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Degger [83]
3 years ago
4

A boy does 465 J of work pulling an empty wagon along level ground with a force of 111 N [31° below the horizontal]. A frictiona

l force of 155 N opposes the motion and is actually slowing the wagon down from an initial high velocity. The distance the wagon travels is _____."
W = Fdcosθ
465 J = (111 N)d(cos31°)
d = (465 J)/(111 N)(cos31°)
d = 4.89 m

Where does the frictional force come in here? Does it actually affect the distance travelled or does it just slow down the wagon?
Physics
2 answers:
Agata [3.3K]3 years ago
8 0
It will slow down the wagon and produces heat. we call 'lost' which makes efficiency less than 100%

total energy that the boy exerts = work done + heat
... we consider only the work done so friction force is not included.

how much heat is produced?
heat = friction force x distance
mote1985 [20]3 years ago
8 0

Answer:

Explanation:

When an object is moved over rough floor then during their motion the net work done is

W = 465 J

now the displacement of the wagon can be found by formula of work done as

W = F d cos\theta

465 = 111(d)cos31

d = \frac{465}{111 cos31}

d = 4.89 m

now here this is the energy which is lost by the boy in order to move the wagon

here frictional force is opposite force while wagon is moving on the rough floor

so here we can say that friction force only oppose the motion due to which object slow down and work done by friction force is loss of energy in form of heat

so here we will say that the work done by boy on the wagon is his energy loss and that energy is used to compensate the energy loss due to friction force.

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Troyanec [42]

Since it was stated that it must move at constant velocity, so the only force it must overpower is the frictional force.

So the equation is:

F cos θ = Ff

F cos 36 = 65 N

F = 80.34 N

 

<span>So the nurse must exert 80.34 N of force</span>

4 0
3 years ago
Following a collision between a large spacecraft and an asteroid, a copper disk of radius 28.0 m and thickness 1.20 m, at a temp
coldgirl [10]

Answer:

A. 9.31 x10^10J

B. -8.47x10 ^ 12J

C. 8.38x 10^12J

Explanation:

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3 years ago
Action reaction forces are in which one of newtons laws
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Answer:

the third law

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4 years ago
An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force
sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = 10ms^{-2}, F = ma = 15*10 = 150N

2) m = 3kg, v = 4ms^{-1}, r = 4m, F = \frac{mv^{2} }{r}

\frac{3*4^{2} }{4} = 12N

3) a = 10ms^{-2}, r = 10m, v=?

F = \frac{mv^{2} }{r} and F = ma

equating the two equations and cancelling a, we have:

\frac{v^{2} }{r} = a

making v the subject of formula, we have:

v = \sqrt{ar}

= \sqrt{100}

= 10ms^{-1}

4) r = 10m, v = 5ms^{-1}, a = ?

F = \frac{mv^{2} }{r}

F = ma

equating the above equations and making a subject of formula, we get:

a = \frac{v^{2} }{r}

a = 25/10 = 2.5ms^{-2}

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6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is tripled

F = (3v)^{2}

F = 9v^{2}

We can see that the force will be 9X greater than it was.

9) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is doubled

F = (2v)^{2}

F = 4v^{2}

We can see that the force will be 4X greater than it was.

10) F = \frac{mv^{2} }{r}

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

7 0
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Tpy6a [65]

Answer:

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Explanation:

We can find its velocity when it reaches the buoy by applying one of Newton's equations of motion:

v^2 = u^2 + 2as

where v = final velocity

u = initial velocity

a = acceleration

s = distance traveled

From the question:

u = 28 m/s

a = -4 m/s^2

s = 91 m

Therefore:

v^2 = 28^2 + 2 * (-4) * 91\\\\v^2 = 784 + -728 = 56\\\\v = \sqrt{56}\\ \\v = 7.5 m/s

The velocity of the boat when it reaches the buoy is 7.5 m/s.

6 0
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