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3 years ago

Johnathan wants to lift a 200N box onto a 5m ledge. How much work is done?

Physics

2 answers:

Y_Kistochka [10]3 years ago

8 0

Answer:

1000 J which agrees with answer a in your list of options

Explanation:

Recall the formula for work (W) as the force (F) applied times the distance (d) covered in the same direction of the force:

W = F * d

W = 200 N * 5 m = 1000 J

lara31 [8.8K]3 years ago

3 0

The answer is - a - 100 j

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3 years ago

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Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

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3 years ago

Pairs of magnets are shown in the diagram.

belka [17]

The answer is X. I hope this helped

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3 years ago

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LARGER vibration produces a higher ............

balandron [24]

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4 years ago

Light passes through a 0.15 mm-wide slit and forms a diffraction pattern on a screen 1.25 m behind the slit. The width of the ce

Elena L [17]

Answer:

The value is $\lambda = 900 \ nm$

Explanation:

From the question we are told that

The width of the slit is $a = 0.15 \ mm = 0.00015 \ m$

The distance of the screen from the slit is D = 1.25 m

The width of the central maximum is $y = 0.75 \ cm = 0.0075 \ m$

Generally the width of the central maximum is mathematically represented as

$y = \frac{m * D * \lambda}{a}$

Here m is the order of the fringe and given that we are considering the central maximum, the order will be m = 1 because the with of the central maximum separate's the and first maxima

$\lambda = \frac{a y}{ m * D }$

=> $\lambda = \frac{ 0.000015 * 0.0075}{ 1 * 1.2 }$

=> $\lambda = 900 *10^{-9} \ m$

=> $\lambda = 900 \ nm$

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3 years ago