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3 years ago

There is a very slight change in gravity at the top of a mountain. Which property

Physics

1 answer:

son4ous [18]3 years ago

8 0

Answer:

The entity that decreases is the Weight.

Explanation:

Every object on the surface of the earth is usually pulled downwards by the action of gravity. The force or weight is W = mg

Now, the farther one gets away from the surface of the earth by climbing a mountain, the further the person is from the Earth's centre of gravity.

We know that the formula for Force of gravity away from the Earth's surface is given by;

F_g = GM1M2/r²

So the greater the distance, the lesser the gravitational force.

Now, the gravitational force equals the weight of the individual.

Thus, the entity that decreases is the Weight of the individual.

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When a 4.60-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.30 cm. (a) If

Alex787 [66]

Answer:

a) y = 0.0075 m

b) W = 1.569 J

Explanation:

See attachment for the solution

6 0

3 years ago

The speed of sound through diamond is about 12,000 m/s. The speed of sound through wood is about 3,300 m/s. Which statement expl

Damm [24]

Answer: They have different rigidities.

Explanation:

7 0

3 years ago

What is the elevation of Y and Z? Y=3250, Z=2950 Y=3200, Z=2900 Y=3250, Z=2900 Y=3000, Z=2850

TiliK225 [7]

I think the elevation of Y and Z are the following:
<span>Y=3200,
Z=2900 </span>

8 0

3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

8 0

4 years ago

If we put negative charge between two similar positive charges then what is it's equilibrium? And how?

Gnesinka [82]

Your question has been heard loud and clear.

Well it depends on the magnitude of charges. Generally , when both positive charges have the same magnitude , their equilibrium point is towards the centre joining the two charges. But if magnitude of one positive charge is higher than the other , then the equilibrium point will be towards the charge having lesser magnitude.

Now , a negative charge is placed in between the two positive charges. So , if both positive charges have same magnitude , they both pull the negative charge towards each other with an equal force. Thus the equilibrium point will be where the negative charge is placed because , both forces are equal , and opposite , so they cancel out each other at the point where the negative charge is placed. However if they are of different magnitudes , then the equilibrium point will be shifted towards the positive charge having less magnitude.

Thank you

5 0

3 years ago

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