Answer:
Height of the rocket be one minute after liftoff is 40.1382 km.
Explanation:

v = velocity of rocket at time t
g = Acceleration due to gravity =
= Constant velocity relative to the rocket = 2,900m/s.
m = Initial mass of the rocket at liftoff = 29000 kg
r = Rate at which fuel is consumed = 170 kg/s
Velocity of the rocket after 1 minute of the liftoff =v
t = 1 minute = 60 seconds'
Substituting all the given values in in the given equation:


Height of the rocket = h



Height of the rocket be one minute after liftoff is 40.1382 km.
Answer:
The right solution is:
(a) 2.87 eV
(b) 1.4375 eV
Explanation:
Given:
Wavelength,
= 433 nm
Potential difference,
= 1.43 V
Now,
(a)
The energy of photon will be:
E = 
= 
or,
= 
= 
(b)
As we know,
⇒ 
By substituting the values, we get
⇒ 
⇒ 
or,
⇒ 
⇒ 
Answer:
B = 0.8 T
Explanation:
It is given that,
Radius of circular loop, r = 0.75 m
Current in the loop, I = 3 A
The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.
When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.
We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

B is magnetic field

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.