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A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5\times 10^{-

joja [24]

Answer:

$5.25\cdot 10^{40} kg m^2/s$

Explanation:

The angular momentum of the pulsar is given by:

$L=m\omega r^2$

where

$m=2.8\cdot 10^{30} kg$ is the mass of the pulsar

$r = 10.0 km = 1\cdot 10^4 m$ is the radius

$\omega$ is the angular speed

Given the period of the pulsar, $T=33.5\cdot 10^{-3} s$ , the angular speed is given by

$\omega=\frac{2\pi}{T}=\frac{2 \pi}{33.5\cdot 10^{-3}s}=187.5 rad/s$

And so, the angular momentum is

$L=m\omega r^2=(2.8\cdot 10^{30}kg)(187.5 rad/s)(1\cdot 10^4 m)^2=5.25\cdot 10^{40} kg m^2/s$

8 0

3 years ago

A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.

NemiM [27]

Answer:

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, $\omega=2.8\ rad/s$

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

$K_t=\dfrac{1}{2}mv^2$

$K_t=\dfrac{1}{2}m(r\omega)^2$

$K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2$

The rotational kinetic energy of the ball is :

$K_r=\dfrac{1}{2}I \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2$

Ratio of translational to the rotational kinetic energy as :

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0

3 years ago

During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa

maw [93]

Answer:

Explanation:

Given

Initial speed $u=60\ mi/hr\approx 88\ ft/s$

distance traveled before coming to rest $d_1=120\ ft$

using equation of motion

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$0-(88)^2=2\times a\times 120---1$

for $u_2=80\ mi/hr\approx 117.33\ ft/s$

using same relation we get

$0-(117.33)^2=2\times a\times (d_2)----2$

divide 1 and 2 we get

$(\frac{88}{117.33})^2=\frac{120}{d_2}$

$d_2=213.32\ ft$

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

8 0

3 years ago

A 48.9 kg meteor is moving in outer space. If a 8.6 N force is applied opposite the direction of motion, what is the deceleratio

nalin [4]

Answer:

The deceleration is 0.18 m/s²

Explanation:

Hi there!

Using Newton´s second law, we can calculate the deceleration:

∑F = m · a

Where:

∑F = the sum of all forces in a given direction.

m = mass of the object.

a = acceleration.

Solving for a:

∑F/m = a

The only force acting on the meteor is the applied force of 8.6 N. So, the acceleration will be:

8.6 N / 48.9 kg = a

a = 0.18 m/s²

The deceleration is 0.18 m/s² or, in other words, the acceleration is -0.18 m/s²

Have a nice day!

3 0

3 years ago

While anchored in the middle of a lake, you count exactly five waves hitting your boat every 10 s. you raise anchor and start mo

777dan777 [17]

<span>Every 10s 5 waves; t1 = 2s for each wave
When v = 1.5m/s, 3 waves in 10s t2 = 10 / 3s
Calculating the frequency in first case f1 = 5 / 10 = 0.5
Calculating the frequency in second case f2 = 3 / 10 = 0.3
Using the Doppler formula f = (1-v/c) f0
For the formula f = f2, v = velocity of boat= 1.5 m/s, f0 = f1, c is velocity of wave 0.3 = 0.5 x (1 - 1.5/c) => 1.5/c = 1 - 0.6 => 1.5/c = 0.4 => c = 1.5/0.4 Velocity of the wave = 3.75 m/s</span>

7 0

3 years ago