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Salsk061 [2.6K]
2 years ago
6

You shove a chair with wheels(and let it go)and it starts moving across the floor. However,the chair soon shows down and comes t

o a complete stop. To keep the chair moving,you need to keep pushing on it.
This simple experiment implies that ;

A) Newton's first law,or the law of inertia,does not apply to a chair moving horizontally
B) The action-reaction forces act on the chair and cancel each other
C) A net force in the direction opposite to the motion of the chair acts to slow it down
D) The weight of the chair slows it down ​
Physics
2 answers:
raketka [301]2 years ago
5 0
I would say *B* because *D*dose not matter if it was heavier or lighter. And I’m pretty sure in reality the thing that’s stopping it is friction ‘laughing emoji’
GarryVolchara [31]2 years ago
4 0

Answer:

Personally I think, that the answer is B.

Explanation:

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Diego is trying to lift a piano to the second floor of his house. Diego uses a pulley system and gives a big lift to the piano.T
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3 0
3 years ago
A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the fri
omeli [17]

Answer:

4500 N

Explanation:

When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:

a = v^2/r

where v is the velocity of the body and r is the radius of the circumference:

Therefore, a body with mass m, will feel a force f:

f = m v^2/r

Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:

fs = μN = μmg

The car will not slide if     f = fs,   i.e.

fs = μmg =  m v^2/r

That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration

fs = (1000 kg)  * (30m/s)^2 / (200 m) = 4500 N

7 0
3 years ago
Read 2 more answers
A steady beam of alpha particles (q = + 2e, mass m = 6.68 × 10-27 kg) traveling with constant kinetic energy 22 MeV carries a cu
cluponka [151]

Answer:

Explanation:

q = 2e = 3.2 x 10^-19 C

mass, m = 6.68 x 10^-27 kg

Kinetic energy, K = 22 MeV

Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A

(a) time, t = 2.8 s

Let N be the alpha particles strike the surface.

N x 2e = q

N x 3.2 x 10^-19 = i t

N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8

N = 2.36 x 10^12

(b) Length, L = 16 cm = 0.16 m

Let N be the alpha particles

K = 0.5 x mv²

22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²

v² = 1.054 x 10^15

v = 3.25 x 10^7 m/s

So, N x 2e = i x t

N x 2e = i x L / v

N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)

N = 4153.85

(c) Us ethe conservation of energy

Kinetic energy = Potential energy

K = q x V

22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V

V = 1.17 x 10^7 V

5 0
3 years ago
Bronco the skydiver, whose mass is 100 kg experiences 200 N of air resistance. What is the acceleration of his fall? Show Work
bekas [8.4K]

Explanation:

air resistance (f) =mg

200=100g

g=200/100

g=2m/s^2

5 0
2 years ago
Convert 278000 into scientific notation
nirvana33 [79]

2.78 X 10^5 is the best answer to be responded to

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