All categories

3 years ago

Which of Newton’s Laws involves mass and acceleration? Question 1 options: 3rd 1st 2nd All of them

Physics

2 answers:

Oksana_A [137]3 years ago

7 0

Newton's second law states the following:

The change of movement is directly proportional to the force applied to the body and occurs in the direction where the force is applied.

In other words, we have:

F = ma

Where,

F: net force acting on the body

m: body mass (proportionality constant)

a: body acceleration

Answer:

Newton’s Law that involves mass and acceleration is:

2nd Law

Feliz [49]3 years ago

6 0

The answer would be Newton’s Second Law

You might be interested in

A lifeguard on a beach observes that waves have a speed of 7 m/s and a distance of 4.6m between wave crests. what is the period

Pepsi [2]

Answer:

0.657 seconds

Explanation:

speed of wave= wavelength / time period

time period= wavelength / speed

= 4.6/7

=0.657 sec

4 0

3 years ago

A taxi is travelling at 15 m/s. Its driver accelerates with

Hitman42 [59]

If we assume that the acceleration is constant, we can use on the kinematic equations:

Vf = Vi + a*t = 15 + 3*4 = 27 m/s

6 0

3 years ago

How are sublimation and deposition different from each other

yanalaym [24]

There are six distinct changes of phase which happens to different substances at different temperatures. ... Sublimation: the substance changes directly from a solid to a gas without going through the liquid phase. Deposition: the substance changes directly from a gas to a solid without going through the liquid phase.

Hope u find this helpful

3 0

3 years ago

I am having trouble with writing a 2 paragraph summary about this vidio can someone help me

sveticcg [70]

Answer:

hmmm ok

Explanation:

3 0

2 years ago

You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind and you perceive

Irina-Kira [14]

Answer:

Explanation:

In the first case you can use the expression for the Doppler effect when the source is getting closer and getting away

$f'=f(\frac{v}{v-v_{s}})$ ( 1 )

$f''=f(\frac{v}{v+v_s})$ ( 2 )

f' = perceived frequency when the source is getting closer

f'' = perceived frequency when the source is getting away

f = source frequency

v = relative speed

vs = sound speed

by dividing (1) and (2) you have

$\frac{f'}{f''}=\frac{f}{f}\frac{\frac{v}{v-v_s}}{\frac{v}{v+v_s}}=\frac{v+v_s}{v-v_s}\\\\f'v-f'v_s=f''v+f''v_s\\\\v(f'-f'')=v_s(f''+f')\\\\v=v_s\frac{f''+f'}{f'-f''}=(340\frac{m}{s})\frac{1370Hz+1330Hz}{1370Hz-1330Hz}=67.5\frac{m}{s}$

but this is the relative velocity, you have that

$v=v_{sir}-v_{car}\\v_{sir}=v+v_{car}=67.5\frac{m}{s}+35\frac{m}{s}=102.5\frac{m}{s}$

a. hence, the speed of the police car is 102.5m/s

6 0

3 years ago