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MrMuchimi
3 years ago
6

A ball hits a wall. What is true about the magnitude of the force experienced by the ball compared with the force experienced by

the wall?
A. The ball experiences more force than the wall.
B. The ball experiences less force than the wall.
The ball and the wall experience the same force.
D. The ball experiences half the force of the wall.
Reset
Next
Physics
1 answer:
Natalka [10]3 years ago
6 0

Answer:

The ball and the wall experience the same force.

Explanation:

According to the third law of Newton, which states that "for every action, there is an equal and opposite reaction", this means that when an object 1 acts on object 2 with a certain force, object 2 also acts on object 1 with the same magnitude of force but in an opposite direction.

According to this question, a ball hits a wall with a certain force. This means that the wall will react on the ball with the same force magnitude, but in an opposite manner. Hence, the ball and the wall experience the same force.

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Is it true that playing badmenton help you to become a better person?
Anuta_ua [19.1K]

Answer:

There is no scientific evidence that playing specifically l badminton makes you a better person, but sport and exercise in general release hormones which can make you feel more happy therefore making you nicer to the people around you and 'a better person'.

8 0
2 years ago
Read 2 more answers
The Sun has a mass of 1.99x10^30 kg and a radius of 6.96x10^8 m. Calculate the acceleration due to gravity, in meters per second
just olya [345]

Answer:

g=274\ m/s^2

Explanation:

Mass of the Sun, M=1.99\times 10^{30}\ kg

The radius of the Sun, r=6.96\times 10^8\ m

We need to find the acceleration due to gravity on the surface of the Sun. It is given by the formula as follows :

g=\dfrac{GM}{r^2}\\\\g=\dfrac{6.67\times 10^{-11}\times 1.99\times 10^{30}}{(6.96\times 10^8)^2}\\\\g=274\ m/s^2

So, the value of acceleration due to gravity on the Sun is 274\ m/s^2.

8 0
3 years ago
A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon
Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

n=\frac{m}{M_m}

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

N=n N_A

where

n is the number of moles

N_A=6.022\cdot 10^{23} mol^{-1} is the Avogadro number

Substituting,

N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18} atoms

The energy emitted by each atom (the energy of one photon) is

E_1 = \frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda=508 nm=5.08\cdot 10^{-7}nm is the wavelength

Substituting,

E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J

And so, the total energy emitted by the sample is

E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J

4 0
3 years ago
If a charge at 60c flow in a conductor for 30 second then the current that flow in a conductor is​
saw5 [17]

Explanation:

<h3>Given</h3>

- Charge = 60c

time = 30 sec

<h3>To find -</h3>

current

<h3>Solution </h3>

Current = Charge/time

I = V/T

I = 60/30

I = 2 ampere

More to know -

I = Current

V = Charge

T = Time

3 0
3 years ago
A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th
ankoles [38]
<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 3 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 3 + 0.5 x 9.81 x 3²

                      s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0
3 years ago
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