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IrinaVladis [17]
3 years ago
15

A bicycle starts at 2.5m/s and accelerates along a straight path to a speed of 12.5m/s in a time of 4.5 seconds. What is the bic

yclist’s acceleration to the nearest tenth of a m/s^2 ?
Physics
1 answer:
zalisa [80]3 years ago
3 0

Answer:

The bicyclist's acceleration is 2.2m/s^2

Explanation:

Given

u = 2.5m/s ---- Initial Velocity

v = 12.5m/s ---- Final Velocity

t = 4.5s ---- Time

Required

Determine the acceleration

This will be solved using the first equation of motion

v = u + at

Substitute values for v, u and a

12.5 = 2.5 + a * 4.5

12.5 = 2.5 + 4.5a

Collect Like Terms

4.5a = 12.5 - 2.5

4.5a = 10.0

Solve for a

a = 10.0/4.5

a = 2.2m/s^2 ---- (approximated)

<em>Hence, the bicyclist's acceleration is 2.2m/s^2</em>

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Answer:

Yes

Explanation:

There is a position that works better than this and that is switching the sides of the forks.

7 0
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Two objects exert a gravitational force of 3 N on one another when they are 10 m apart. What would that force be if the distance
jeka94
Gravitational force = G ( m1 m2 ) / r²
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A body travels the first half of the total distance with velocity v and second half with v2 calculate avg velocity
LuckyWell [14K]

Answer:

v = 2 v₁ v₂ / (v₁ + v₂)

Explanation:

The body travels the first half of the distance with velocity v₁.  The time it takes is:

t₁ = (d/2) / v₁

t₁ = d / (2v₁)

Similarly, the body travels the second half with velocity v₂, so the time is:

t₂ = (d/2) / v₂

t₂ = d / (2v₂)

The average velocity is the total displacement over total time:

v = d / t

v = d / (t₁ + t₂)

v = d / (d / (2v₁) + d / (2v₂))

v = d / (d/2 (1/v₁ + 1/v₂))

v = 2 / (1/v₁ + 1/v₂)

v = 2 / ((v₁ + v₂) / (v₁ v₂))

v = 2 v₁ v₂ / (v₁ + v₂)

8 0
3 years ago
A tuning fork with a frequency of 512 Hz is used to tune a violin. When played together, beats are heard with a frequency of 4 H
Ganezh [65]

Answer:

<em> 508Hz</em>

Explanation:

A tuning fork with a frequency of 512 Hz is used to tune a violin. When played together, beats are heard with a frequency of 4 Hz. The string on the violin is tightened and when played again, the beats have a frequency of 2 Hz. The original frequency of the violin was ______.

When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - this phenomenon is beat production

frequency is the number of oscillation a wave makes in one seconds.

f1-f2=beats

therefore f1=512Hz

f2=?

beats=4Hz

512Hz-f2=4Hz

f2=512-4

f2=508Hz

the original frequency of the violin is 508Hz

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11111nata11111 [884]
Entropy measures disorder and nothing else
8 0
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Read 2 more answers
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