Explanation:
initial height, yo = 2 m
initial velocity, u = 20 m/s
angle of projection,θ = 5 degree
distance of net = 7 m
height of net = 1 m
Let it covers a vertical distance y in time t .
Use Second equation of motion for vertical motion
As it hits the ground in time t, so put y = 0
Taking positive sign, t = 0.84 s
The ball travels a horizontal distance x in time t
X = 20 Cos5 x t
X = 16.76 m
As this distance is more than the distance of net, so it clears the net.
Let t' be the time taken to travel a horizontal distance equal to the distance of net
7 = 20 cos5 x t'
t' = 0.35 s
Let the vertical distance traveled by the ball in time t' is y'.
So,
y' = 2.008 m
So, it clears the net which is 1 m high.
It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m
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Answer:
Uncertainty in position of the bullet is 
Explanation:
It is given that,
Mass of the bullet, m = 35 g = 0.035 kg
Velocity of bullet, v = 709 m/s
The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :
Uncertainty in momentum is,
We need to find the uncertainty in position. It can be calculated using Heisenberg uncertainty principal as :
Hence, this is the required solution.
The rate of fuel burning in grams per hour if the DT reaction is used is 1.08 ×
J/g per hour
<h3>How is the rate of fuel burning in grams per hour calculated when the D-T reaction is used?</h3>
- D + T → He + n
- The D-T fusion reaction results in a Helium (He) and neutron (n)
E = 17.59 MeV
Mass = 2.014u + 3.016u
= 5.030u
Energy per Kg = (17.59×
×1.6×
) ÷ ( 5.030×1.66×
)
= 3.37×
J/Kg
= 3.0× 
J/g
Rate of fuel burning in grams per hour = 3.0× × 3600
= 3.6×3.0×
= 1.08 × J/g per hour
To learn more about fusion reactor and energy production, refer
brainly.com/question/13399644
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Answer:
pressure = density x g x height
= 1000 x 10 x 6 Pascal
=60000 Pascal
OR 60 kP
Answer:
When the platform rotates, the rotating mass will travel in a circular path due to the force exerted on it by the string (by way of the tension in the spring). Since it is not possible to have an instantaneous readout of this tension force while the platform is rotating, an indirect measurement of this force will be made using the weight of the static mass as shown and explained below.
