Answer:
The acceleration of the rocket is 10 m/s².
Explanation:
Let the acceleration of the rocket be m/s².
Given:
Mass of the rocket is,
Thrust force acting upward is,
Acceleration due to gravity is,
Now, force acting in the downward direction is due to the weight of the rocket and is given as:
Now, net force acting on the rocket in upward direction is given as:
Therefore, from Newton's second law, net force acting on the rocket is equal to the product of mass and acceleration.
Therefore, the acceleration of the rocket is 10 m/s².
Given :
Displacement , y = 0.75 m .
Angular acceleration , .
Initial angular velocity , .
To Find :
The value of vertical velocity after time t = 0.25 s .
Solution :
By equation of circular motion is given by :
Putting all given values we get :
Now , vertical velocity is given by :
Therefore , the numerical value of the vertical velocity of the car at time t=0.25 s is 4.90 m/s .
Hence , this is the required solution .
<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>
<span> </span>
Answer:
(a) Magnetic moment will be
(b) Torque will be
Explanation:
We have given dimension of the rectangular 5.4 cm × 8.5 cm
So area of the rectangular coil
Current is given as
Number of turns N = 25
(A) We know that magnetic moment is given by
(b) Magnetic field is given as B = 0.350 T
We know that torque is given by
Answer:
The magnitude of the torque on the loop due to the magnetic field is .
Explanation:
Given that,
Diameter = 10 cm
Current = 0.20 A
Magnetic field = 0.30 T
Unit vector
We need to calculate the torque on the loop
Using formula of torque
Where, N = number of turns
A = area
I = current
B = magnetic field
Put the value into the formula
Hence, The magnitude of the torque on the loop due to the magnetic field is .