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gladu [14]
3 years ago
9

You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after

pushing it across some distance d, and the block leaves your hand with a velocity of 0.85 m/s. While you are pushing, the work done by friction between the ice and the ground is 3 Nm (3 J). Assuming that the ice block was stationary before you push it, find d.
Physics
1 answer:
kotykmax [81]3 years ago
7 0

Answer: d = 33 cm or 0.33 m

Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:

W = F.d.cosθ

F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.

For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:

W = F.d.cos(0)

W = F.d

To determine d:

d = \frac{W}{F}

d = \frac{3}{9}

d = 0.33 m

The distance d the block ice moved is 33 cm.

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Suppose a baseball pitcher throws the ball to his catcher.
amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

\Delta v = |v-u|

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

\Delta v=|v-0|=|v|

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

\Delta v =|0-v|=|-v|

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

\Delta p = m \Delta v

where

m is the mass of the object

\Delta v is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

|\Delta p|=m|\Delta v|

- In situation 1 (pitcher throwing the ball), the change in momentum is

\Delta p = m|\Delta v|=m|v|=mv

- In situation 2 (catcher receiving the ball), the change in momentum is

\Delta p = m\Delta v = m|-v|=mv

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

I=\Delta p

where

I is the impulse

\Delta p is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it  doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

I=F\Delta t

where

I is the impulse

F is the force applied

\Delta t is the time of impact

This can be rewritten as

F=\frac{I}{\Delta t}

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- \Delta t when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to \Delta t, this means that the force is larger when the ball is catched.

6 0
3 years ago
suppose that a large cargo truck needs to cross a bridge. the truck is 30 m long and 3.2 m wide. the cargo exerts a force nof 54
Tamiku [17]
<span>No, because the truck applies more pressure than the bridge can support.</span>
7 0
3 years ago
A 100 N force is applied to move an object a horizontal distance of 5 meters at constant speed in 10 seconds. How much power is
Tpy6a [65]

Answer:

50 W

Explanation:

<h3><u>Given :</u></h3>

  • Force applied = 100 N
  • Distance covered = 5 metres
  • Time = 10 seconds

<h3><u>To find :</u></h3>

Power

<h3><u>Solution :</u></h3>

For calculating power, we first need to know about the work done.

\bf \boxed{Work = Force \times displacement}

Now, substituting values in the above formula;

Work = 100 × 5

= 500 Nm or 500 J

We know that,

\bf \boxed{Power=\dfrac{Work\:done}{Time\: taken}}

Substituting values in above formula;

Power = 500/ 10

= 50 Nm/s or 50 W

Hence, power = 50 W .

5 0
3 years ago
a college student produces about 100 kcal of heat per hour on the average what is the rate of energy production and joules
Bond [772]

Given:

Amount of heat produced = 100 kcal per hour

Let's find the rate of energy production in joules.

We know that:

1 calorie = 4.184 Joules

1 kcal = 4.184 Joules

To find the rate of energy production in Joules, we have:

\begin{gathered} Rate=100\ast4.184 \\  \\ \text{Rate}=418.4\text{ KJ/hour} \end{gathered}

Therefore, the rate of energy production in joules is 418.4 kJ/h which is equivalent to 418400 Joules

ANSWER:

418.4 kJ/h

6 0
11 months ago
A passenger in a train accelerating smoothly away from a station observes that a child’s yo-yo hanging by its string from a lugg
olchik [2.2K]

Answer:

1.73 m/s²

3.0 cm

Explanation:

Draw a free body diagram of the yo-yo.  There are two forces: weight force mg pulling down, and tension force T pulling up 10° from the vertical.

Sum of forces in the y direction:

∑F = ma

T cos 10° − mg = 0

T cos 10° = mg

T = mg / cos 10°

Sum of forces in the x direction:

∑F = ma

T sin 10° = ma

mg tan 10° = ma

g tan 10° = a

a = 1.73 m/s²

Draw a free body diagram of the sphere.  There are two forces: weight force mg pulling down, and air resistance D pushing up.  At terminal velocity, the acceleration is 0.

Sum of forces in the y direction:

∑F = ma

D − mg = 0

D = mg

½ ρₐ v² C A = ρᵢ V g

½ ρₐ v² C (πr²) = ρᵢ (4/3 πr³) g

3 ρₐ v² C = 8 ρᵢ r g

r = 3 ρₐ v² C / (8 ρᵢ g)

r = 3 (1.3 kg/m³) (100 m/s)² (0.47) / (8 (7874 kg/m³) (9.8 m/s²))

r = 0.030 m

r = 3.0 cm

3 0
3 years ago
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