Https://phet.colorado.edu/sims/html/balloons-and-static-electricity/latest/balloons-and-static-electricity

What objects can be seen from earth because they producde there own light? 

Lena [83]

- neon signs
- fireworks
- rockets during the ascent
- meteors
- fireflies
- stars

7 0

3 years ago

1. You serve a volleyball with a mass of 2.1kg. The ball leaves your

Darina [25.2K]

The kinetic energy is 945 joules.

Kinetic energy is the energy that an object has as a result of motion. It is defined as the effort required to accelerate a mass-determined body from rest to the indicated velocity.

The speed of an object or particle, which is a scalar quantity, is the size of the change in its location over time or the size of the change in its position per unit of time.

The mass of the volleyball is 2.1 kg.

The speed of the ball when the ball leaves the hand is 30 m/s.

m = 2.1 kg

v = 30 m/s

The kinetic energy of an object is given as:

KE = (1/2 ) × m × v²

KE = (1 / 2) × 2.1 kg × ( 30 m/s)²

KE = (1 / 2) × 2.1 kg × 30 m/s × 30 m/s

KE = 2.1 kg × 15 m/s × 30 m/s

KE = 945 J

Learn more about kinetic energy here:

brainly.com/question/8101588

#SPJ9

6 0

1 year ago

A 1400 kg car is moving at 33.8 m/s when a force is applied the opposite direction of the car's motion. The car slows down to 21

zubka84 [21]

Solve for acceleration:

a = (21.4 m/s - 33.8 m/s) / (4.7 s)

a ≈ -2.6 m/s²

Solve for force:

F = (1400 kg) a ≈ -3700 N

The minus sign tells you the force points in the opposite direction of the car's motion. Its magnitude is always positive, so F = 3700 N.

3 0

3 years ago

A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torqu

seropon [69]

Answer:

$\theta=50\ revolution$

Explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone, $\omega_i=600\ rpm=62.83\ rad/s$

Finally it shuts off, $\omega_f=0$

Time taken, t = 10 s

Let $\alpha$ is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{0-62.83}{10}$

$\alpha =-6.283\ rad/s^2$

Let $\theta$ is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

$\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }$

$\theta=\dfrac{-62.83^2}{2\times -6.283}$

$\theta=314.15\ radian$

$\theta=49.99\ revolution$

or

$\theta=50\ revolution$

So, the number of revolutions of the grindstone after the power is shut off is 50.

7 0

3 years ago

In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas

Anni [7]

Answer:

The phase difference is $\Delta \phi = 180^o$

Explanation:

From the question we are told that

The distance between the slits is $d = 0.2 \ mm = \frac{0.2}{1000} = 0.2 *10^{-3} \ m$

The distance to the screen is $D = 100 cm = \frac{100}{100} = 1 \ m$

The wavelength is $\lambda = 400nm$

The distance of the wave from the central maximum is $L = 5mm = 5*10^{-3} m$

Generally the path difference of this waves is mathematically represented as

$y = d sin \theta$

Here $\theta$ is the angle between the the line connecting the mid-point of the slits with the screen and the line connecting the mid-point of the slits to the central maximum

This implies that

$tan \theta = \frac{L}{D}$