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2 years ago

A bulldozer does 45,000 joules of work to push a boulder a distance of 30 meters.

Physics

1 answer:

Andrew [12]2 years ago

7 0

Answer:

1500N

Explanation:

we have ,

w= f*d

45,000 =f * 30

f=45000/30

f=1500N..

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In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1

lara31 [8.8K]

Answer:

Part a)

$f = \frac{8}{9}$

Part b)

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 500 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$2v_o = 2v_{1f} + v_o + v_{1f}$

now we have

$v_{1f} = \frac{v_o}{3}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{8}{9}$

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 300 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$10v_o = 10v_{1f} + 3(v_o + v_{1f})$

now we have

$v_{1f} = \frac{7v_o}{13}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

8 0

2 years ago

A 1.0 × 10^3 kg sports car is initially traveling at 15 m/s. The driver then applies the brakes for several seconds so that -25

algol13

Answer:

Ea = 112500[J]

Eb = 87500[J]

Explanation:

To solve this problem we must use the principle of energy conservation which tells us that the energy of a body plus the work done or applied by the body equals the final energy of a body.

This can be easily visualized by the following equation:

$E_{A}+E_{friction}=E_{B}$

Now we must define the energies at points A & B.

<u>For point A</u>

At point A we only have kinetic energy since it moves at 15 [m/s]

So the kinetic energy

$E_{A}=\frac{1}{2}*m*v_{A}^{2} \\E_{A}=\frac{1}{2} *1000*(15)^{2} \\E_{A}=112500[J]$

The final kinetic energy can be calculated as follows:

$112500-25000=E_{B}\\E_{B}=87500[J]$

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2 years ago

Colors of light that combine to produce white light PLEASE COMMENT WITH A B C D

Oduvanchick [21]

A- additive colors

Because if colors are added it gets brighter but if colors are subtracted it makes black

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2 years ago

A pumpkin is dropped after 5 seconds its velocity is 47m/s what is its acceleration

ANEK [815]

9.81m/s Hope i helped

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3 years ago

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The magnetic field around a current carrying wire at a distance of 1cm is twice as strong as 2cm . How does the Ford strength at

masha68 [24]

Answer:

It is half the field strength at 0.5cm

Explanation:

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2 years ago