex]T_{i}[/tex] - Initial and final temperatures, measured in [tex]^{\circ}C[/tex].
Then,
[tex]Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)[/tex]
[tex]Q_{needed} = 180748800\,J[/tex]
The energy needed in kilowatt-hours is:
[tex]Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)[/tex]
[tex]Q_{needed} = 50.208\,kWh[/tex]
The electric energy required to heat up the water is:
[tex]E = \frac{50.208\,kWh}{0.75}[/tex]
[tex]E = 66.944\,kWh[/tex]
Lastly, the cost of heating a hot tub is: (USD - US dollars)
[tex]E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)[/tex]
[tex]E = 6.024\,USD[/tex]
The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.
b) The current required for the electric heater is:
[tex]i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}[/tex]
[tex]i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}[/tex]
[tex]i = 88.200\,A[/tex]
510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the squirrel can be approximated as a rectanglar prism with cross-sectional area of width 11.6 cm and length 23.2 cm. Note, the squirrel may not reach terminal velocity by the time it hits the gr
Answer:
The terminal velocity is [tex]v_t =17.5 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the squirrel is [tex]m_s = 50\ g = \frac{50}{1000} = 0.05 \ kg[/tex]
The surface area is [tex]A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2[/tex]
The height of fall is h =4.8 m
The length of the prism is [tex]l = 23.2 = 0.232 \ m[/tex]
The width of the prism is [tex]w = 11.6 = 0.116 \ m[/tex]
The terminal velocity is mathematically represented as
[tex]v_t = \sqrt{\frac{2 * m_s * g }{\dho_s * C * A } }[/tex]
Where [tex]\rho[/tex] is the density of a rectangular prism with a constant values of [tex]\rho = 1.21 \ kg/m^3[/tex]
[tex]C[/tex] is the drag coefficient for a horizontal skydiver with a value = 1
A is the area of the prism the squirrel is assumed to be which is mathematically represented as
[tex]A = 0.116 * 0.232[/tex]
[tex]A = 0.026912 \ m^2[/tex]
substituting values
[tex]v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }[/tex]
[tex]v_t =17.5 \ m/s[/tex]
A 1,269 kg rocket is traveling at 413 m/s with 2,660 kg of fuel on board. If the rocket fuel travels at 1,614 m/s relative to the rocket, what is the rockets final velocity after it uses half of its fuel?
Answer:
About 2104m/s
Explanation:
[tex]F=ma \\\\F=\dfrac{2660kg}{2}\cdot 1614m/s=2,146,620N \\\\2,146,620N=1,269kg\cdot a \\\\a\approx 1691m/s \\\\v_f=v_o+at=413m/s+1691m/s=2104m/s[/tex]
Hope this helps!
Your electric drill rotates initially at 5.35 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.331 rad/s2 for 4.81 s. What is the drill's angular displacement during that time interval?
Answer:
The angular displacement is [tex]\theta = 29.6 \ rad[/tex]
Explanation:
From the question we are told that
The initial angular speed is [tex]w = 5.35 \ rad/s[/tex]
The angular acceleration is [tex]\alpha = 0.331 rad /s^2[/tex]
The time take is [tex]t = 4.81 \ s[/tex]
Generally the angular displacement is mathematically represented as
[tex]\theta = w * t + \frac{1}{2} \alpha * t^2[/tex]
substituting values
[tex]\theta = 5.35 * 4.81 + \frac{1}{2} * 0.331 * (4.81)^2[/tex]
[tex]\theta = 29.6 \ rad[/tex]
In each pair, select the body with more internal energy.
Answer:
rt
Explanation:
Someone please helpp me out thanks !
Answer:
Silver.
Explanation:
To determine the identity of the metal, we need to calculate the density of the metal. This is illustrated below:
Mass of metal (m) = 18.15g
Length (L)= 1.2cm
Volume (V) = L³ = 1.2³ = 1.728cm³
Density =.?
The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Mathematically, it is expressed as:
Density = Mass /volume
With the above formula, we can obtain the density of the metal as follow:
Mass = 18.15g
Volume = 1.728cm³
Density =.?
Density = Mass /volume
Density = 18.15g/1.728cm³
Density of the metal = 10.50g/cm³
Comparing the density of metal obtained with the densities given in the table above, we can see that the density of the metal is the same with that of silver.
Therefore, the metal is silver.
While her kid brother is on a wooden horse at the edge of a merry-go-round, Sheila rides her bicycle parallel to its edge. The wooden horses have a tangential speed of 6 m/s. Sheila rides at 4 m/s. The radius of the merry-go-round is 8 m. At what time intervals does Sheila encounter her brother, if she rides opposite to the direction of rotation of the merry-go-round?
a. 5.03 s
b. 8.37 s
c. 12.6 s
d. 25.1 s
e. 50.2 s
Answer:
t = 5.03 s
Explanation:
To find the time interval when Sheila encounter her brother, you first calculate the angular speed of both Sheila and her brother.
You use the following formula:
[tex]\omega = \frac{v}{r}[/tex]
w: angular speed
v: tangential speed
r: radius of the trajectory = 8 m
For you have:
[tex]\omega=\frac{4m/s}{8m}=0.5\frac{rad}{s}[/tex]
For her brother:
[tex]\omega'=\frac{6m/s}{8m}=0.75\frac{rad}{s}[/tex]
Next, they will encounter to each other when the angular distance of the Brother of sheila equals the angular distance of Sheila in the opposite direction. This can be written as follow:
[tex]\theta=\omega t\\\\\theta'=\omega ' t[/tex]
They encounter for θ = 2π-θ':
[tex]\omega t=2\pi-\omega' t[/tex]
You replace the values of the parameters in the previous equation and solve for t:
[tex]0.5t=2\pi-0.75t\\\\1.25t=2\pi\\\\t=5.026\approx5.03[/tex]
Hence, Sheila encounter her brother in 5.03 s
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?
Sort the following quantities as known or unknown. Take the horizontal direction to be along the x axis.
mQ: the mass of the quarterback
mB: the mass of the football
(vQx)i: the horizontal velocity of quarterback before throwing the ball
(vBx)i: the horizontal velocity of football before being thrown
(vQx)f: the horizontal velocity of quarterback after throwing the ball
(vBx)f: the horizontal velocity of football after being thrown
Answer:
vBxf = 0.08625m/s
Explanation:
This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.
[tex]p_f=p_i[/tex]
pf: final momentum
pi: initial momentum
The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.
Then, you have:
[tex]m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}[/tex] (1)
mQ: the mass of the quarterback = 80kg
mB: the mass of the football = 0.43kg
(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s
(vBx)i: the horizontal velocity of football before being thrown = 0m/s
(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?
(vBx)f: the horizontal velocity of football after being thrown = 15 m/s
You replace the values of the variables in the equation (1), and you solve for (vBx)f:
[tex]0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}[/tex]
Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.
Hence, the speed of the quarterback after he throws the ball is 0.08625m/s
Organ pipe A, with both ends open, has a fundamental frequency of 475 Hz. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. Use 343 m/s for the speed of sound in air. How long are (a) pipe A and (b) pipe B?
Answer:
The length of organ pipe A is [tex]L = 0.3611 \ m[/tex]
The length of organ pipe B is [tex]L_b = 0.2708 \ m[/tex]
Explanation:
From the question we are told that
The fundamental frequency is [tex]f = 475 Hz[/tex]
The speed of sound is [tex]v_s = 343 \ m/s[/tex]
The fundamental frequency of the organ pipe A is mathematically represented as
[tex]f= \frac{v_s}{2 L}[/tex]
Where L is the length of organ pipe
Now making L the subject
[tex]L = \frac{v_s}{2f}[/tex]
substituting values
[tex]L = \frac{343}{2 *475}[/tex]
[tex]L = 0.3611 \ m[/tex]
The second harmonic frequency of the organ pipe A is mathematically represented as
[tex]f_2 = \frac{v_2}{L}[/tex]
The third harmonic frequency of the organ pipe B is mathematically represented as
[tex]f_3 = \frac{3 v_s}{4 L_b }[/tex]
So from the question
[tex]f_2 = f_3[/tex]
So
[tex]\frac{v_2}{L} = \frac{3 v_s}{4 L_b }[/tex]
Making [tex]L_b[/tex] the subject
[tex]L_b = \frac{3}{4} L[/tex]
substituting values
[tex]L_b = \frac{3}{4} (0.3611)[/tex]
[tex]L_b = 0.2708 \ m[/tex]