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timofeeve [1]
3 years ago
10

The man fire a 50-g arrow that moves at an unknown speed. It hits and embeds in a 350-g block that slides on an air track. At th

e end, the block runs into and compresses a 4000-N/m spring 0.10 mm.
Part A
How fast was the arrow traveling?

Part B
Indicate the assumptions that you made and discuss how they affect the result.
Physics
1 answer:
Natasha_Volkova [10]3 years ago
7 0

Answer:

a)  vAix = 80 m/s

b) The assumptions and implications were:

Assume that friction between the block and the surface it rests on does not change the momentum of the system during the collision.

Assume that friction is negligible throughout the process and the system’s internal energy does not change.

Assume all the system’s kinetic energy is converted into elastic potential energy at the end of the process.

If any of the assumptions are invalid then the arrow must have been travelling initially, vAix > 80 m/s

Explanation:

Arrow embeds into block

Take in the first instance the system to be the arrow + block (isolated). Establish reference coordinate system with the +x axis running horizontally in the direction of the arrow’s motion. The initial state (i) is the arrow travelling with velocity vAix and the final state (f) is the arrow embedded in the block. Now, apply the component form of the Generalized Impulse Momentum Equation to this system:

pAi + pBi + JonA + JonB = pAf + pBf

pAix + pBix + Jx = pAfx + pBfx

mA*vAix + mB*vBix + 0 = (mA + mB)*vfx

0.05*vAix + 0 = (0.05 + 0.35)*vfx

vAix = 8*vfx        (1)

Arrow embeds into block

Now consider the next phase of motion and take as the system the arrow + block + spring. The initial state (i) is the arrow and block travelling with velocity equivalent to the final velocity from equation 1 (final state velocity in first phase becomes initial velocity in next phase);  vix’ = vfx  and the final state (f) is the arrow + block brought to rest and the spring compressed an amount, Δx = 0.1 m. Now, apply the Generalized Work Energy Principle to the system

Ei + W = Ef

Ki + Usi + W = Kf + Usf

0.5*(mA + mB)*vix’² = 0.5*k*Δx²

(0.05 + 0.35)*vfx² = 4000*(0.1)²

vfx = √(40/0.4) = 10 m/s

Substituting above back into equation 1:

vAix = 8* 10 m/s = 80 m/s

Arrow embeds into block

The assumptions and implications were:

Assume that friction between the block and the surface it rests on does not change the momentum of the system during the collision.

Assume that friction is negligible throughout the process and the system’s internal energy does not change.

Assume all the system’s kinetic energy is converted into elastic potential energy at the end of the process.

If any of the assumptions are invalid then the arrow must have been travelling initially, vAix > 80 m/s

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The gravitational field on Saturn is 10.4 times its value on Earth.
nalin [4]

The weight of 100 kg mass on Saturn is 10,192 N.

The weight of 7.0 kg mass on Pluto is 2.01 N.

The given parameters;

  • <em>acceleration due to gravity, = 10.4g</em>
  • <em>mass of the Saturn, m = 100 kg</em>
  • <em>acceleration due to gravity on mercury, = 3.59 m/s²</em>

The weight of 100 kg mass on Saturn is calculated as follows;

W = mg

W = 100 x 10.4 x 9.8

W = 10,192 N

The weight of 7.0 kg mass on Pluto is calculated as follows;

W = 7 x 0.08 x 3.59

W = 2.01 N

Learn more here:brainly.com/question/18603430

5 0
3 years ago
A 1.7-kg firework is fired from the ground straight up on a planet with whose acceleration due to gravity is 4.8 m/s/s. You want
Sunny_sXe [5.5K]

Answer:

41.0 m/s, 10.6 s

Explanation:

Given:

a = -4.8 m/s²

Δy = 165 m

v = -10.0 m/s

Find: v₀ and t

v² = v₀² + 2aΔy

(-10.0 m/s)² = v₀² + 2(-4.8 m/s²) (165 m)

v₀ = 41.0 m/s

Δy = vt − ½ at²

165 m = (-10.0 m/s) t − ½ (-4.8 m/s²) t²

165 = -10t + 2.4t²

0 = 2.4t² − 10t − 165

Solve with quadratic formula:

t = [ -(-10) ± √((-10)² − 4(2.4)(-165)) ] / 2(2.4)

t = (10 ± √1684) / 4.8

t = 10.6 s

4 0
3 years ago
The magnitude of the force associated with the gravitational field is constant and has a value F. A particle is launched from po
Goryan [66]

Answer:

Energy gained by the second particle = 12Uo

Explanation:

Given Data;

Resistant force = 12F

Initial kinetic energy = Uo

Calculating the kinetic energy gained, we have;

u = f *r

where f= resistant force = 20F

r = initial kinetic energy = Uo

Therefore,

U = 12 * uo

  = 12 Uo

Therefore, energy gained by the second particle = 12Uo

4 0
4 years ago
Suppose you increase your walking speed from 4 m/s to 15 m/s in a period of 1 s. What is your acceleration?
hichkok12 [17]

Answer:

The acceleration is: 11\,\frac{m}{s^2}

Explanation:

Recall that acceleration is defined as the change of velocity over the time it took to produce this change. This is expressed mathematically as:

a=\frac{v_f-v_i}{t_f-t_i}

with v_i being the initial velocity of the person (in our case 4 m/s);

v_f being his final velocity (in our case 15 m/s);  

and the difference t_f-t_i the time the change in velocity took (in our case 1 second).

Therefore in our example, the person's acceleration is:  

a=\frac{v_f-v_i}{t_f-t_i}\\a=\frac{15-4}{1}\,\frac{m}{s^2} \\a=11\,\frac{m}{s^2}

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State the objects in the universe that reflect light from stars
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Everything in the universe that is not a star reflects light from Stars. Otherwise you can't see it.
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