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baherus [9]
3 years ago
10

What is a Super Massive Black Hole?

Physics
1 answer:
Serggg [28]3 years ago
4 0

Answer:

I dont really know much but i know that it swallow anything it comes across in space.

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A 1,400 kg car accelerates from rest to 30 m/s in 6.0 seconds. what is the net force on the car?
Allushta [10]
So your finding acceleration first which is 30m/s divides by 6 seconds equals 5m/s^s and then multiply that by 1,400 kg and you have net force which is 7,000N
4 0
3 years ago
How much force is needed to stretch a spring 1. 2 m if the spring constant is 8. 5 N/m? 7. 1 N 7. 3 N 9. 7 N 10. 2 N.
statuscvo [17]

The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N). Force is needed to stretch spring is 10.2 N.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =?

K is the spring constant= 8.5 N/m

x is the length by which spring got stretched = 1.2m

\rm F_S=Kx \\\\ \rm F_S=8.5 \times 1.2 \\\\ \rm F_S=10.2 N

Hence the force is needed to stretch the spring is 10.2 N.

To learn more about the spring force refer to the link;

brainly.com/question/4291098

6 0
2 years ago
an object at rest starts accelerating if it travels 30 meters to end up going 10 m/s what was it’s acceleration
vfiekz [6]

First we have to calculate the time taken to travel the distance 30 m, is

t = \frac{distance}{velocity} = \frac{30 \ m}{10 \ m/s } =  3 \ s.

Now from equation of motion,

v = u + at

Given, v = 10 \ m/s .

As object starts from rest, so  u = 0.

Substituting these values in above equation, we get

 10 \ m/s = 0 + a \times 3 \ s \\\\ a = \frac{10}{3}  = 3.33 \ m/s^2.

Thus, the acceleration is 3.33 \ m/s^2

3 0
3 years ago
Miswer
quester [9]

Answer: 4.0

Explanation:

5 0
3 years ago
Radar uses radio waves of a wavelength of 2.4 \({\rm m}\) . The time interval for one radiation pulse is 100 times larger than t
blondinia [14]

Answer:

120 m

Explanation:

Given:

wavelength 'λ' = 2.4m

pulse width 'τ'= 100T ('T' is the time of one oscillation)

The below inequality express the range of distances to an object that radar can detect

τc/2 < x < Tc/2 ---->eq(1)

Where, τc/2 is the shortest distance

First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'

f = c/λ (c= speed of light i.e 3 x 10^{8} m/s)

f= 3 x 10^{8} / 2.4

f=1.25 x  10^{8} hz.

As, T= 1/f

time of one oscillation T= 1/1.25 x  10^{8}

T= 8 x 10^{-9} s

It was given that pulse width 'τ'= 100T

τ= 100 x 8 x 10^{-9} => 800 x 10^{-9} s

From eq(1), we can conclude that the shortest distance to an object that this radar can detect:

x_{min}= τc/2 =>  (800 x 10^{-9} x 3 x 10^{8})/2

x_{min}=120m

8 0
3 years ago
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