<h2>
Answer:</h2>
<em><u>Velocity of throwing arrow = 43.13 m/s.</u></em>
<h2>
Explanation:</h2>
In the question,
Let us say the height from which the arrow was shot = h
Distance traveled by the arrow in horizontal = 61 m
Angle made by the arrow with the ground = 2°
So,
From the <u>equations of the motion</u>,
Now,
Also,
Finally, the angle made is 2 degrees with the horizontal.
So,
Final horizontal velocity = v.cos20°
Final vertical velocity = v.sin20°
Now,
u = v.cos20° (No acceleration in horizontal)
Also,
So,
We can say that,
<em><u>Therefore, the velocity with which the arrow was shot by the archer is 43.13 m/s.</u></em>
Answer:
Explanation:
Given
Plane is initially flying with velocity of magnitude
at angle of with North towards west
Velocity of plane airplane can be written as
Now wind is encountered with speed of at angle of
resultant velocity
for direction
west of North
Answer with Explanation:
We are given that
Charge on alpha particle=q=2 e=
1 e=
Mass of alpha particle=m=
Potential difference,V=
Magnetic field,B=3.49 T
a.Speed of alpha particle=v=
By using the formula
b.Magnetic force,F
F=
c.Radius of circular path, r=
r=0.084 m