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3 years ago

A tennis ball is hit into the air with a racket. When is the balls kinetic energy the greatest?

Physics

1 answer:

madam [21]3 years ago

3 0

Answer:

Kinetic energy is maximum when the player hits the ball.

Explanation:

Kinetic energy $=\frac{1}{2} mv^2$ , where m is the mass and v is the velocity.

So kinetic energy is proportional to square of velocity.

Velocity is maximum when the player hits the ball.

So kinetic energy is maximum when the player hits the ball.

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A can bavelling at 90kma/hr

Readme [11.4K]

Answer:

20 mins

Explanation:

To go 90 km it takes =60 mins(1 hr)

" " 1 km it takes =60/90 mins

" " 30 km it takes =60/90 multiplied by 30

= 20 mins

5 0

3 years ago

State the objects in the universe that give out their own light

Valentin [98]

All of those are stars.

3 0

3 years ago

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts help.

GrogVix [38]

Answer:

$H = 532 m$

Explanation:

When teacher falls into the cliff then she shout for Help at the same time

so here we know that sound will go down and reflect back up

so here in 3 s distance traveled by the sound

$d = vt$

$d = 340 \times 3$

$d = 1020 m$

now in the same time the distance that teacher will fall down is given as

$d_1 = \frac{1}{2}gt^2$

$d_1 = \frac{1}{2}(9.81)(3^2)$

$d_1 = 44.1 m$

now total distance traveled by teacher and sound in 3 s

$d_{total} = d + d_1$

$d_{total} = 1020 + 44.1$

this total distance must be equal to twice the height of the cliff

$2H = 1064.1 m$

$H = 532 m$

7 0

3 years ago

Read 2 more answers

Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the

nydimaria [60]

Answer: $F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

$F_1=\frac{kq(-q)}{(L\sqrt{2})^2}$

where $L\sqrt{2}=$ Distance between the two charges

$F_1=-\frac{kq^2}{2L^2}$

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

$F_2=\frac{kq(-q)}{(L)^2}$

The magnitude of force by both the charge is same but at an angle of $90^{\circ}$

thus combination of two forces at 2 and 3 will be

$F'=\sqrt{2}\frac{kq^2}{2L^2}$

Now it will add with force due to 1 charge

Thus net force will be

$F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

6 0

3 years ago

A car traveling with constant speed travels 150 km in 7200 s what is speed of the car

spayn [35]

The speed of the car is exactly 150/7200 km/sec, or 125/6 meters/sec.

In more familiar units, that speed is equivalent to ...

-- (20 and 5/6) meters/sec

-- 75 km/hour

8 0

3 years ago