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nexus9112 [7]
3 years ago
15

A car has wheels of radius of 0.380 m and they rotate once every 0.100 s. What is the linear velocity of the wheels? (Units = m/

s)
Physics
2 answers:
quester [9]3 years ago
6 0

Answer:

v = 23.87 m/s

Explanation:

Given that,

The radius of wheels, r = 0.380 m

The wheels rotate once every 0.1 s

We need to find the linear velocity of the wheels. The linear velocity of a wheel is given by :

v=\dfrac{2\pi r}{t}\\\\=\dfrac{2\pi \times 0.38}{0.1}\\\\=23.87\ m/s

So, the required linear velocity of the wheel is equal to 23.87 m/s.

AlekseyPX3 years ago
4 0

Answer:

23.87

Explanation:

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The law of reflection establishes a definite relationship between the angle of incidence of a light ray striking the boundary be
sladkih [1.3K]

Answer:

The law of reflection states that the angle of incidence = the angle of reflection.

Explanation:

Reflection is the phenomenon that occurs when a ray of light hits the boundary between two media and it is reflected back into the first medium.

In such a situation, we call:

- angle of incidence: it is the angle between the direction of the incident ray and the normal to the surface

- angle of reflection: it is the angle between the direction of the reflected ray and the normal to the surface

There is a precise relationship between the angle of incidence and the angle of reflection. In fact, the Law of Reflection states that:

- The incident ray, the reflected ray and the normal to the surface all lie within the same plane

- The angle of reflection is equal to the angle of incidence

4 0
3 years ago
So what would the answer for D be?
Mademuasel [1]
For what, exactly? XD
8 0
3 years ago
Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.
Fudgin [204]

Answer:

g'_h=1.096\times 10^{-5}\ m.s^{-2}

Explanation:

We know that the gravity on the surface of the moon is,

  • g'=\frac{g}{6}
  • g'=1.63\ m.s^{-2}

<u>Gravity at a height h above the surface of the moon will be given as:</u>

g'_h=\frac{G.m}{(r+h)^2} ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

  • G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}
  • m=7.35\times 10^{22}\ kg
  • r=1.74\times 10^6\ m
  • h=384.4\times 10^6\ m is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}

g'_h=1.096\times 10^{-5}\ m.s^{-2} is the gravity on the moon the earth-surface.

4 0
3 years ago
1 A thing ring has a mass of 6kg and a radius of 20cm. calculate the rotational inertia. ​
marissa [1.9K]

Answer:

2400kgm²

Explanation:

Rotational inertia=mass x radius²

7 0
3 years ago
Invader Zim’s spaceship is sitting at rest in outer space. The ship then accelerates at a uniform rate of 12 m/s2 for 10 seconds
melisa1 [442]

Answer:

120 m/s

Explanation:

Given:

v₀ = 0 m/s

a = 12 m/s²

t = 10 s

Find: v

v = at + v₀

v = (12 m/s²) (10 s) + 0 m/s

v = 120 m/s

6 0
3 years ago
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