Answer: No, the student does not do any work on the wall.
Explanation: The work is defined as the movement of an object done/caused by a force.
And it can be calculated as:
W = F*d
where W is work, F is force and d is the displacement of the object
In this case, the student is applying force on the wall, so we have the first part, but we also know that the wall does not move, so in this case, d = 0, which would imply that the work applied to the wall must be equal to zero
(because W = F*0 = 0)
So the answer is no, the student does not do any work on the wall.
0.77 m/s2 directed 35° south of west
net force = (-17,-12)
net force = mass * acceleration
(-17,-12) = 27 * (x-acceleration,y-acceleration)
(x-acceleration,y-acceleration) = (-17/27,-12/27) = (-0.629629629..., -0.444...)
angle of acceleration = tan^-1 (-0.444.../-0.629629...) = 35.21759 degrees below negative x-axis.
magnitude of acceleration = sqrt((-0.629629...)^2 + (-0.444...)^2) = 0.77069 (5dp)
A scalar is something that only has speed without a DIRECTION while a vector has direction. acceleration is a vector and mass is a scalar.
Answer:
Diagrammatic representation
<span>a. We can find the velocity when the camera hits the ground.
v^2 = (v0)^2 + 2ay = 0 + 2ay
v = sqrt{ 2ay }
v = sqrt{ (2)(3.7 m/s^2)(239 m) }
v = 42 m/s
The camera hits the ground with a velocity of 42 m/s
b. We can find the time it takes for the camera to hit the ground.
y = (1/2) a t^2
t^2 = 2y / a
t = sqrt{ 2y / a }
t = sqrt{ (2)(239 m) / 3.7 m/s^2 }
t = 11.4 seconds
It takes 11.4 seconds for the camera to hit the ground.</span>