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3 years ago

A boy throws a ball vertically up and catches it after 3 s. What height did the ball reach?

Physics

2 answers:

Crazy boy [7]3 years ago

7 0

Answer:

11.025 meters

Explanation:

The gravity’s acceleration on the surface of earth is approximately 9.8m/s

If it took 3 s to go up and come back, that means it took (1/2) × 3 to go from his hand to its highest point, and that amount of time again (1.5 seconds) to go back down to his hand.

This means the ball accelerated from Vi (initial Velocity) to 0 in 1.5 s

To find g in 1.5 i.e speed =

1.5 × 9.8 m/s = 14.7 m/s.

To find the height the ball reached=

We will use d = rt for this. Distance = Rate × Time

Since, it went 14.7 m/s at the start, and 0 m/s at the top, so the average speed of the ball going upward was (0+14.7)/2 = 7.35 m/s

7.35 m/s was the average speed = rate.

Time was 1.5 seconds, so we multiply.

7.35 m/s × 1.5seconds = 11.025 meters

Darya [45]3 years ago

3 0

The top height the ball had is (9/8)*g
g being the strength of the earth's attraction. so if g is about 9.8 then the ball went up about 11 meters

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3 years ago

Lucy and her bike together have a mass of 120kg. She slows down from 4.5m/s to 3.5m/s. How much kinetic energy does she lose?

vovangra [49]

The kinetic energy of a moving object is given by
$K= \frac{1}{2}mv^2$
where m is the object's mass and v its velocity.

In our problem, the initial kinetic energy is:
$K_i = \frac{1}{2} m v_i^2 = \frac{1}{2}(120 kg) (4.5 m/s)^2=1215 J$

while the final kinetic energy is:
$K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(120 kg)(3.5 m/s)^2= 735 J$

So, the kinetic energy lost by Lucy and her bike is
$\Delta K = K_i - K_f = 1215 J - 735 J = 480 J$

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3 years ago

A car moves at speed v across a bridge made in the shape of a circular arc of radius r. (a) Find an expression for the normal fo

inna [77]

Answer:

(a) FN = m (g - $\frac{v^{2} }{r}$ )

(b) vmin = 17.146 m/s

Explanation:

The radius of the arc is

r = 30m

The normal force acting on the car form the highest point is

FN = m (g - $\frac{v^{2} }{r}$ )

If the normal force become 0 we have

m (g - $\frac{v^{2} }{r}$ ) = 0

g - $\frac{v^{2} }{r}$ = 0

This way, when FN = 0, then v = vmin, so

g - $\frac{vmin^{2} }{r}$ = 0

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3 years ago

Read 2 more answers

Find the exponent A in the equation

hodyreva [135]

Start by swapping the variables in the equations with their units (leave A alone for now): $v=\frac{a^2t^A}{x} \Rightarrow \frac{L}{T} = \frac{ (\frac{L}{T^2})^2 T^A }{L}$

Multiply by L and square the fraction: $\frac{L^2}{T} = \frac{L^2}{T^4} T^A \Rightarrow \frac{L^2}{T} \div \frac{L^2}{T^4}=T^A$

Divide the fraction by multiplying by the reciprocal: $\frac{L^2}{T} \times \frac{T^4}{L^2} = T^A \Rightarrow T^3=T^A$

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