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3 years ago

Explain why a 10 kg stone and 1 kg stone dropped from the same height

1 answer:

3 0

The acceleration of a body in a gravitational field is independent of its mass. Both the stones will fall with the same acceleration through the same height and hence they will strike the ground simultaneously.
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There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force, $F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2$

$F_1=F_3=F$

$F_{net}=\sqrt{F^2+F^2+0}-F_2$

$F_{net}=\sqrt 2F-F_2$

$F=\frac{kq^2}{d^2}$

$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

$F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})$

$F_{net}=0.208 N$

3 0

3 years ago

The length of the mercury thread is found to be 4cm and 24cm at ice point and steam point respectively on an ungraduated thermom

BabaBlast [244]

Answer:

The difference between ice and steam in Celsius (Centigrade) is 100 deg.

So the difference between and 4 cm and 24 cm of the thread corresponds to 100 deg C.

So 8 cm is 4 cm greater than the ice point

4 cm / 20 cm = 1/5 since the steam point and the ice point are 20 cm apart

Then 1/5 * 100 deg C = 20 deg C the requested temperature

6 0

3 years ago

7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N

artcher [175]

Answer:

20.4 $m/s^{2}$

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get: