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Alona [7]
3 years ago
6

A car accelerates at a rate of 6m/s^2 for 4 seconds until it has traveled a total of 40m what was the initial velocity of the ca

r
Physics
1 answer:
s344n2d4d5 [400]3 years ago
7 0

Using the 2nd equation of motion;

= s = ut + ½at²

= 40 = u×2 + ½ × 6 × 2²

= 40 = 2u + 3 × 4

= 40 - 12 = 2u

= 28/2 = u

= 14 m/s = u

And its done! Simple isn't? :P

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For any object, the greater the force that's applied to it, the greater its ____ will be.
Masteriza [31]

Answer:

For any object, the greater the force that's applied to it, the greater its <em>acceleration</em> will be.

Explanation:

The statement can be explained using Newton's second law of motion that states, "A net Force 'F' applied on a body produces acceleration 'a' in it which is directly proportion to the Force applied and inversely proportional to the mass of the object."

We can conclude:

F ∝ a

6 0
3 years ago
Read 2 more answers
two vehicles have a head on collision. one vehicle has a mass of 3000 kg and moves at 25 m/s while the second vehicle has a mass
emmasim [6.3K]

Answer:

The speed of the combined vehicles is 6.82m/s

Explanation:

Using the law of conservation of momentum which stayed that the sum of momentum of bodies before collision is equal to their sum of momentum after collision. After collision, both object moves with the same velocity.

Momentum = mass×velocity

Before collision:

Momentum of vehicle or mass 3000kg moving with velocity 25m/s

= 3000×25

= 75000kgm/s

Pa = 75000kgm/s

Momentum of vehicle with mass 2500kg moving with velocity of -15m/s

= 2500×-15

= -37500kgm/s

After collision:

Momentum = (3000+2500)V

Where v is their common velocity

Momentum after collision = 5500V

Based on the law:

75000+(-37500) = 5500V

75000-37500 = 5500V

37500 = 5500V

V = 37500/5500

V = 6.82m/s

7 0
3 years ago
Why can't you trust the law of gravity? (RIDDLE)
alukav5142 [94]
Because it will always let you down
8 0
3 years ago
Read 2 more answers
Two particles A and B start from rest at the origin x = 0 [ft] and move along a straight line such that a = (613) Ift/s) and ag
Amanda [17]

Answer:

Distance between them after 3 s is 2695.5 ft.

Total distance traveled by A in 3 s is 2758.5 ft.

Total distance traveled by B in 3 s is 5454 ft.

Explanation:

For particle A:

u = 0, a = 613 ft/s

Let the distance traveled by particle A in 3 seconds is Sa.

Use second equation of motion

S = u t + 1/2  at ^2

Sa = 0 + 1/2 x 613 x 3 x 3 = 2758.5 ft

For particle B:

u = 0, a = 1212.8 ft/s

Let the distance traveled by particle B in 3 seconds is Sb.

Use second equation of motion

S = u t + 1/2  at ^2

Sb = 0 + 1/2 x 1212 x 3 x 3 = 5454 ft

Thus, the difference in the distance traveled by A and B is 5454 - 2758.5 = 2695.5 ft.

3 0
3 years ago
In which type of wave do air particles move together or apart parallel to the direction of the wave?
Phoenix [80]
The best and most correct answer among the choices provided by the question is <span>B.sound waves</span><span>.
</span>

<span>Particles move together or apart parallel to the direction of the sound wave.
</span>
Hope my answer would be a great help for you.    
If you have more questions feel free to ask here at Brainly.
8 0
4 years ago
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