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3 years ago

6

A box moves 1000 m horizontally as a force F 2000 N is applied downward. What is the work done on the box by F during the displa

cement?

1 answer:

stiks02 [169]3 years ago

5 0

Answer:

D. OJ

Explanation:

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A 5kg object accelerates from 3m/s to 7m/s in 5 seconds. Calculate the force required

Ad libitum [116K]

Answer:

4N

Explanation:

a = (7-3)/5 = 0.8m/s^2

F = ma = (5)(0.8) = 4 Newtons

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3 years ago

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Marysya12 [62]

Answer:

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3 years ago

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8) Calculate the kinetic energy of a truck that has a mass of 6100 kg and is moving at 55 m/s.

Inessa [10]

Answer:

<h2>9,226,250 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

v is the velocity

m is the mass

From the question we have

$k = \frac{1}{2} \times 6100 \times {55}^{2} \\ = 3050 \times 3025 \\ = 9226250$

We have the final answer as

<h3>9,226,250 J</h3>

Hope this helps you

7 0

3 years ago

Two 2.0 kg bodies, A and B, collide. The velocities before the collision are ~vA = (15ˆi + 30ˆj) m/s and ~vB = (−10ˆi + 5.0ˆj) m

AleksandrR [38]

Answer:

Part a)

$10\hat i + 15\hat j = \vec v$

Part b)

$\Delta K = 500 J$

Explanation:

As we know that there is no external force on the system of two masses so here total momentum of the system will remains conserved

so we can say

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

$(2kg)(15\hat i + 30 \hat j) + (2 kg)(-10\hat i + 5\hat j) = 2kg(-5\hat i + 20\hat j) + 2\vec v$

$5\hat i + 35\hat j = (-5\hat i + 20\hat j) +\vec v$

$10\hat i + 15\hat j = \vec v$

Part b)

magnitude of the initial speed of A = $\sqrt{15^2 + 30^2} = 33.54 m/s$

magnitude of the initial speed of B = $\sqrt{10^2 + 5^2} = 11.18 m/s$

magnitude of final speed of A = $\sqrt{5^2 + 20^2} = 20.61 m/s$

magnitude of final speed of B = $\sqrt{10^2 + 15^2} = 18.03 m/s$

Now change in total kinetic energy is given as

$\Delta K = (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) - (\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2)$

$\Delta K = (\frac{1}{2}2(33.54)^2 + \frac{1}{2}2(11.18)^2) - (\frac{1}{2}2(20.61)^2 + \frac{1}{2}2(18.03)^2)$

$\Delta K = 500 J$

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3 years ago

Which name is given to the type of friction that objects falling through air experience?

Pani-rosa [81]

Some call it "air resistance", and others just call it "drag".

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3 years ago

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