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artcher [175]
3 years ago
5

Which of the following frictionless ramps (A, B, or C) will give the ball the greatest speed at the bottom of the ramp? Explain.

Physics
1 answer:
Elenna [48]3 years ago
8 0
Let the ball is initially at point P1. Finally it reaches at point P2.( bottom most). Since only conservative force is acting on the ball throughout the motion, mechanical energy will be conserved. Therefore conserving energy from P1 to P2, Initially ball is at rest(P1). So initial kinetic energy is 0. Let ball is at height h1 initially, so initial potential energy = mgh ( m= mass of the ball). At bottom most point (P2), potential energy of ball changes into kinetic energy. So final kinetic energy =1/2 mv^2. Final potential energy = 0. Equating energies at P1 and P2, Mgh= 1/2 mv^2. Therefore, v= (2gh)^1/2. Since all the initial and final points are at same height, velocity will be same for all the paths(irrespective of the path taken by ball). So velocity will be same.
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An 85-kg man plans to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that h
Lelu [443]

Answer:

The greatest acceleration the man can give the airplane is 0.0059 m/s².

Explanation:

Given that,

Mass of man = 85 kg

Mass of airplane = 109000 kg

Distance = 9.08

Coefficient of static friction = 0.77

We need to calculate the greatest friction force

Using formula of friction

F=\mu mg

Where, m = mass of man

g = acceleration due to gravity

Put the value into the formula

F = 0.77\times85\times9.8

F= 641.41\ N

We need to calculate the acceleration

Using formula of newton's second law

F = ma

a=\dfrac{F}{m}

Put the value into the formula

a=\dfrac{ 641.41}{109000}

a=0.0059\ m/s^2

Hence, The greatest acceleration the man can give the airplane is 0.0059 m/s².

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3 years ago
What is the student's kinetic energy at the bottom of the hill if he is moving
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Answer:

KE = 10530 J or 10.53 KJ

Explanation:

The formula for kinetic energy is KE = 1/2 mv^2

Let's apply the formula:

KE = 1/2 mv^2

KE = 1/2 (65kg) (18m/s)^2

KE = 10530 J or 10.53 KJ

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3 years ago
A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas
Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

\varepsilon' =\dfrac{S_y}{E}

\varepsilon' =\dfrac{240}{110\times 1000}

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

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A university is planning to send an optical telescope into orbit. stuents attending the university will be able to view which of
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Where are the answers? If there's anything about a white light coming from space I would choose that one. 
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