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pashok25 [27]
2 years ago
15

Which statement best describes how light behaves with liquids, gases, and solids?

Physics
1 answer:
juin [17]2 years ago
3 0

Answer:

C number is write i think

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A car rounds a curve. The radius of curvature of the road is R, the banking angle with respect to the horizontal is θ and the co
exis [7]

Answer:

v = √[gR (sin θ - μcos θ)]

Explanation:

The free body diagram for the car is presented in the attached image to this answer.

The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.

Resolving the weight into the axis parallel and perpendicular to the inclined plane,

N = mg cos θ

And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ

Frictional force = Fr = μN = μmg cos θ

Centripetal force responsible for keeping the car in circular motion = (mv²/R)

So, a force balance in the plane parallel to the inclined plane shows that

Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)

(mv²/R) = (mg sin θ - μmg cos θ)

v² = R(g sin θ - μg cos θ)

v² = gR (sin θ - μcos θ)

v = √[gR (sin θ - μcos θ)]

Hope this Helps!!!

5 0
3 years ago
A body starting from rest moves with uniform acceleration until it attains a speed of 100m/s after 20 s. It maintained this spee
densk [106]

Answer:by 10

Explanation:research

4 0
3 years ago
An aluminum rod and a nickel rod are both 5.00 m long at 20.0 degree Celsius. The temperature of each is raised to 70.0 degrees
vitfil [10]

Answer:

0.002925 m

Explanation:

Lt = LO(1 +α Δt ) here Lt is total length Lo is original length α is coefficient of linear expansion and Δt is change in temperature

<h2>for aluminium</h2>

α=25×10^-6

Lt = 5(1+25×10^-6×(70-20))

Lt = 5 (1+25×10^-6×50)

Lt = 5 ( 1+0.00125)

Lt = 5×1.00125

Lt =5.00625 m

<h2>for nickel </h2>

α=13.3×10^-6

Lt =5(1+13.3×10^-6×50)

Lt = 5(1+0.000665)

Lt =5.003325 m

hence difference in length =5.00625-5.003325

                                           = 0.002925 m

3 0
3 years ago
Suppose a magnetic field of 0.20 T is applied to the area within a circular loop of wire. The strength of the field is then chan
Mazyrski [523]
0.1T...........................................................
8 0
3 years ago
A body moving with an initial velocity of 30m/s accelerates uniformly at the rate of 10m/s . what is the distance covered during
nikdorinn [45]

Answer:

The distance covered by the body is, S = 800 m

Explanation:

Given data,

The initial velocity of the body, u = 30 m/s

The acceleration of the body, a = 10 m/s²

Let the time period of travel be, t = 10 s

Using the II equations of motion,

                       S = ut + ½ at²

Substituting the given values,

                        S = 30 x 10 + ½ x 10 x 10²

                         S = 800 m

Hence, the distance covered by the body is, S = 800 m

5 0
3 years ago
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