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IrinaK [193]
3 years ago
12

In which direction does a bag at rest move when a force of 20 newton's is applied from the right?

Physics
1 answer:
Olegator [25]3 years ago
8 0

The bag moves to the left.

This is because of Newton's third law of motion that states:

For every action force on a body, there is an opposite and equal reaction force.

Thus pushing the bag from the right makes it move to the left.


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makvit [3.9K]

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hygrometer is a weather instrument used to measure the amount of humidity in the atmosphere. There are two main types of hygrometers -- a dry and wet bulb psychrometer and a mechanical hygrometer.

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3 years ago
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Vlad [161]

Answer:

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4 0
2 years ago
3 light-years =________.
KATRIN_1 [288]

               ( 3 yr) · (186,282.397 mile/s) · (86,400 s/day) · (365 day/yr)

           =  (3 · 186,282.397 · 86,400 · 365)  mile

           =      1.762380502 x 10¹³  miles

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6 0
3 years ago
Two plane mirrors are separated by 120°, as the drawing illustrates. If a ray strikes mirror M1 at a θ1 = 64° angle of incidence
tino4ka555 [31]

Angle, θ2 at which the light leaves mirror 2 is 56°

<u>Explanation:</u>

Given-

θ1 = 64°

So, α will also be 64°

According to the figure:

α + β = 90°

So,

β = 90° - α

  = 90° - 64°

  = 26°

β + γ + 120° = 180°

γ = 180° - 120° - β

γ = 180° - 120° - 26°

γ = 34°

γ + δ = 90°

δ = 90° - γ

δ = 90° - 34°

δ = 56°

According to the law of reflection,

angle of incidence = angle of reflection

θ2 = δ = 56°

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8 0
2 years ago
A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i
erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

\mu_k = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring =  Kinetic energy of the mass 2

or

\frac{1}{2}kx^2=\frac{1}{2}mv_3^2

on substituting the values, we get

\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2

or

x = 2.86 m

8 0
3 years ago
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