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Anna11 [10]
3 years ago
12

If you push on a wall with a force of 200 N, with what force does the wall push back?

Physics
1 answer:
PSYCHO15rus [73]3 years ago
8 0
200 N, that is if the force is balanced and the wall doesn't move
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Sledding down a hill, you are traveling at 10 m/s when you reach the bottom. You (inertia 80 kg ) then move across horizontal sn
RSB [31]

Answer:

V = 0.32 m /s

Explanation:

Momentum of the man on the sledge along with sledge

= (80+8) x 10 = 880 kg. m/s When the man jumps off the sledge , the velocity of the sledge will remain intact at 10 m/s

When the sledge hits the boulder and bounces back the momentum of the sledge - boulder system will remain conserved .

change in momentum of sledge = 8 x (10 +6 )

= 128  kg m/s

Change in the momentum of boulder

= 400 V -0

= 400V

400V = 128

V = 0.32 m /s

5 0
3 years ago
An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro
valkas [14]

Answer:

Final velocity of electron, v=6.45\times 10^5\ m/s    

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, u=4.52\times 10^5\ m/s

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

v^2=u^2+2as........(1)

a is the acceleration, a=\dfrac{F}{m}

We know that electric force, F = qE

a=\dfrac{qE}{m}

Use above equation in equation (1) as:

v^2=u^2+\dfrac{2qEs}{m}

v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m

v = 647302.09 m/s

or

v=6.45\times 10^5\ m/s

So, the final velocity of the electron when it reaches point B is 6.45\times 10^5\ m/s. Hence, this is the required solution.

3 0
3 years ago
Which of the following changes will increase the period of an oscillating spring mass system?
Thepotemich [5.8K]

Answer:

a. an increase in the mass on the spring.

Explanation:

An increase in the mass on the spring will increase the period of an oscillating spring mass system.

Mathematically, the period of an oscillating spring mass system is given by the formula;

T = 2π √(m/k)

Where;

T is the period.

m is the mass of the spring.

k is the spring constant.

Hence, the mass of a spring is directly proportional to the period of oscillation of the spring.

This ultimately implies that, as the mass of the spring increases, the period of oscillation will increase. Similarly, the period of oscillation will decrease with an increase in the spring constant.

7 0
3 years ago
why was it intresting that democritus came up with atomism given that he was a materialist philosopher
enyata [817]

Answer:

Democritus was the first person to say that everything is made of Atoms, a very long time ago

Hope I was helpful

Explanation:

7 0
3 years ago
Help me please Ima give brainiest
julsineya [31]
Sand. You can identify tiny sugar crystals.


4 0
3 years ago
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