A-200 kg
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The hang time of the ball is 4.08 s
Explanation:
The ball is in free fall motion: this means that it is acted upon gravity only, so its acceleration is the acceleration of gravity,

downward (the negative sign refers to the downward direction).
Since this is a uniformly accelerated motion, we can solve the problem by using the following suvat equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
First we calculate the time it takes for the ball to reach the maximum height, where the velocity is zero:
v = 0
Substituting:
u = +20 m/s

we find t

The motion of the ball is symmetrical, so the total time of flight is just twice the time needed to reach the maximum height, therefore:

Learn more about free fall:
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Explanation:
It is given that,
When the front wheels are over the scale, the weight recorded by the scale is 5800 N, F₁ = 5800 N
When the rear wheels are over the scale, the scale reads 6500 N, F₂ = 6500
The distance between the front and rear wheels is measured to be 3.20 m, x₂ = 3.2 m
We need to find the location of center of mass behind the front wheels. Let the center of is located at a distance of x₁. Thus balancing the torques we get :

On solving the above equation we get, x₂ = 1.69 m
So, the center of mass is located at a distance of 1.69 meters behind the front wheels.
Answer:
The height of the image of the candle is 20 cm.
Explanation:
Given that,
Size of the candle, h = 12 cm
Object distance from the candle, u = -6 cm
Focal length of converging lens, f = 15 cm
To find,
The height of the image of the candle.
Solution,
Firstly, we will find the image distance of the candle. Let it is equal to v. Using lens formula to find the image distance.

v is image distance

If h' is the height of the image. Magnification is given by :


So, the height of the image of the candle is 20 cm.