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dimaraw [331]
2 years ago
7

Consider a point on a bicycle wheel as the wheel turns about a fixed axis, neither speeding up nor slowing down. Compare the lin

ear and angular accelerations of the point.
a. Both are zero.
b. Only the angular acceleration is zero.
c. Only the linear acceleration is zero.
d. Neither is zero

Explain your choice
Physics
1 answer:
Natalija [7]2 years ago
4 0

Answer:

c. Only the linear acceleration is zero.

Explanation:

The linear acceleration is defined as the rate of change of linear velocity. Since the bicycle is moving in the same direction, with the same speed, without speeding up or slowing down. Therefore, there will be no change in linear velocity and as a result, linear acceleration will be zero.

The angular acceleration is the rate of change of angular velocity. Since the angular velocity is changing its direction constantly. Therefore, it has a certain component of acceleration at all times called centripetal acceleration.

Therefore, the correct option is:

<u>c. Only the linear acceleration is zero.</u>

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NO LINKS!!!!!! Help its urgent, PLEASE HELP!!!!!!!!!!!!!!!!
Likurg_2 [28]

Answer:

I think ur answer would be B

Explanation:

srry if wrong

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7 0
2 years ago
4 (a) (1) Which lens is a converging lens with the greatest power?
SOVA2 [1]

Answer:

Answer

Explanation:

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3 0
2 years ago
66. Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown st
kozerog [31]

Answer:

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:  

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t  Equation 2

y: The vertical distance the ball moves at time t  

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t  

Known information

We know the following data:

Vi=15 m / s

g =9.8 \frac{m}{s^{2} }

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)  

a) t=1s

y = 15*1 - ½ 9.8*1^{2}=15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

y = 15*1.5 - ½ 9.8*1.5^{2}=22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

y = 15*2 - ½ 9.8*2^{2}= 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

3 0
3 years ago
What is the force needed to move an 225 kg car a distance of 150 meters
Kobotan [32]

Physics

mass = 225 kg

weight = m × a = 2250 N

distance = 150 m

force : ______?

Answers :

W = F × s

W = 2250 × 150

W = 337500 ✅

4 0
3 years ago
Read 2 more answers
20% Part (a) Use an "E Field Sensor" and move it along either equipotential. What can you say about the E field along an equipot
Alex

Answer:

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

Explanation:

As we know that the relation between electric field and electric potential is given as

\Delta E = -\frac{dV}{dr}

here if we say that potential is constant because electric field sensor is moving along equi-potential line.

Then we will say

V = constant

so we have

\Delta E = 0

so electric field will remain constant always in magnitude and always remains perpendicular to the surface

so we have

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

6 0
3 years ago
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