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DochEvi [55]
3 years ago
11

Two uniform spheres,each of mass 0.260kg are fixed at points A and B

Physics
1 answer:
Ivan3 years ago
3 0

Answer:

a) 2.76x10^-9 m/s^2

b) downward

Explanation:

a) In order to do this, we need to use the newton's law of gravitation which is:

F = G.m/r²

Where:

G: Constant of gravitation (6.67x10^-11 N/m)

m: mass of the object

r: distance or radius of the object.

In this case, we have two spheres, so this law becomes:

F = G*m1*m2 / r²

Now, you are not providing a picture with this, however, this problem is very similar to one I solved before, so, I'm gonna use the same values to solve this. If, you have different values for the distance, just replace them in this procedure and should get the correct and more accurate answer.

In this case, the distance that separates sphere in P from sphere A and B is 10 cm. As both spheres have the same mass and they are at the same distance from P, we can assume that F1 = F2 so:

F1 = (6.67x10^-11) * (0.01) * (0.26) / (0.1)² = 1.73x10^-11 N

sphere at P, will roll downward to the middle of sphere A and B. The distance between these two points it's 8 cm (See picture attached)

Now, according to my data, they are exerting a force in the y axis, so:

F1y = Fy * senФ

senФ using trigonometry is:

senФ = opposite lenght / hipotenuse

senФ = 8/10 = 0.8

Replacing:

F1y = 1.73x10^-11 * 0.8 = 1.38x10^-11 N

This force it's exactly the same as F2y, so the total force is:

F = 1.38x10^-11 * 2 = 2.76x10^-11 N

Finally, using the second law of Newton:

F = m*a

Solving for a:

a = F/m

replacing:

a = 2.76x10^-11 / 0.01

a = 2.76x10^-9 m/s²

According to the picture and the value of this magnitude, it's going downward.

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When does the transfer of thermal energy end?​
Dahasolnce [82]

Answer:

I think when the object transferring the thermal energy reaches the same temp as the object absorbing it

Explanation:

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3 years ago
What current is used to power the Untied States power grid? O alternating current O direct current​
Alina [70]

Answer:

Alternating

Explanation:

It is alternating because it is easy to distribute long distance.

Direct current is found in batteries and have large voltage drop over distance.

3 0
3 years ago
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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
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kykrilka [37]

Answer:

C would be the answer

Explanation:

Hope this helps! :)

5 0
2 years ago
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In a race, Usain Bolt accelerates at
jeka94

Answer:

65.87 s

Explanation:

For the first time,

Applying

v² = u²+2as.............. Equation 1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance

From the question,

Given:  u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m

Substitute these values into equation 1

v² = 0²+2(1.99)(60)

v² = 238.8

v = √238.8

v = 15.45 m/s

Therefore, time taken for the first 60 m is

t = (v-u)/a............ Equation 2

t = (15.45-0)/1.99

t = 7.77 s

For the final 40 meter,

t = (v-u)/a

Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²

Substitute into the equation above

t = (0-15.45)/-0.266

t = 58.1 seconds

Hence total time taken to cover the distance

T = 7.77+58.1

T = 65.87 s

3 0
3 years ago
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