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2 years ago

A student pushes a 50-N box across the floor a distance of 15 m. How much work was done to move the box?

Physics

2 answers:

andrezito [222]2 years ago

7 0

Answer:

750 j

Explanation:

I took the test in k12

irinina [24]2 years ago

4 0

Answer:750

Explanation:

50 times 15

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Sodium and phosphorus combine to form Na3P. The name of this compound is .

g100num [7]

Binary ionic compound

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2 years ago

Read 2 more answers

A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a fina

Komok [63]

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

Temperature = 22°C

Internal energy $U=4.30\times10^{3}\ J$

Final temperature = 16°C

We need to calculate the approximate change in entropy

Using formula of the entropy

$\Delta S=\dfrac{\Delta U}{T}$

Where, $\Delta U$ = internal energy

T = average temperature

Put the value in to the formula

$\Delta S=\dfrac{-4.30\times10^{3}}{\dfrac{22+273+16+273}{2}}$

$\Delta S=-14.72\ J/K$

Hence, The approximate change in entropy is -14.72 J/K.

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3 years ago

The work done on a box is 532 joules. The force applied to the box was 48 N. What was the displacement of the box? *

Semenov [28]

Explanation:

Work = force × displacement

532 J = 48 N × d

d ≈ 11 m

8 0

3 years ago

A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to h

timofeeve [1]

To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

PART A) Normal Force.

$F_n = \frac{mv^2}{r}$

$N+mgcos\theta = \frac{mv^2}{r}$

Here,

Normal reaction of the ring is N and velocity of the ring is v

$N+mgcos\theta = \frac{mv^2}{r}$

$N+Wcos\theta = \frac{W}{g} (\frac{v^2}{r})$

$N+2cos30\° = \frac{2}{32.2}*\frac{10^2}{2}$

$N = 1.374lb$

PART B) Acceleration

$F_t = ma_t$

$-mgsin\theta = ma_t$

$-W sin\theta = \frac{W}{g} a_t$

$-2Sin30\° = (\frac{2}{32.2})a_t$

$a_T = -16.10ft/s^2$

Negative symbol indicates deceleration.

<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>

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3 years ago

Astronauts wear liquid cooled space suits to keep their body temperature moderate. One

Anton [14]

Answer:

D. Because they are using space technology on a shirt so people can wear it on earth as well

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2 years ago