How to calculate maximum height, when i know only V<br><br> Pleeeaaassseee as fast as you can

A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular

ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

$\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0$

Where,

$\theta =$ Angular Displacement

$\alpha =$ Angular Acceleration

$\omega_0 =$ Angular velocity

$\theta_0 =$ Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

$\theta = \frac{1}{2}\alpha t^2$

Re-arrange for $\alpha,$

$\alpha = \frac{2\theta}{t^2}$

Replacing our values,

$\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}$

$\alpha = 0.139rad/s$

Therefore the ANgular acceleration of the mass is $0.139rad/s^2$

4 0

3 years ago

A daring 510-N swimmer dives off a cliff with a running horizontal leap. What must her minimum speed be just as she leaves the t

oee [108]

Answer:

$v_x = 1.26 m/s$

Explanation:

given,

weight of swimmer = 510 N

length of ledge, L = 1.75 m

vertical height of the cliff, h = 9 m

speed of the swimmer = ?

horizontal velocity of the swimmer should be that much it can cross the wedge.

distance = speed x time

d = v_x × t

1.75 = v_x × t ........(1)

now,time taken by the swimmer to cover 9 m

initial vertical velocity of the swimmer is zero.

using equation of motion for time calculation

$s = ut +\dfrac{1}{2}gt^2$

$9= 0+\dfrac{1}{2}\times 9.8\times t^2$

t² = 1.938

t = 1.39 s

same time will be taken to cover horizontal distance.

now, from equation 1

1.75 = v_x × 1.39

$v_x = 1.26 m/s$

horizontal speed of the swimmer is equal to 1.26 m/s

4 0

3 years ago

The CERN particle accelerator is circular with a circumference of 7.0 km.

Contact [7]

Answer:

$a_c=2.0196\times 10^{13}\ m/s^2$

$F=3.37273\times 10^{-14}\ N$

Explanation:

m = Mass of proton = $1.67\times 10^{-27}\ kg$

v = Speed of proton = 0.5c = $0.5\times 3\times 10^8=1.5\times 10^8\ m/s$

Circumference of the colider is 7 km

$P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m$

$a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2$

Centripetal acceleration is $2.0196\times 10^{13}\ m/s^2$

$F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N$

Force on protons is $3.37273\times 10^{-14}\ N$

8 0

3 years ago

1pt Which is an observation? O A. A person hears the song of a bird. OB. A person suggests that bird calls are a means of commun

Doss [256]

Answer:

D

Explanation:

he describes as he writes them down

3 0

3 years ago

What kind of energy change occurs when a battery is operating a remote control toy?

solniwko [45]

Answer:

Explanation:

The correct answer is option C.

When the battery is operating a remote control toy the energy is converted from potential energy to the kinetic energy.

A battery stores electrical potential from the chemical reaction.

When the circuit is connected to the potential energy of the battery helps in the movement of the toy.

The energy produced by the movement of the control toy is kinetic energy.

Hence, we can say that Potential energy is changed to kinetic energy

7 0

4 years ago

How to calculate maximum height, when i know only V Pleeeaaassseee as fast as you can