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AlladinOne [14]
3 years ago
10

A car travels in a flat circle of radius R. At a certain instant the velocity of the car is 24 m/s west and the total accelerati

on of the car is 2.5m/s2 53 degrees north of west. Find the radial and tagential components of the acceleration of the car at that moment. If the cars tangential acceleration is constant how long will it take for it to make one full circle from the point at which its velocity is 24m/s west?
Physics
1 answer:
lisov135 [29]3 years ago
8 0

Answer with Explanation:

We are given that

Velocity,v=24 m/s west

Acceleration,a=2.5m/s^2

\theta=53^{\circ}(N of W)

Horizontal component of acceleration=Tangential acceleration

Tangential acceleration,a_x=acos\theta=2.5cos53=1.505 m/s^2

Radial acceleration=Vertical acceleration=a_y=asin\theta

a_y=2.5sin53=1.998 m/s^2

Radial acceleration,a_y=\frac{v^2}{R}

1.998=\frac{(24)^2}{R}

R=\frac{(24)^2}{1.998}

R=288.29 m

Time,t=\frac{2\pi R}{velocity}

t=\frac{2\pi(288.29)}{24}

t=75.47 s

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Inertia is directly proportional to mass.

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7 0
1 year ago
Two ice skaters stand together. They push off and travel directly away from each other, the boy with a velocity of v = +0.35 m/s
DIA [1.3K]

Answer:

-0.55m/s

Explanation:

Given that: For the boy

Weight = 745N

Velocity = +0.35 m/s

Mass of the boy = ?

g = 9.81m/s^2

W = mg

745 = m×9.81

m = 75.94kg

For the girl

Given that:

Weight = 477 N

g = 9.81m/s^2

m = ?

W = mg

477 = m×9.81m/s^2

m = 48.62kg

To solve for the v of the girl, the two has to add up

48.62kg×v + 75.94kg×+0.35 m/s = 0

48.62v + 26.579 = 0

48.62v = - 26.579

v = -26.579/48.62

v = -0.5466

v = -0.55m/s

Hence, the velocity of the girl is -0.55m/s.

The negative sign is as a result of the two of them moving is opposite direction.

5 0
3 years ago
A mass-spring system has k = 56.8 N/m and m = 0.46 kg.
andriy [413]

Answer:

A. \omega=11.1121\ rad.s^{-1}

B. f=1.7685\ Hz

C. T=0.5654\ s

Explanation:

Given:

  • spring constant, k=56.8\ N.m^{-1}
  • mass attached, m=0.46\ kg

A)

for a spring-mass system the frequency is given as:

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{56.8}{0.46}}

\omega=11.1121\ rad.s^{-1}

B)

frequency is given as:

f=\frac{\omega}{2\pi}

f=\frac{11.1121}{2\pi}

f=1.7685\ Hz

C)

Time period of a simple harmonic motion is given as:

T=\frac{1}{f}

T=0.5654\ s

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3 years ago
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Hes going 180 mph
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