Of the cliff?
Projectile motion
In the problem we are asked to find a height of certain cliff when a motorcycle stunt driver zoom out horizontally at the end the cliff at an initial velocity. So we will use one of the kinematics equation for projectile motion,
y
=
v
o
y
t
+
1
2
g
t
where
v
o
y
is just equal to zero since we can assume that the driver zooms out horizontally,
g
=
9.8
m
/
s
2
and
t
is time after
Answer:
(a): The car's relative position to the base of the cliff is x= 32.52m.
(b): The lenght of the car in the ir is tfall= 1.78 sec.
Explanation:
Vo= 0
V= ?
d= 50m
h= 30m
a= 4 m/s²
t= √(2*d/a)
t= 5 sec
V= a*t
V= 20 m/s
Vx= V * cos(24º)
Vx= 18.27 m/s
Vy= V* sin(24º)
Vy= 8.13 m/s
h= Vy*t + g*t²/2
clearing t:
tfall= 1.78 sec (b)
x= Vx * tfall
x= 32.52 m (a)
components of the speed of the coin is given as
now the time taken by the coin to reach the plate is given by
now in order to find the height
so it is placed at 1.52 m height
Answer:
A negative charge, if free to move in an electric field, will move from a low potential point to a high potential point. To move a positive charge against the electric field, work has to be done by you or a force external to the field.
Explanation:
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Answer:
Force, F = 20240 N
Explanation:
It is given that,
Pressure exerted by the four tires of an automobile, 
Area of each tire, 
Area of 4 tires, 
We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :
F = 20240 N
So, the weight of the automobile is 20240 N. Hence, this is the required solution.
