Ask question

Ask question

All categories

3 years ago

10

If you are at a train station and a train comes toward you, the train blows its whistle. as the train comes closer the whistle g

ets louder and as the train gets farther the whistle is quieter. What is this effect called?

1 answer:

Wittaler [7]3 years ago

4 0

This is called the Doppler Effect.

You might be interested in

A. An automobile mass is 3.5 x103 kg. If the forward thrust (Fnet)

katrin [286]

Answer:

a = 0.8 m/s^2

Explanation:

Force equation: F = ma

F = ma -> a = F/m = 2.8*10^3 N / 3.5*10^3 kg = 0.8 m/s^2

8 0

3 years ago

Why does the car stop? Where did the energy go?

Viefleur [7K]

Answer:

Because you hit the break?

8 0

3 years ago

How does metamorphic rock change into another type of metamorphic rock?

Marrrta [24]

Metamorphic rocks are formed by tremendous heat, great pressure, and chemical reactions. To change it into another type of metamorphic rock you have to reheat it and bury it deeper again beneath the Earth's surface.

Hope this helped! :)

4 0

3 years ago

What do the law of superposition and the law of inclusion have in common? (1 point) 1. Both laws are about matching fossils in d

r-ruslan [8.4K]

Answer:

its b

i took the test

Explanation:

4 0

3 years ago

A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w

kramer

Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = ?

$v_{1i}$ = initial velocity of the first body before collision = $v$

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

$v_{1f}$ = final velocity of the first body after collision =

using conservation of momentum equation

$m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}$

Using conservation of kinetic energy

$m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35$

b)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = 1.35 kg

$v_{1i}$ = initial velocity of the first body before collision = 4 ms⁻¹

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

$v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67$ ms⁻¹

8 0

3 years ago