All categories

3 years ago

A girl kicks a blue ball with a velocity of 20.0 m/s at 65.0o. How long is it in the air?

Physics

1 answer:

34kurt3 years ago

8 0

Explanation:

t = usin©/g

Where t is the time to reach the maximum height

Time spent in air is T = 2t

Hence, T = 2usin©/g

T = 2 x 20 x sin 65°/ 9.8

T = 3.69s

You might be interested in

Fifty points please help?

Zina [86]

it would be C laminated soda lime glass

6 0

3 years ago

Total charge is uniformly distributed on a spherical surface of radius R. The sphere is centered at the origin and spins around

Flauer [41]

Answer:

Explanation:

i have a A in physics

5 0

3 years ago

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

3 years ago

A 34.0 kg sled is accelerated by a constant force of 45.0 N from 2.00 m/s to 16.00 m/s. For how long was the force acting on the

goblinko [34]

Answer: 10.6 sec

Explanation:

Because I got it right on my quiz :D

also because you can use the impulse momentum formula, Ft=m(triangle)v

so basically u do 16m/s-2m/s=14m/s and thats your triangle v (change in velocity) then multiply 14 times the mass, which is 34. Thats 476, so now you have ft=476, and you know F, the force, so all you have to do is divide 476 by 45, and you get like 10.5777777 which rounds up to 10.6!

4 0

3 years ago

For an RLC circuit in the limit of very high frequency, what is the effective behavior of the capacitor and the inductor in the

Aliun [14]

Answer:

Explanation:

Given

In R-L-C circuit frequency is very high

Inductive Reactance for inductor is given by $X_L=2\pi f\cdot L$