Answer:1.902 m
Explanation:
Given
height of apartment=1.5 m
It takes 0.21 sec to reach the bottom from apartment
So



i.e. if ball is dropped from top its velocity at window is 6.11 m/s
So height of upper floor above window

where s= height of upper floor above window
here u=0


s=1.902 m
Answer:
b) No acceleration in the vertical
c) 35N
d) 35N
e) 
Explanation:
a) The situation can be shown in the free body diagram shown in the figure below where F is the applied force, Fr is the friction force, W is the weight of the book and N is the normal force exerted vertically up from the desk to the book
b) The vertical movement is restrained by the normal force which opposes to the weight. In absence of any other force, they both are in equilibrium and the net force is zero
c) The net horizontal force acting on the book is the vectorial sum of the applied force and the friction force. Since they lie in the same axis and are opposed to each other:

d) The net force acting on the book is the vector sum of all forces in all axes. The normal and the weight cancel each other in the y-axis, so our resulting force is the x-axis net force, computed as above:
in the x-axis
e) Following Newton's second law, the acceleration is calculated as

The velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
<h3>What is Electric field?</h3>
Electric field is the physical field that surrounds a charge.
<h3>How to find final velocity of the electron when it moves some distance in a certain electric field?</h3>
- From Newton's second law, the acceleration the electron will be
a=F/m=qE/m
- where q= charge of electron
E= electric field
m= mass of electron
=(−1.60×10^−19C)(3×10³N/C)/(9.11×10^-31kg)
=10¹⁵×0.526m/s²
- The kinematics equation v²=v0²+2a(Δx)
- where v=final velocity of the electron
v0=initial velocity of the electron =5×10⁶m/s
a=acceleration of the electron =10¹⁵×0.526m/s²
Δx=distance moved by the electron in east direction =1cm=10^-2m
- Now v^2=(5×10⁶)²+2×10¹⁵×0.526×10^-2
=25×10¹²+10.52×10¹²
=35.52×10¹²
- Now velocity of electron=5.95×10⁶m/s.
Thus , we can conclude that the velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
Learn more about electric field here:
brainly.com/question/26199225
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I hope this is true or false
The answer would be TRUE
Hipparcos estimated its distance at roughly 96 parsecs from Earth, or 310 ± 20 light years away.