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
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
Because it will always let you down
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.
The best and most correct answer among the choices provided by the question is <span>B.sound waves</span><span>.
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<span>Particles move together or apart parallel to the direction of the sound wave.
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Hope my answer would be a great help for you.
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