Answer:
F = 51.3°
Explanation:
The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

where,
W = Weight of Wagon = 150 N
θ = Angle of Inclinition = 20°
Therefore,

<u>F = 51.3°</u>
<h2><u>Projectile</u><u> </u><u>motion</u><u>:</u></h2>
<em>If</em><em> </em><em>an</em><em> </em><em>object is given an initial velocity</em><em> </em><em>in any direction and then allowed</em><em> </em><em>to travel freely under gravity</em><em>, </em><em>it</em><em> </em><em>is</em><em> </em><em>called a projectile motion</em><em>. </em>
It is basically 3 types.
- horizontally projectile motion
- oblique projectile motion
- included plane projectile motion
The refractive index for glycerine is

, while for air it is

.
When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:

where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
u= 215 km/hr = 215 * 1000/ 3600 = aprx 60m/s
v=0
t=2.7sec
v= u - at
u= at
60/2.7 = 22.23 m/s^2
Hope it helps
Answer: d. 8.25 m/s
Explanation:
We are given that Current= 5 m/s in j direction
Velocity= 8 m/s i + 3 m/s j
Now, we have to find Jada's speed with respect to the water.
First we find Jada's velocity with respect to water
v= (8 i + 3 j) - (5 j)
v= 8i - 2 j
To find the speed, we take the magnitude of this velocity vector we have
|v|= 
|v|=
= 8.246 m/s
which comes out to be around = 8.25 m/s
So option d is correct.