Answer:
The coefficient of kinetic friction between the sled and the snow is 0.0134
Explanation:
Given that:
M = mass of person = 52 kg
m = mass of sled = 15.2 kg
U = initial velocity of person = 3.63 m/s
u = initial velocity of sled = 0 m/s
After collision, the person and the sled would move with the same velocity V.
a) According to law of momentum conservation:
Total momentum before collision = Total momentum after collision
MU + mu = (M + m)V
Substituting values:
The velocity of the sled and person as they move away is 2.81 m/s
b) acceleration due to gravity (g) = 9.8 m/s²
d = 30 m
Using the formula:
The coefficient of kinetic friction between the sled and the snow is 0.0134
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.
<span>E=hc/wav. len
E = (6.62 x 10^-34 x 3 x 10^8)/0.0275 x 10^-9
E = 7.22182 x 10^-15 J
To convert to eV divide by 1.6 x 10^-19
E = 7.22182 x 10^-15/1.6 x 10^-19 eV
E =45.36 x 10^3 eV
Th energy, E, of a single x-ray photon in eV is = 45.36keV.
Number of photons, n = total energy/ energy of photon
n = 3.85 x 10^-6/7.22182 x 10^-15
n = 5.33 x 10^8 photons </span>
