Answer:
The centripetal acceleration of the stone is 5 m/s²
Explanation:
The length of the string to which the stone is attached, r = 1 m
The speed with which the string is rotated, v = 5 m/s
The centripetal acceleration, 
, is given as follows;
Therefore, the centripetal acceleration of the stone found as follows;
The centripetal acceleration of the stone, = 5 m/s².
Answer:
5.en
6.ex
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Explanation:
<h3>#CARRY ON LEARNING</h3><h3>#BRAINLITS </h3>
Your answer is B. The human eye can only detect color and Tv waves
Let's start with the concept of momentum. What is it? Linear momentum in physics is mathematically written as a product of mass and velocity of an object. Now let us suppose a body of mass m is moving in an inertial frame of reference with velocity v. Consider the fact that no external force is acting on the system. The momentum of this body is given by mv, where m is the mass and v is its velocity. In case of simple real world problems not delving into the realms of relativity, mass is a conserved quantity and it cannot be zero. Hence the velocity of the body must be zero and hence the momentum.
However, photons are considered to have a rest mass zero.
However note the point carefully "rest mass". A body in motion cannot have mass to be zero.
<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em>
Answer:
How far will the electron travel beforehitting a plate is 248.125mm
Explanation:
Applying Gauss' law:
Electric Field E = Charge density/epsilon nought
Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12
Therefore E = 1.0 x 10^-6/8.85× 10^-12
E= 1.13×10^5N/C
Force on electron F=qE
Where q=charge of electron=1.6×10^-19C
Therefore F=1.6×10^-19×1.13×10^5
F=1.808×10^-14N
Acceleration on electron a = Force/Mass
Where Mass of electron = 9.10938356 × 10^-31
Therefore a= 1.808×10^-14 /9.11 × 10-31
a= 1.985×10^16m/s^2
Time spent between plate = Distance/Speed
From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2
Therefore Time = 0.01/2×10^6
Time =5×10^-9s
How far the electron would travel S =ut+ at^2/2 where u=0
S= 1.985×10^16×(5×10^-9)^2/2
S=24.8125×10^-2m
S=248.125mm
