What is the threshold velocity vthreshold(water) (i.e., the minimum velocity) for creating Cherenkov light from a charged partic
1 answer:
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Answer:
<em>Its speed will be 280 m/s</em>
Explanation:
<u>Constant Acceleration Motion</u>
It's a type of motion in which the speed of an object changes by an equal amount in every equal period of time.
If a is the constant acceleration, vo the initial speed, vf the final speed, and t the time, vf can be calculated as:
The object accelerates from rest (vo=0) at a constant acceleration of . The final speed at t=35 seconds is:
Its speed will be 280 m/s
Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
The correct answer to the question is : 29.88 m.
EXPLANATION :
As per the question, the mass of the rock m = 50 Kg.
The rock is rolling off the edges of the cliff.
The final velocity of the rock when it hits the ground v = 24 .2 m/s.
Let the height of the cliff is h.
The potential energy gained by the rock at the top of the cliff = mgh.
Here, g is known as acceleration due to gravity, and g =
When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.
When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.
The kinetic energy of the rock when it touches the ground is given as -
Kinetic energy K.E = .
From above we know that -
Kinetic energy at the bottom of the cliff = potential energy at a height h
⇒
⇒
⇒
⇒
Hence, the height of the cliff is 29.88 m