The correct answer is (B)
Which is (kQ1Q2) / d^2
Answer:
option A is correct
Explanation:
acceleration the time rate of change of velocity or speed so
a=Δv/t
Δv=vf-vi
Δv= final speed-initial speed
now a=final speed-initial speed/time
We know, Potential Energy = m * g * h
Here, mass & gravity would be same, but their height will change so it will be:
ΔU = U₂ - U₁
ΔU = mgh₂ - mgh₁
ΔU = mg (h₂ - h₁)
Hope this helps!
Answer:
Explanation:
mass of car, m = 1000 kg
initial velocity, u = 20 m/s
final velocity, v = 0 m/s
distance, s = 120 m
Let a be the acceleration of motion
use third equation of motion
v² = u² + 2 as
0 = 20 x 20 + 2 x a x 120
a = - 1.67 m/s²
Let F be the force
Force, F mass x acceleration
F = - 1000 x 1.67
F = - 1666.67 N
The direction of force is towards south and the magnitude of force is 1666.67 N.
Answer:
v(t)=3t²+3, and a(t)=6t
v(0)=3cm/sec, and a(0)=0cm/sec²
v(1)=6cm/sec, and a(1)=6cm/sec²
Explanation:
<u>Find velocity and acceleration functions ( v(t) and a(t) )</u>
(Relevant background: Let's say f(t) is a function of y with respect to x. Think of the derivative of this function, f'(t) , as representing the rate at which y changes with respect to x)
s(t) is a function of distance with respect to time. Therefore, s'(t) - the derivative of this - represents the rate at which distance changes with time, which is just the definition of velocity. So we can say velocity v(t) = s'(t).
s(t)=t³+3t+3
s'(t)=v(t)=3t²+3
Similarly, if v(t) is a function of speed with respect to time, then v'(t) represents the rate at which speed changes with time, which is acceleration. So we can say that acceleration a(t)=v'(t)=s''(t)
v(t)=3t²+3
v'(t)=a(t)=6t
<u>Find velocity and acceleration at t=0 and t=1</u>
<u>t=0</u>
v(t)=3t²+3
v(0)=3(0²)+3
v(0)=3 cm/sec
a(t)=6t
a(0)=6(0)
a(0)=0 cm/sec²
<u>t=1</u>
v(t)=3t²+3
v(1)=3(1²)+3
v(1)=3+3
v(t)=6 cm/sec
a(t)=6t
a(1)=6(1)
a(1)=6 cm/sec²
