Answer:
The weight of the block on the moon is 15 kg.
Explanation:
It is given that,
The acceleration of the block, a = 7.5 m/s²
Force applied to the box, F = 70 N
The mass of the block will be, 

m = 9.34 kg
The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :


W = 15.13 N
or
W = 15 N
So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.
v₀ = initial speed of the object = 8 meter/second
v = final speed of the object = 16 meter/second
t = time taken to increase the speed = 10 seconds
d = distance traveled by the object in the given time duration = ?
using the kinematics equation
d = (v + v₀) t/2
inserting the above values in the above equation
d = (16 + 8) (10)/2
d = 120 meter
Answer:
P = 5.22 Kg.m/s
Explanation:
given,
mass of the projectile = 1.8 Kg
speed of the target = 4.8 m/s
angle of deflection = 60°
Speed after collision = 2.9 m/s
magnitude of momentum after collision = ?
initial momentum of the body = m x v
= 1.8 x 4.8 = 8.64 kg.m/s
final momentum after collision
momentum along x-direction
P_x = m v cos θ
P_x = 1.8 x 2.9 x cos 60°
P_x = 2.61 kg.m/s
momentum along y-direction
P_y = m v sin θ
P_y = 1.8 x 2.9 x sin 60°
P_y = 4.52 kg.m/s
net momentum of the body


P = 5.22 Kg.m/s
momentum magnitude after collision is equal to P = 5.22 Kg.m/s
Answer:
B. stearothermophilus and S. ruber
Explanation:
B. stearothermophilus and S. ruber
In solar evaporation ponds the temperature is higher and the salt concentration is also higher because of the water evaporated so sunder such extreme conditions this hybrid bacteria is capable of surviving. B. stearothermophilus is thermophilus bacteria which grows at high temperature and S. ruber is halophilic bacteria which grows in saline environment. So, these two bacteria best suited for the above hybrid condition.
Answer:
Work out = 28.27 kJ/kg
Explanation:
For R-134a, from the saturated tables at 800 kPa, we get
= 171.82 kJ/kg
Therefore, at saturation pressure 140 kPa, saturation temperature is
= -18.77°C = 254.23 K
At saturation pressure 800 kPa, the saturation temperature is
= 31.31°C = 304.31 K
Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.
Thus,
=
= 171.82 kJ/kg
We know COP of heat pump
COP = 
= 
= 6.076
Therefore, Work out put, W = 
= 171.82 / 6.076
= 28.27 kJ/kg