How do you think the weight and the lift should not be one of the same magnitude

A car that brakes suddenly comes to a screeching halt. Is the sound energy produced in this conversion a useful form of energy?

Nezavi [6.7K]

Answer:

No, the sound produced by screeching brakes is not a useful form of energy, because it does not contribute to stopping the car.

7 0

2 years ago

The molecules of a substance become more closely packed and move more quickly.

andrey2020 [161]

When the substance are moved close together and they move more quickly they get compressed.

4 0

3 years ago

Consider the following situations: i. A ball moving at speed v is brought to rest ii. The same ball starts at rest and is projec

tatyana61 [14]

Answer:

c. Case iii

Explanation:

the ball will experience the largest change in case iii

3 0

3 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

natali 33 [55]

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

$v_e=\sqrt{\frac{2GM}{R}}$

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

$v=0.421 v_e$

So its initial kinetic energy will be

$K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}$

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

$K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}$

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius: $0.177 \frac{GMm}{R}=\frac{GMm}{nR}$

And solving the equation we find

$n=\frac{1}{0.177}=5.65$

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

$K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

This time, the projectile has 0.421 times this energy:

$K=0.421 \frac{GMm}{R}$

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

$0.421 \frac{GMm}{R}=\frac{GMm}{nR}$

and by solving for n we find

$n=\frac{1}{0.421}=2.36$

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) $E=U=\frac{GMm}{R}$

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

$E=U=\frac{GMm}{R}$

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

$K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

6 0

2 years ago

A rescue team is searching for Andrew, a geologist who was stranded while conducting research in the mountains of Colorado. The

Sindrei [870]

Answer:

I = 9.82 10⁻⁷ W / m²

Explanation:

The intensity of the sound wave is the energy of the wave between the order per unit area of the same

I = P / A = E / T A

the energy is calculated by integrating the mechanical energy in a period, where the mass is changed by the density and ‘s’ is the amplitude of the sound wave

I = ½ ρ v (w s)²

I = ½ 1.35 328 (2π 530 2.00 10⁻⁸)²

I = 221.4 (4.435 10⁻⁹)

I = 9.82 10⁻⁷ W / m²

4 0

3 years ago

How do you think the weight and the lift should not be one of the same magnitude​

How do you think the weight and the lift should not be one of the same magnitude