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aleksley [76]
3 years ago
6

A car is traveling at 21.0 m/s. It slows to a stop at a constant rate over 5.00s. How far does the car travel during those 5.00

seconds before it stops?
Physics
1 answer:
Doss [256]3 years ago
5 0

Answer:

d = 105 m

Explanation:

Speed of a car, v = 21 m/s

We need to find the distance traveled by the dar during those 5 s before it stops. Let the distance is d. It can be calculated as :

d = v × t

d = 21 m/s × 5 s

d = 105 m

So, it will cover 105 m before it stops.

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dmitriy555 [2]

Answer:

3

Explanation:

it's too because on you measuring mass of something

6 0
3 years ago
Anyone know how to do this?
Gala2k [10]

Answer:

I think, (remember think) it might be 2.0 m/s

Explanation:

If it's wrong I'm truly sorry.

6 0
3 years ago
Would you expect to observe every supernova in our own Galaxy? Why or why not?
Taya2010 [7]

Explanation:

Only few supernova are observed in our galaxy -

Type II supernovae ( i.e. the explosions of the massive stars ) occurred in the Milky Way, and they might be hidden by the intervening dust if they are located in the more distant parts of our Galaxy .

Type Ia supernovae , which need a white dwarf star in the binary star system , are brighter than the type II supernovae , but some of them could also happen in the older parts of Galaxy which are hidden due to the buildup of the dust and gas .

3 0
3 years ago
A car (mass = 1090 kg) is traveling at 30.4 m/s when it collides head-on with a sport utility vehicle (mass = 2880 kg) traveling
Thepotemich [5.8K]

Answer:

The sport utility vehicle was traveling at V2= 11.5 m/s.

Explanation:

m1= 1090 kg

V1= 30.4 m/s

m2= 2880 kg

V2= ?

m1*V1 = m2*V2

V2= (m1*V1)/m2

V2= 11.5 m/s

7 0
3 years ago
A flywheel of mass 182 kg has an effective radius of 0.62 m (assume the mass is concentrated along a circumference located at th
musickatia [10]

Answer:

A)5524J,

B) 29.2Nm

Explanation:

This question can be treated using work- energy theorem

Work= change in Kinectic energy

W= Δ KE

Work= difference between the final Kinectic energy and intial Kinectic energy.

We know that

Kinectic energy= 1/2 mv^2 .............eqn(1)

This can be written in term of angular velocity, as

KE= 1/2 I

4 0
3 years ago
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