Answer:
Explanation:
We have the following data:
- distance covered by the child: d = 2 m (length of the slide)
- time taken to cover this distance: t = 3 s
- initial velocity of the child: 0 m/s (he starts from rest)
So we can find the acceleration by using the equation:
Where a is the acceleration.
Substituting the values and solving for a,
Longitudinal waves transfer energy parallel to the direction of the wave motion
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
Answer:
15193.62 m/s
Explanation:
t = Time taken = 6.5 hours
u = Initial velocity = 0 (Assumed)
m = Mass of rocket = 1380 kg
F = Thrust force = 896 N
v = Final velocity
a = Acceleration of the rocket
Force
Equation of motion
The velocity of the rocket after 6.5 hours of thrust is 15193.62 m/s
