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andreev551 [17]
3 years ago
12

a 0.04kg ball tied to a string moves in a circle that has a radius of 0.70 m. If the ball is accelerating 43.2m/s, what is the t

angential velocity of the ball?​
Physics
1 answer:
Luba_88 [7]3 years ago
4 0

Answer:

Tangential Velocity = 30.24 m/s

Explanation:

Given that,

Mass of ball, m = 0.04 Kg

Length of the string, r = 0.70 m

Acceleration of the ball, a = 43.2 m/s²

The tangential velocity of ball, V = ?

The centripetal force is given by the relation

        Fc = mV²/r  newton

where,   m - mass of body

              V - tangential velocity of body

               r - radius of the trajectory

Force applied on the ball to rotate on a circular path

                F = m x a newton

The applied force is equal to centripetal force.

So, equalizing the force equations

                     m x a = m V²/r

Therefore

                     V² = a x r

                      V = \sqrt{a X r}

Substituting the values

                       V = \sqrt{43.2 X 0.70}

                        V = 30.24 m/s

So, the tangential velocity of the ball is 30.24 m/s

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The excess charge on earth's surface was calculated to be 4.56 × 10⁵ C

Explanation:

Using the formula for an electric field;

E = kQ/r²

k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²

E = 100N/C

r = radius of the earth = 6400 km = 6400000m

Q = Er²/k = 100 × (6400000)²/(8.99 × 10⁹)

Q = 455617.4 C = 4.56 × 10⁵ C

Hope this helps!!!

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A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

\omega=-\alpha t

\alpha=-\dfrac{\omega}{t}

\alpha=-\dfrac{50.0}{20.0}

\alpha=-2.5\ rad/s^2

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

\theta=\omega_{0}t+\dfrac{1}{2}\alpha t

Put the value into the formula

\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2

\theta=500\ rad

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

\vec{\tau}=\vec{r}\times\vec{f}

\tau=r\times f\sin\theta

Put the value into the formula

\tau=2.5\times4\times 9.8\sin60

\tau=84.87\ N-m

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0
3 years ago
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