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3 years ago

10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of

Physics

1 answer:

Rina8888 [55]3 years ago

4 0

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

$W=Fd cos \theta$

where :

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

$\theta=22^{\circ}$ is the angle between the direction of the force and the displacement

Substituting, we find the work done

$W=(87)(44)(cos 22^{\circ})=3549 J$

Learn more about work:

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A piston–cylinder device initially contains 0.6 m3 of saturated water vapor at 250 kPa. At this state, the piston is resting on

AysviL [449]

Answer:

T2 = 662°c Q(in) = 911kj W(out) = 180kj

Explanation:

1. Firstly, we find the mass from the initial volume and specific volume from steam table A-5 for given pressure and state.

M = V(1)/a(1) = 0.6/0.7183(kg)

Where a(1) = specific volume

M = 0.835kg

The work is determined from the final pressure and the volume change that occurs when it is reached.

W = P2(V2 - V1)

W = P2V2

= 300×0.6kg = 180kg

Therefore, to get the final temperature and final pressure, the final temperature is determined fro steam table A-6 using interpolation.

T2 = T1* + T2*-T1*/a2*-a1* (a2-a1*)

T2 = 600 + 700-600/1.4958-1.34139 (1.43746-1.34139)

T2 = 662°c

2. The final internal energy can be determined in the same way just like the initial ones were done. From A-5 on steam table given the pressure

The heat Q(in) = ∆u + W(out)

= m(U2-U1) + W(out)

= 0.835(3412.3-2536.8)kj

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3 years ago

A charged particle (Q = 6 C, m = 0.02 kg) moves at a speed of 70 m/s perpendicular to the direction of a uniform magnetic field

san4es73 [151]

Answer:

Explanation:

When a charged particle moves in a magnetic field , a force acts on it perpendicular to its velocity.

Force = Bqv , B is magnetic field , q is charge of the particle and v is velocity

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3 years ago

What is the acceleration of a 113 kg object exposed to a 110 N force. (Round to one decimal place.)

Norma-Jean [14]

Answer:

$\boxed {\boxed {\sf a \approx 1.0 \ m/s^2}}$

Explanation:

We are asked to find the acceleration of an object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F= m \times a$

We know the object has a mass of 113 kilograms and it is exposed to a 110 Newton force.

Let's convert the units of force to make the problem easier later. 1 Newton is equal to 1 kilogram meter sper second squared. So, the force of 110 Newtons is equal to 110 kilogram meters per second squared.

F= 110 kg*m/s²
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Substitute the values into the formula.

$110 \ kg*m/s^2=113 \ kg * a$

We are solving for a, the acceleration, so we isolate this variable. It is being multiplied by 113 kilograms. The inverse operation of multiplication is division, so we divide both sides by 113 kg.

$\frac {110 \ kg*m/s^2 }{113 \ kg}= \frac{113 \ kg * a}{113 \ kg}$

$\frac {110 \ kg*m/s^2 }{113 \ kg}=a$

The units of kilograms (kg) cancel.

$\frac {110 m/s^2 }{113 }=a$

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We are asked to round to one decimal place or the tenths place. The 7 in the hundredths place tells us to round the 9 up to a 0, but then we must also round the 0 in the ones place to a 1.

$1.0 \ m/s^2 \approx a$

The acceleration of the object is approximately <u>1.0 meter per second squared.</u>

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2 years ago