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Scorpion4ik [409]
3 years ago
8

Under the doppler effect, if the source of a wave is moving, it must be changing the frequency of the wave. true false

Physics
1 answer:
Ede4ka [16]3 years ago
4 0
True, a source coming towards you has an increased frequency, a source moving away from you has a reduced frequency.
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the rate of work done or doing work is usually measured by the power of a machine .

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A white dwarf is not continuing in nuclear fusion, yet it still shines brightly for thousands of years because...
Afina-wow [57]

Explanation:

Understanding that stars are naturally quite hot, imagine I pulled a piece of hot iron from a furnace. It would glow a bright red, and then slowly fade to black as the iron cools.

White dwarfs glow for the same reason - they are HOT!

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3 years ago
What change does this cause concerning weather?
Vika [28.1K]

Answer:

More extreme weather.

Explanation:

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Here's a picture I found on it:

6 0
3 years ago
Read 2 more answers
Can someone please help me with these physics problems? I just don’t even know where to start.
KIM [24]

#1

for the block of mass 5 kg normal force is given as

F_n = mg

F_n = 5*9.8 = 49 N

friction force is given as

F_f = \mu F_n

F_f = 0.1*49 = 4.9 N

Net force is given as

F_{net} = ma

F_{net} = 5*2 = 10 N

now we know that

F_{net} = F_{app} - F_f

10 = F_{app} - 4.9

F_{app} = 14.9 N

#2

Normal force is given as

F_n = mg

F_n = 6*9.8

F_n = 58.8 N

now we know that

F_{net} = F_{app} - F_f

F_{net} = 0

as object moves with constant velocity

F_{app} = F_f = 15 N

now for coefficient of friction we can use

F_f = \mu F_n

15 = \mu * 58.8

\mu = 0.255

#3

net force upwards is given as

F = 1.2 * 10^{-4} N

mass is given as

m = 7 * 10^{-5} kg

now as per newton's law we can say

F = ma

1.2 * 10^{-4} = 7 * 10^{-5} * a

a = 1.71 m/s^2

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

F_{net} = F_{app} - F_f

ma = F_{app} - F_f

5*6 = 40 - F_{f}

F_f = 40 - 30 = 10 N

part b)

for coefficient of friction we can use

F_f = \mu F_n

10 = \mu * F_n

here normal force is given as

F_n = mg = 5*9.8 = 49 N

now we have

\mu = \frac{10}{49} = 0.204

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

d = v_i*t + \frac{1}{2}at^2

20 = 0 + \frac{1}{2}a(5^2)

a = 1.6 m/s^2

now net force is given as

F_{net} = ma

F_{net} = 10*1.6 = 16 N

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

a = \frac{v_f - v_i}{t}

a = \frac{0 - 25}{1.5} = -16.67 m/s^2

now the force is gievn as

F = ma

F = 5*16.67 = 83.3 N

3 0
4 years ago
Who would benefit of your own handled problem?why?​
Bezzdna [24]
You would benefit from your own handled problem because you solved a problem you had.

i’m pretty sure that’s it, going off the info i had.
5 0
2 years ago
Read 2 more answers
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