-- What's the volume of a cylinder with radius=1m and height=55m ?
( Volume of a cylinder = π R² h )
-- How much does that volume of water weigh ?
1 liter of water = 1 kilogram of mass
Weight = (mass) x (acceleration of gravity)
-- What's the area of the bottom of that 1m-radius cylinder ?
Pressure = (force) / (area)
Answer:
27.1 m/s
Explanation:
Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.
Using third equation of motion,
V^2 = U^2 + 2aS
Since the car is decelerating, the final velocity V = 0
Substitute all the parameter into the equation above,
0 = U^2 - 2 * 40.52 * 9.06
U^2 = 734.22
U = 
U = 27.096
U = 27.1 m/s approximately
Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid
Answer:
Resultant force, R = 10 N
Explanation:
It is given that,
Force acting along +x direction, 
Force acting along +y direction, 
Both the forces are acting on a point object located at the origin. Let the resultant force of the object is given by R. So,

Here 


R = 10 N
So, the resultant force on the object is 10 N. Hence, this is the required solution.
The work done by the applied force on the block against the frictional force is 15.75 J.
<h3>
Work done by the applied force</h3>
The work done by the applied force is calculated as follows;
W = Fd
F - Ff = ma
where;
- F is applied force
- Ff is frictional force
Fcos(37) - μmgsin(37) = ma
Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)
0.799F - 7.077 = 0.8
F = 9.86 N
W = Fdcosθ
W = 9.86 x 2 x cos(37)
W = 15.75 J
Thus, the work done by the applied force on the block against the frictional force is 15.75 J.
Learn more about work done here: brainly.com/question/25573309
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I’m so sorry, I need more information. Good luck and I’m sorry I couldn’t help you :(