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aliina [53]
4 years ago
7

Why is the motion of an athlete moving along the circular path with Constant speed considered to be an accelerated motion?

Physics
1 answer:
Strike441 [17]4 years ago
5 0

The speed is changing its direction all the time. There is an acceleration which changes the direction of the speed – that is called centripetal acceleration. Only uniform linear motions are considered to have no acceleration.

This is the general formula for acceleration

a = dv/dt

When calculating dv, you should keep in mind the change in the velocity vector’s direction. You can easily see in a graph that with dt tending to 0 (so the length of the arc covered is also tending to 0), the difference between vectors Vf and V0 has a direction which is perpendicular to velocity (the shorter the arc, the closest the angle is to 90 degrees).

There is a formula (which can be deducted from the previous formula) which allows you to calculate the acceleration:

a = v^2/r

Let’s talk about the units:

v is in m/s

r is in m

so v^2/r

is in (m/s)^2/m = (m^2/s^2)/m = m/s^2

which is the same unit as dv/dt:

dv/dt = (m/s)/s= m/s^2

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As light shines from air to water, the index of refraction is 1.02 and the angle of incidence is 38.0 °. What is the light's an
muminat

Answer:

Light's angle of refraction = 37.1° (Approx.)

Explanation:

Given:

Index of refraction = 1.02

Base of refraction = 1

Angle of incidence = 38°

Find:

Light's angle of refraction

Computation:

Using Snell's law;

Sin[Angle of incidence] / Sin[Light's angle of refraction] = Index of refraction / Base of refraction

Sin38 / Light's angle of refraction = 1.02 / 1

Sin[Light's angle of refraction] = Sin 38 / 1.02

Sin[Light's angle of refraction] = [0.6156] / 1.02

Sin[Light's angle of refraction] = 0.6035

Light's angle of refraction = 37.1° (Approx.)

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3 years ago
A 1400 kg car traveling at 17.0 m/s to the south collides with a 4700 kg truck that is at rest. The car and truck stick together
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Answer:

Final velocity = 7.677 m/s

KE before crash = 202300 J

KE after crash = 182,702.62 J

Explanation:

We are given;

m1 = 1400 kg

m2 = 4700 kg

u1 = 17 m/s

u2 = 0 m/s

Using formula for inelastic collision, we have;

m1•u1 + m2•u2 = (m1 + m2)v

Where v is final velocity after collision.

Plugging in the relevant values;

(1400 × 17) + (4700 × 0) = (1400 + 1700)v

23800 = 3100v

v = 23800/3100

v = 7.677 m/s

Kinetic energy before crash = ½ × 1400 × 17² = 202300 J

Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J

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Answer:

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