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inna [77]
2 years ago
12

Two points are located on a rigid wheel that is rotating with decreasing angular velocity about a fixed axis. Point A is located

in the rim of the wheel and pint B is halfway between the rim and the axis. Which one of the following statements concerning this situation is true?
1. The angular velocity at point A is greater than that of point B
2. Both points have the same centripetal acceleration
3. Both points have the same tangential acceleration4. Both points have the same instantaneous angular velocity
5. Each second, point A turns through a greater angle than point B
Physics
1 answer:
Nata [24]2 years ago
4 0

Answer:

4. Both points have the same instantaneous angular velocity

Explanation:

Angular velocity is a measure of the the number of rotations per unit time. This does not depend on the radius of the wheel. Hence, all points on the wheel have the same angular velocity. This invalidates option 1.

The centripetal acceleration is given by the product to the square of the angular velocity and the radius or distance from the centre. A and B are located at different distances from the centre. Hence, they have different centripetal acceleration. This invalidates option 2.

The tangential acceleration depends on the linear velocity which itself is a product of the angular velocity and the distance from the centre. Hence, it is different for both points because they are at different distances from the centre.

Since both A and B are fixed points on the wheel, they move through equal angles in the same time. In fact, for any other fixed point, they all move through the same angle in the same time. This invalidates option 5.

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Let the displacement function of a particle is x(t)=(20t^2-15t+200). Find the total displacement, instantaneous velocity and ins
irga5000 [103]

Answer:

A.) 39.5 m

B.) 0

C.) 60m/s^2

Explanation:

Given that a displacement function of a particle is x(t)=(20t^2-15t+200).

To Find the total displacement,

Reduce everything by dividing them by 5

X(t) = 4t^2 - 3t + 40 ...... (1)

For instantaneous velocity, differentiate x(t). That is,

dy/dt = 60t - 15 ...... (2)

But dy/dt = velocity.

If dy/dt = 0, then

60t - 15 = 0

60t = 15

t = 15/60

t = 0.25s

Substitutes t in equation (1)

Total displacement will be

X(t) = 4(0.25)^2 - 3(0.25) + 40

X(t) = 0.25 - 0.75 + 40

Total displacement = 39.5 m

To calculate instantaneous velocity, substitute t into equation (2)

V = 60 (0.25) - 15

V = 0.

and to find instantaneous acceleration, differentiate dv/dt

dv/dt = 60

Therefore, acceleration = 60 m/s^2

4 0
2 years ago
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on
leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

q_1=50\ nC at y=6\ m

q_2=-80\ nC at x=-4\ m

q_3=70\ nC at y=-6\ m

Electric Potential is given by

V=\frac{kQ}{r}

Distance of q_1 from x=8\ m

d_1=\sqrt{6^2+8^2}

d_1=\sqrt{36+64}

d_1=10\ m

similarly d_2=8-(-4)

d_2=12\ m

d_3=\sqrt{(-6)^2+8^2}

d_3=\sqrt{36+64}

d_3=10\ m

Potential at x=8\ m is

V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}

V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]

V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}

V_{net}=9\times 5.33

V_{net}=47.97\approx 48\ V

5 0
3 years ago
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