OMG PLZ PLZ HELP!!

How does the height from which you drop the ball relate to the height that the ball bounces back up?

Stels [109]

The higher you go the more potential energy there is, and the lower it is the more kinetic energy there is, so the more kinetic energy there is the higher the ball will bounce.

7 0

3 years ago

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

yaroslaw [1]

Answer:

K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

p₀ = 0

final moment

p_f = m_a v_a + m_b v_b

p₀ = p_f

0 = m_a v_a + m_b v_b

tells us that

m_a = 8 m_b

0 = 8 m_b v_a + m_b v_b

v_b = - 8 v_a (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

Em₀ = K₀ = 73 J

final point. After separating the body

Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

K₀ = K_f

73 = ½ m_a (v_a² + v_b² / 8)

we substitute equation 1

73 = ½ m_a (v_a² + 8² v_a² / 8)

73 = ½ m_a (9 v_a²)

73/9 = ½ m_a (v_a²) = K_a

K_a = 8,111 J

3 0

3 years ago

Which simple machine belongs to the inclined plane family

Vlad [161]

The wedge and screw simple machines

3 0

3 years ago

When the acceleration of a mass on a spring is zero, the velocity is at a

Sergeu [11.5K]

1) Maximum

2) Maximum

Explanation:

The force acting on a mass on a spring is given by Hooke's law; in magnitude:

$F=kx$

where

F is the force

k is the spring constant

x is the displacement

Also we know from Newton's second law that we can write

$F=ma$

where

m is the mass

a is the acceleration

So we can write the equation as

$ma=kx$ (1)

From this relationship, we see that the acceleration is directly proportional to the displacement.

On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2=const.$ (2)

where the first term is the elastic potential energy while the second term is the kinetic energy, and where

v is the velocity of the mass

From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.

Therefore this means that:

- When acceleration increases, velocity decreases

- When acceleration decreases, velocity increases

Therefore, the two answers here are:

- When the acceleration of a mass on a spring is zero, the velocity is at a maximum

When the velocity of a mass on a spring is zero, the acceleration is at a maximum

6 0

2 years ago

For a certain optical medium the speed of light varies from a low value of 1.90 × 10 8 m/s for violet light to a high value of 2

Dmitry_Shevchenko [17]

Answer:

a. The refractive index ranges from 1.5 - 1.56

b. 18.7° for violet light and 19.5° for red light.

c. 33.7° for violet light and 35.3° for red light.

Explanation:

a. The refractive index of an object is the ratio of the speed of light in a vacuum and the speed of light in the object.

Mathematically,

$n = \frac{c}{v}$

The speed of violet light in the object is $1.9 * 10^8 m/s$ .

The speed of red light in the object is $2 * 10^8 m/s$

Hence, the refractive index for violet light is:

$n = \frac{3 * 10^8 }{1.9 * 10^8} \\\\n = 1.56$

and for red light, it is:

$n = \frac{3 * 10^8 }{2 * 10^8} \\\\n = 1.5$

Hence, the refractive index ranges from 1.5 - 1.56.

b. The refractive index is also the ratio of the sine of the angle of incidence to the sine of the angle of refraction.

$n = \frac{sin(i)}{sin(r)}$

The angle of incidence is 30°.

The angle of refraction for violet light will be:

$1.56 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.56} = \frac{0.5}{1.56} \\\\sin(r) = 0.3205\\\\r = 18.7^o$

And the angle of refraction for red light will be:

$1.5 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.5} = \frac{0.5}{1.5} \\\\sin(r) = 0.3333\\\\r = 19.5^o$

The angle of refraction for red light is larger than that of violet light when the angle of incidence is 30°.

c. The angle of incidence is 60°.

The angle of refraction for violet light will be:

$1.56 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.56} = \frac{0.8660}{1.56} \\\\sin(r) = 0.5551\\\\r = 33.7^o$

And the angle of refraction for red light will be:

$1.5 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.5} = \frac{0.8660}{1.5} \\\\sin(r) = 0.5773\\\\r = 35.3^o$

The angle of refraction for red light is still larger than that of violet light when the angle of incidence is 60°.

6 0

3 years ago