Answer:
Explanation:
For a convex mirror, the value of its image distance and its focal length are negative.
using the mirror formula 1/f = 1/u+1/v
f is the focal length = Radius of curvature/2 = 0.560/2
f= 0.28m
u is the object distance = 10.6m
v is the position of the image = ?
On substitution;
1/0.28 = 1/10.6 + 1/-v
3.57 = 0.094 - 1/v
3.57 - 0.094 = -1/v
3.476 = -1/v
v = -1/3.476
v = -0.2877m
B) Since the image distance is negative, this means that the image is an upright and a virtual image. All Upright images has their image distance to be negative.
C) Magnification = Image distance/object distance
Magnification = 0.2877/10.6
Magnification = 0.0271
A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2. The car makes it one quarter of the way around the circle before it skids off the track. From these data, determine the coefficient of static friction between the car and track.
Required:
Determine the coefficient of static friction between the car and the track.
Answer:
Approximately [tex]0.608[/tex] (assuming that [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex].)
Explanation:
The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.
Let [tex]m[/tex] represent the mass of this car.Let [tex]r[/tex] represent the radius of the circular track.This answer will approach this question in two steps:
Step one: determine the centripetal force when the car is about to skid.Step two: calculate the coefficient of static friction.For simplicity, let [tex]a_{T}[/tex] represent the tangential acceleration ([tex]1.90\; \rm m \cdot s^{-2}[/tex]) of this car.
Centripetal Force when the car is about to skidThe question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to [tex]90^\circ[/tex] or [tex]\displaystyle \frac{\pi}{2}[/tex] radians.
The angular acceleration of this car can be found as [tex]\displaystyle \alpha = \frac{a_{T}}{r}[/tex]. ([tex]a_T[/tex] is the tangential acceleration of the car, and [tex]r[/tex] is the radius of this circular track.)
Consider the SUVAT equation that relates initial and final (tangential) velocity ([tex]u[/tex] and [tex]v[/tex]) to (tangential) acceleration [tex]a_{T}[/tex] and displacement [tex]x[/tex]:
[tex]v^2 - u^2 = 2\, a_{T}\cdot x[/tex].
The idea is to solve for the final angular velocity using the angular analogy of that equation:
[tex]\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta[/tex].
In this equation, [tex]\theta[/tex] represents angular displacement. For this motion in particular:
[tex]\omega(\text{initial}) = 0[/tex] since the car was initially not moving.[tex]\theta = \displaystyle \frac{\pi}{2}[/tex] since the car travelled one-quarter of the circle.Solve this equation for [tex]\omega(\text{final})[/tex] in terms of [tex]a_T[/tex] and [tex]r[/tex]:
[tex]\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}[/tex].
Let [tex]m[/tex] represent the mass of this car. The centripetal force at this moment would be:
[tex]\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}[/tex].
Coefficient of static friction between the car and the trackSince the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, [tex]m\, g[/tex].
Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.
Let [tex]\mu_s[/tex] denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:
[tex]F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g[/tex].
The size of this force should be equal to that of the centripetal force when the car is about to skid:
[tex]\mu_s\, m\, g = \pi\, m\, a_{T}[/tex].
Solve this equation for [tex]\mu_s[/tex]:
[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g}[/tex].
Indeed, the expression for [tex]\mu_s[/tex] does not include any unknown letter. Let [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex]. Evaluate this expression for [tex]a_T = 1.90\;\rm m \cdot s^{-2}[/tex]:
[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608[/tex].
(Three significant figures.)
The red giant Betelgeuse has a surface temperature of 3000 K and is 600 times the diameter of our sun. (If our sun were that large, we would be inside it!) Assume that it radiates like an ideal blackbody.a) If Betelgeuse were to radiate all of its energy at the peak-intensity wavelength, how many photons per second would it radiate?b) Find the ratio of the power radiated by Betelgeuse to the power radiated by our sun (at 5800 K).
Answer:
Explanation:
a )
Radius of the sun = .69645 x 10⁹ m .
600 times = 600 x .69645 x 10⁹ m
= 4.1787 x 10¹¹ m .
surface area A = 4π (4.1787 x 10¹¹)²
= 219.317 x 10²²
energy radiated E = σ A Τ⁴
= 5.67 x 10⁻⁸ x 219.317 x 10²² x (3000)⁴
= 100695 x 10²⁶ J
To know the wavelength of photon emitted
[tex]\lambda_mT= b[/tex]
[tex]\lambda_m= \frac{b}{T}[/tex]
= 2.89777 x 10⁻³ / 3000
= 966 nm
= 1275 /966 eV
1.32 x 1.6 x 10⁻¹⁹ J
= 2.112 x 10⁻¹⁹ J
No of photons radiated = 100695 x 10²⁶ / 2.112 x 10⁻¹⁹
= 47677.5 x 10⁴⁵
= .476 x 10⁵⁰ .
b )
energy radiated by our sun per second
E₂ = σ A 5800⁴
energy radiated by Betelgeuse per second
E₁ = σ x 600²A x 3000⁴
E₁ / E₂ = σ x 600²A x 3000⁴ / σ A 5800⁴
= 36 X 10⁴ x 3⁴ x 10¹² / 58⁴ x 10⁸
= 25.76 x 10⁸ x 10⁻⁵
= 25760 times .
Consider a weather balloon floating in the air. There are three forces acting on this balloon: the force of gravity is FG, the force from lift towards balloon is FL, and the force from the wind is labeled Fw. The orientation of these forces along with a coordinate system is given below:
Assume that || FG || = 20 N, || FL ||= 25 N, and || Fw ll = 15 N.
Required:
Find the magnitude of the resultant force acting on the weather balloon and round your answer to two decimal places.
A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal
length lens. Calculate the size of the image formed on flim
Answer:
18.89cm
Explanation:
As we know that the person is standing 5m in front of the camera
[tex]d_0=5m=500cm[/tex]
The focal length of the lens =50cm
f=50 cm
By Lens formula we have:
[tex]\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm[/tex]
By the formula of magnification
[tex]\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm[/tex]
The height of the image formed is 18.89cm.
A student has derived the following nondimensionally homogeneous equation: a=x/t2-vt+F/m where v is a velocity's magnitude , a is an acceleration's magnitude, t is a time, m is a mass, F is a force's magnitude , and x is a distance (or length). Which terms are dimensionally homogeneous? .
a) x/t
b) vt
c) a
d) F/m
Answer:
Letter C) and D) is the correct answer.
Explanation:
We know that the a is an acceleration's magnitude, so the units of a are m/s².
Now, let's analyze each terms. If we want that each term will be dimensionally homogeneous, all of them must have the same units of a.
[tex][\frac{x}{t}]=[\frac{m}{s}][/tex]
[tex][vt]=[m][/tex]
[tex][\frac{F}{m}]=[\frac{N}{kg}]=[kg\frac{m}{s^{2}kg}]=[\frac{m}{s^{2}}][/tex]
Therefore, the term F/m is the correct answer.
I hope it helps you!
We can see that a and F/M are dimensionally homogeneous.
In solving dimensions, we try to express a quantity in terms of the fundamental quantities;
MassLengthTimeFor the term a, its dimension is LT^-2
For the term F/m, its dimension is LT^-2
Hence, it follows that a and F/M are dimensionally homogeneous.
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A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s
an aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. the hole is 30mm in diameter and is 30mm and is 100mm long. if modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180KN
Answer:
ΔL = 1.011 mm
Explanation:
Let's begin by listing out the given information:
Length (L) = 600 mm = 0.6 m,
Diameter (D) = 40 mm = 0.04 m ⇒ Radius (r) = 20 mm = 0.2 m,
Area (cross sectional) = πr² = 3.14 x .02² = 0.001256 m²,
Modulus of Elasticity (E) = 85 GN/m²,
Compressive load (F) = 180 KN
Using the formula, Stress = Load ÷ Area
Mathematically,
σ = F ÷ A = 180 x 10³ ÷ 0.001256
σ = 143312.1 KN/m²
Modulus of elasticity = stress ÷ strain
E = σ ÷ ε
ε = ΔL/L
85 x 10⁹ = 143312.1 x 10³ ÷ (ΔL/L)
ΔL = 143312.1 x 10³ ÷ 85 X 10⁹ = 1686.02 * 10⁻⁶
ΔL = L x 1686.02 * 10⁻⁶
ΔL = 0.6 * 1686.02 * 10⁻⁶ = 1011.61 x 10⁻⁶
ΔL = 1.011 x 10⁻³ m
ΔL = 1.011 mm
∴The bar contracts by 1.011 mm
4. A neutrally charged conductor has a negatively charged rod brought close to it, and thus has an induced positive charge on the surface closest to the rod. What can we say about the overall charge on the conductor
Answer:
Overall charge still remains zero on conductor until touched by charged rod.
Explanation:
Here, we want to know what has happened to the overall charge on the conductor.
Since the conductor is neutral, the overall charge on the conductor must remain zero because positive charge is induced on close end to rod then equal and negaitve charge is induced on far end to rod.
Thus, overall charge still remains zero on conductor until touched by charged rod.
A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible. Part A What is the speed of the rock when the string passes through the vertical position
Answer:
v = 3.33 m/s
Explanation:
In the position of 45 degrees, all the energy of the rock is gravitational, then we have:
E = m*g*L*cos(angle)
and in the vertical position of the string, all the energy is kinetic, so we have:
E = m*v^2/2
If there is no dissipation, both energies are equal, so we have:
m*g*L*cos(45) = m*v^2/2
9.81 * 0.8 * 0.7071 * 2 = v^2
v^2 = 11.0986
v = 3.33 m/s
A river flows due south with a speed of 5.00 m/s. A man steers a motorboat across the river; his velocity relative to the water is 4.00 m/s due east. The river is 780 m wide. Part A What is the magnitude of his velocity relative to the earth
Answer:
6.4 m/s
Explanation:
From the question, we are given that
Speed of the river, v(r) = 5 m/s
velocity relative to the water, v(w) = 4 m/s
Width of the river, d = 780 m
The magnitude of his velocity relative to the earth is v(m)
v(m) can be gotten by using the relation
[v(m)]² = [v(w)]² + [v(r)]²
[v(m)]² = 4² + 5²
[v(m)]² = 16 + 25
[v(m)]² = 41
v(m) = √41
v(m) = 6.4 m/s
thus, the magnitude of the velocity relative to earth is 6.4 m/s
Dacia asks Katarina why it is important to learn a new coordinate system, because they have been using the Cartesian coordinate system and it seems to Dacia that it works fine. Which of Katarina's replies to Dacia are correct?
Answer: A and D
a. Many objects move in arcs of circles or complete circles at times, and polar coordinates allowthe motion of such objects to be comprehended more easily. Some of Newton's laws in certaincases, such as calculation ofg from first principles, are much easier to calculate using polar coordinates.
b. "These coordinates are just used to confuse students.
c. ""Physics teachers are helping math teachers by getting students to practice theirtrigonometry.
d. ""In some cases, such as addition of forces, where a force magnitude is specified, it is simpler todescribe the forces in polar coordinates and be able to convert to the xy representation.
Explanation:
(A). Any force possessing a fixed magnitude and direction, it's always important to describe the for exactly as it is and also be able to convert it from one coordinate representation to another.
(D). Anything which involves radial motion(this is a motion along a radius) or motion along an arc of a circle or ellipse, this kind of motion is best explained and easily understood when in polar coordinates.
Tech A says that as engines gain miles, the spark plug gap increases, which raises the ignition system’s available voltage. Tech B says that misfire occurs when required voltage is higher than available voltage. Who is correct? Group of answer choices
Answer: Tech A is correct
Explanation:
Every vehicle has ignition system and without this system,it will not work. The battery of everything vehicle contain energy that start the vehicle and ignore it to start working. Electrical current move from the vehicle's battery and get to the induction coil, the induction coil increases the voltage in it so that the plug will be ignited. The spark plugs produce fire. The spark plug is connected to the ignition system. Once voltage is produced from the induction coil, electrical impulses move from induction coil to insulated plug wires. The spark plug need a very high voltage from the small voltage battery. Once the high voltage exceed the dielectric strength of the gases, spark jump the gap between the plug's fire end.
n astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a time interval 11.2 s . What are the average acceleration in each interval? Assume that the positive direction is to the right.
Answer:-
-1 m/s^2
Explanation:
The average acceleration is given by dividing the change in velocity by change in time;
[tex]a_f=\frac{v_f-v_i}{t_f-t_i}[/tex]
[tex]=\frac{(0-11.2)}{(11.2-0)}=-1 m/s^2[/tex]
the point to be noted here is if the velocity is to the left we substitute it with a negative sign and if it is to the right we substitute it with a positive sign.
the heat capacity of 0.125Kg of water is measured to be 523j/k at a room temperature.Hence, calculate the heat capacity of water
(a) per unit mass
(b) per unit volume
Answer:
A. 4148 J/K/Kg
B. 4148 J/K/L
Explanation:
A. Heat capacity per unit mass is known as the specific heat capacity, c.
C = Heat capacity/mass(kg)
C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg
B. Volume of water = mass/density
Density of water = 1 Kg/L
Volume of water = 0.125 Kg/ 1Kg/L
Volume of water = 0.125 L
Heat capacity per unit volume = (523 J/K) / 0.125 L
Heat capacity per unit volume = 4148 J/K/L
1. In 214 BC, Archimedes invented a large spherical-type mirror used to focus the sun's intense rays onto far away enemy boats, which would eventually light them on fire. If the boats were travelling in a nearby channel approximately 1,000 m from the river bank, what would the radius of curvature of his mirror need to be? Show your work.
Answer:
2000 m
Explanation:
since the boat is 1000 m from the river bank, the beam must be focused at this point. This indicates that the focal length is 1000 m
for a spherical mirror, the focal length is given by
f = R/2
where R is the radius of curvature
1000 = R/2
R = 2000 m
R = 2000 m
this means that the radius of curvature must be 2000 m
Find another example of separation that is used to extract a material made useful by humans. Describe the process of separation and what we use the separated component for. (4-6 sentences)
If anyone would answer this I’ll answer ur questions for return!
Please and thank you!
Answer:
Salt
Explanation:
Salt plays a crucial role in maintaining human health. It is the main source of sodium and chloride ions in the human diet. Sodium is essential for the nerve and muscle function and is involved in the regulation of fluids in the body. Sodium also plays a role in the body's control of blood pressure and volume. Salt is harvested by seawater or brine is fed into large ponds of water and is drawn out through natural evaporation which allows the salt to be subsequently harvested.
Have a good day and stay safe!
A proton moving along the x axis has an initial velocity of 4.0 × 106 m/s and a constant acceleration of 6.0 × 1012 m/s2. What is the velocity of the proton after it has traveled a distance of 80 cm? Group of answer choices
Answer:
5.06*10^6 m/s
Explanation:
Given that
Initial velocity, u = 4*10^6 m/s
Acceleration, a = 6*10^12 m/s²
Distance traveled, s = 80 cm
Final velocity, v = ?
We can find the final velocity by using one of the equations of motion.
v² = u² + 2as
On substituting the values, we have
v² = (4*10^6)² + 2 * 6*10^12 * 0.8
v² = 2.56*10^13
v = √2.56*10^13
v = 5.06*10^6 m/s
Therefore, the final velocity of the proton is adjudged to be 5.06*10^6 m/s
The final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex].
The given parameters;
initial velocity of the proton, u = 4 x 10⁶ m/sacceleration of the proton, a = 6 x 10¹² m/s²distance traveled by the proton, s = 80 cm = 0.8 mThe final velocity of the proton over the given distance is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v^2 = (4\times 10^6)^2 \ + \ 2(6.0 \times 10^{12})(0.8)\\\\v^2 = 2.56 \times 10^{13} \\\\v = \sqrt{2.56 \times 10^{13} } \\\\v = 5.06 \times 10^6 \ m/s[/tex]
Thus, the final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex]
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The froghopper, a tiny insect, is a remarkable jumper. Suppose a colony of the little critters is raised on Rhea, a moon of Saturn, where the acceleration due to gravity is only 0.264 m/s2 , whereas gravity on Earth is =9.81 m/s2 . If on Earth a froghopper's maximum jump height is ℎ and its maximum horizontal jump range is R, what would its maximum jump height and range be on Rhea in terms of ℎ and R? Assume the froghopper's takeoff velocity is the same on Rhea and Earth.
Answer:
Maximum height of jump on Rhea is 37.16 times of that on Earth, i.e 37.16h
Maximum range of jump on Rhea is 37.16 of times that on Earth, i.e 37.16R
Explanation:
The acceleration due to gravity on Rhea = 0.264 m/s^2
Acceleration due to gravity on earth here = 9.81 m/s^2
this means that the acceleration due to gravity g on earth is 9.81/0.264 = 37.16 times that on Rhea.
maximum height that can be achieved by the froghopper is given by the equation;
h = [tex]\frac{u^{2}sin^{2} \alpha}{2g}[/tex]
let us put all the numerator of the equation as k, since the velocity of take off is the same for Earth and Rhea. The equation is simplified to
h = [tex]\frac{k}{2g}[/tex]
for earth,
h = [tex]\frac{k}{2*9.81}[/tex] = [tex]\frac{k}{19.62}[/tex]
for Rhea,
h = [tex]\frac{k}{2*0.264}[/tex] = [tex]\frac{k}{0.528}[/tex]
therefore,
h on Rhea is [tex]\frac{k}{0.528}[/tex] ÷ [tex]\frac{k}{19.62}[/tex] = 37.16 times of that on Earth, i.e 37.16h
Equation for range R is given as
R = [tex]\frac{u^{2}sin 2\alpha}{g}[/tex]
following the same approach as before,
R on Rhea will be [tex]\frac{k}{0.264}[/tex] ÷ [tex]\frac{k}{9.81}[/tex] = 37.16 of times that on Earth, i.e 37.16R
A 2.8 kg block slides with a speed of 2.4 m/s on a frictionless horizontal surface until it encounters a spring. Part A If the block compresses the spring 5.6 cm before coming to rest, what is the force constant of the spring
Answer:
5,142.86Explanation:
The kinetic energy possessed by the block when sliding will be equal to the energy needed to compress the string.
Kinetic energy = 1/2 mv² and energy stored in the spring = 1/2 ke²
m = mass of the block (in kg) = 2.8 kg
v = speed of the block (in m/s) = 2.4 m/s
k = force constant of the spring
e = extension (in metres) = 0.056m
Since KE = energy stored in the spring
1/2 mv² = 1/2 ke²
mv² = ke²
2.8(2.4)² = k(0.056)²
16.128 = 0.003136k
k = 16.128/0.003136
k = 5,142.86
The force constant of the spring is 5,236.36
The force that constant of the spring is 5,142.86.
Calculation of the force:The kinetic energy that should be possessed by the block at the time when sliding will be equivalent to the energy required to compress the string.
Here
Kinetic energy = 1/2 mv² and energy stored in the spring = 1/2 ke²
m = mass of the block (in kg) = 2.8 kg
v = speed of the block (in m/s) = 2.4 m/s
k = force constant of the spring
e = extension (in metres) = 0.056m
Since KE = energy stored in the spring
So,
1/2 mv² = 1/2 ke²
mv² = ke²
Now
2.8(2.4)² = k(0.056)²
16.128 = 0.003136k
k = 16.128/0.003136
k = 5,142.86
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A potential difference of 71 mV is developed across the ends of a 12.0-cm-long wire as it moves through a 0.27 T uniform magnetic field at a speed of 6.0 m/s. The magnetic field is perpendicular to the axis of the wire.
Required:
What is the angle between the magnetic field and the wire's velocity?
Answer:
Explanation: please see attached file I attached the answer to your question.
The angle between the magnetic field and the wire's velocity is 33.2 degrees.
Calculation of the angle:Since the potential difference = 71mv = 71 *10 ^-3 V
The length is 12 cm = 0.12m
The magnetic field i.e. B = 0.27T
The speed or v = 4 m/s
here we assume [tex]\theta[/tex] be the angle
So,
e = Bvl sin[tex]\theta[/tex]
So,
[tex]Sin\theta[/tex] = e/bvl
= 71*10^-3 / 0.27 *4*0.12
= 0.5478
= 33.2 degrees
Therefore, the angle should be 33.2 degrees
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Escaping from a tomb raid gone wrong, Lara Croft (m = 62.0 kg) swings across an alligator-infested river from an 8.80-m-long vine. If her speed at the bottom of the swing is 6.30 m/s and she makes it safely across the river, what is the minimum breaking strength of the vine?
Answer:
887.2 N
Explanation:
Given that
Mass of Lara, m = 62 kg
Length of the vine, l = 8.8 m
Speed of the swing, v = 6.3 m/s Then,
We start by calculating her weight.
w = mass * acceleration
w = mg
w = 62 * 9.8
w = 607.6 N
F(c) = mv²/l, on substituting
F(c) = (62 * 6.3²) / 8.8
F(c) = 2460.78 / 8.8
F(c) = 279.6 N
T - 607.6 N = 279.6 N
T = 607.6 N + 279.6 N
T = 887.2 N
Thus, the minimum breaking strength of the vine is 887.2 N
The minimum breaking strength of the vine is about 900N.
TensionTension is a force developed by a rope, string, or cable when stretched under an applied force.
Given:
Mass (m) = 62 kg, length (l) = 8.8 m, velocity (v) = 6.3 m/s, g = 10 m/s².
Weight(W) = m * g = 62 * 10 = 620N
centripetal force = F(c) = mv²/l = 62 * 6.3² / 8.8 = 279.6N
T - W = F(c)
T - 620 = 279.6
T = 900N
The minimum breaking strength of the vine is about 900N.
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A student throws a set of keys vertically upward to his fraternity brother, who is in a window 3.60 m above. The brother's outstretched hand catches the keys 1.60 s later. (Take upward as the positive direction. Indicate the direction with the sign of your answer.)With what initial velocity were the keys thrown?
Answer:
[tex]v_{i}=10.10 m/s[/tex]
Explanation:
The equation of the position is:
[tex]y=y_{i}+v_{i}t-0.5gt^{2}[/tex]
Where:
v(i) is the initial velocity
The initial position y(i) will be zero and the final position y = 3.60 m.
So, we just need to solve this equation for v(i).
[tex]v_{i}=\frac{y+0.5gt^{2}}{t}[/tex]
[tex]v_{i}=\frac{3.6+0.5*9.81*1.6^{2}}{1.6}[/tex]
[tex]v_{i}=10.10 m/s[/tex]
Therefore, the initial velocity is 10.10 m/s upwards.
I hope it helps you!
HELPP ?Air at a temperature of 27 C and 1 atm pressure in a 4 liter cylinder of a diesel engine There. By pushing the piston, the volume of air shrinks 16 times and the pressure increases 40 times. a) How many moles of air are in the cylinder. b) What is the final temperature of the air?
Answer:
a. 0.16240664737515434 moles
b. 67.5 degrees Celcius
Explanation:
a. Use Ideal Gas Equation
PV=nRT
Where P = pressure in pascals, V=Volume in cubic meters, n=number of moles, R is a constant=8.314 J/mol.K and T is temperature in Kelvin.
27C = 273+27=300Kelvin
volume 4L = 0.004m^3
Pressure = 1atm = 101325 Pascal
PV=nRT
101325Pa*0.004m^3=n*8.314J/mol.K*300K
Solving for n from the above you get n=0.16240664737515434 moles
b.Use combined gas law equation
P1*V1/T1=P2*V2/T2
P1= 1atm
V1=4L
T1=27C
P2= 4/16 L =0.25L
P=1*40 atm = 40atm
We do not know T2
USING THE FORMULA
(1atm*4L)/27C = (40atm*0.25L)/T2
(1*4)/27=(40*0.25)/T2
IF you simplify for T2, you get 67.5
Hence final temperature = 67.5 degrees Celcius
Suppose that 7.4 moles of a monatomic ideal gas (atomic mass = 1.39 × 10-26 kg) are heated from 300 K to 500 K at a constant volume of 0.74 m3. It may help you to recall that CV = 12.47 J/K/mole and CP = 20.79 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.
1) How much energy is transferred by heating during this process?2) How much work is done by the gas during this process?3) What is the pressure of the gas once the final temperature has been reached?4) What is the average speed of a gas molecule after the final temperature has been reached?5) The same gas is now returned to its original temperature using a process that maintains a constant pressure. How much energy is transferred by heating during the constant-pressure process?6) How much work was done on or by the gas during the constant-pressure process?
Answer:
Explanation:
1 ) Since it is a isochoric process , heat energy passed into gas
= n Cv dT , n is no of moles of gas , Cv is specific heat at constant volume and dT is rise in temperature .
= 7.4 x 12.47 x ( 500 - 300 )
= 18455.6 J.
2 ) Since there is no change in volume , work done by the gas is constant.
3 ) from , gas law equation
PV = nRT
P = nRT / V
= 7.4 x 8.3 x 500 / .74
= .415 x 10⁵ Pa.
4 ) Average kinetic energy of gas molecules after attainment of final temperature
= 3/2 x R/ N x T
= 1.5 x 1.38 x 10⁻²³ x 500
= 1.035 x 10⁻²⁰ J
1/2 m v² = 1.035 x 10⁻²⁰
v² = 2 x 1.035 x 10⁻²⁰ / 1.39 x 10⁻²⁶
= 1.49 x 10⁶
v = 1.22 x 10³ m /s
5 ) In this process , pressure remains constant
gas is cooled from 500 to 300 K
heat will be withdrawn .
heat withdrawn
= n Cp dT
= 7.4 x 20.79 x 200
= 30769.2 J .
6 )
gas will have reduced volume due to cooling
reduced volume = .74 x 300 / 500
= .444 m³
change in volume
= .74 - .444
= .296 m³
work done on the gas
= P x dV
pressure x change in volume
= .415 x 10⁵ x .296
= 12284 J.
At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball's acceleration is 950 m/s2 and the vertical or y component of its acceleration is 750 m/s2. The ball's mass is 0.35 kg. What is the magnitude of the net force acting on the soccer ball at this instant?
Answer:
F = 423.63 N
Explanation:
Since, the x-component and y-components of the acceleration of ball are given. Therefore, we need to find the resultant or net acceleration of the soccer ball first. For that purpose we use to the formula for the resultant of rectangular components of a vector:
a = √(ax² + ay²)
where,
a = net acceleration = ?
ax = x - component of acceleration = 950 m/s²
ay = y - component of acceleration = 750 m/s²
Therefore,
a = √[(950 m/s²)² + (750 m/s²)²]
a = 1210.4 m/s²
Now, from Newton's Second Law, we know that:
F = ma
where,
m = mass of ball = 0.35 kg
F = Net force acting on ball = ?
F = (0.35 kg)(1210.4 m/s²)
F = 423.63 N
An automobile being tested on a straight road is 400 feet from its starting point when the stopwatch reads 8.0 seconds and is 550 feet from the starting point when the stopwatch reads 10.0 seconds.
A. What was the average velocity of the automobile during the interval from t = 10.0 seconds to t = 8.0 seconds
B. What was the average velocity of the automobile during the interval from t - Ostot - 10.0 s? (Assume that the stopwatch read t = 0 and started at the same time as the auto.)
C. If the automobile averages 100 ft/s from t - 10.0 stot - 20.0 s, what distance does it travel during this interval?
D. The automobile has a special speedometer calibrated in feet/s instead of in miles/hour. Att 85 the speedometer reads 65 ft/s; and at t = 10 s it reads 80 ft/s. What is the average acceleration during this interval?
Answer:
a) v = 75 ft / s , b) v = 55 ft / s , c) Δx = 1000 ft
Explanation:
We can solve this exercise with the expressions of kinematics
a) average speed is defined as the distance traveled in a given time interval
v = (x₂-x₁) / (t₂-t₁)
v = (550 - 400) / (10 -8)
v = 75 ft / s
b) we repeat the calculations for this interval
v = (550 - 0) / (10 -0)
v = 55 ft / s
c) we clear the distance from the average velocity equation
Δx = v (t₂ -t₁)
Δx = 100 (20-10)
Δx = 1000 ft
A turntable rotates with a constant 1.85 rad/s2 clockwise angular acceleration. After 4.00 s it has rotated through a clockwise angle of 30.0 rad . Part A What was the angular velocity of the wheel at the beginning of the 4.00 s interval?
Answer: The angular velocity of the wheel at the beginning of the 4.00 s interval is 3.8 rad/s
Explanation: Please see the attachment below
The angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.
The given parameters:
Angular speed of the turn table = 1.85 rad/s²Time of motion, t = 4.0 sAngular displacement, θ = 30.0 radThe angular velocity of the wheel at the beginning of the 4.0 s time is calculated as follows;
[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2[/tex]
where;
[tex]\omega_i[/tex] is the initial angular velocity
[tex]30 = \omega_i (4) \ + \frac{1}{2}(1.85)(4)^2\\\\30 = 4\omega _i + 14.8\\\\4\omega _i = 30 - 14.8\\\\ 4\omega _i = 15.2\\\\\omega _i = \frac{15.2}{4} \\\\\omega _i = 3.8 \ rad/s[/tex]
Thus, the angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.
Learn more about angular velocity here: https://brainly.com/question/540174
The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor.
a. What is the electric field strength and direction between the plates?
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules
A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 592 N. As the elevator later stops, the scale reading is 400 N. Assume the magnitude of the acceleration is the same during starting and stopping. (a) Determine the weight of the person. N (b) Determine the person's mass. kg (c) Determine the magnitude of acceleration of the elevator. m/s2
Answer:
a) 496Nb) 50.56kgc) 1.90m/s²Explanation:
According to newton's secomd law, ∑F = ma
∑F is the summation of the force acting on the body
m is the mass of the body
a is the acceleration
Given the normal force when the elevator starts N1 = 592N
Normal force after the elevator stopped N2 = 400N
When the elevator starts, its moves upward, the sum of force ∑F = Normal (N)force on the elevator - weight of the person( Fg)
When moving up;
N1 - Fg = ma
N1 = ma + Fg ...(1)
Stopping motion of the elevator occurs after the elevator has accelerates down. The sum of forces in this case will give;
N2 - Fg = -ma
N2 = -ma+Fg ...(2)
Adding equation 1 and 2 we will have;
N1+N2 = 2Fg
592N + 400N = 2Fg
992N 2Fg
Fg = 992/2
Fg = 496N
The weight of the person is 496N
\b) To get the person mass, we will use the relationship Fg = mg
g = 9.81m/s
496 = 9.81m
mass m = 496/9.81
mass = 50.56kg
c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;
N1-N2 = 2ma
592-400 = 2(50.56)a
192 = 101.12a
a = 192/101.12
a = 1.90m/s²
A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is at a radial distance of 1.8 m from the central axis. Which is true?A- Boy and girl have zero tangential and angular accelerations.B- The girl has a larger angular acceleration than the boy.C- The boy has a larger tangential acceleration than the girl.D- The boy has a larger angular acceleration than the girl.E- The girl has a larger tangential acceleration than the boy.
Answer:
E) True. The girl has a larger tangential acceleration than the boy.
Explanation:
In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.
Angular and linear quantities are related
v = w r
a = α r
the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m
as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different
v₁ = w 1,2 (boy)
v₂ = w 1.8 (girl)
whereby
v₂> v₁
reviewing the claims we have
a₁ = α 1,2
a₂ = α 1.8
a₂> a₁
A) False. Tangential velocity is different from zero
B) False angular acceleration is the same for both
C) False. It is the opposite, according to the previous analysis
D) False. Angular acceleration is equal
E) True. You agree with the analysis above,