Ask question

Ask question

All categories

3 years ago

12

If there is an object in top of another object, why does the upper object exert a downward normal force?

1 answer:

rusak2 [61]3 years ago

4 0

Answer:

This is because normal force is exerted perpendicularly to the point of contact between the upper and lower objects.

Explanation:

This is because the upper object is still subject to gravitational pull. Therefore, the amount of force it exerts on the lower object due to gravity will be equal to the normal force that acts in the negative direction of gravitational force. Additionally, normal force is evident because the upper object will not go into the lower object.

You might be interested in

When finding the upper bound of the density, you put what number in the denominator?

lilavasa [31]

Answer: The result of "the upper bound of the density" does not go on the denominator.
So simplified, no. The answer is no.

3 0

3 years ago

A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.

kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight= $\sqrt{\frac{2H}{g}}$

substituting value

= $\sqrt{\frac{2*25}{9.8}}$
=2.26seconds

<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

we have

Horizontal range=u* $\sqrt{\frac{2H}{g}}$

=8.25*2.26
=18.63m

<h3><u>The ball strikes 18.63m far from building</u>. </h3>

7 0

3 years ago

You make an observation that two objects attract. Which object below indicates the claim(s) that could be supported by this obse

wel

4. All of the above.

4 0

3 years ago

a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?

Verdich [7]

The distance between the two charges is $0.3 m$

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

$q_1 = 8\cdot 10^{-5} C$

$q_2 = 3\cdot 10^{-5} C$

$F=2.4\cdot 10^2 N$

Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

$r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m$

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

BRAINLIEST!!!!!

Serggg [28]

Answer:

A., an action which causes movement and an opposite reaction

Explanation:

I don't really know how to explain it, but this helped hopefully! :) Have a bless day

5 0

3 years ago