What happens when electromagnetic waves cause a disturbance in electric

What is the kinetic energy of a 0.50 kg ball traveling at 120 m/s

MatroZZZ [7]

Answer:

3600 J

Explanation:

Kinetic energy of an object is the energy posses by an object solely due to its motion.

It can also be defined as the amount of work that can be done when an moving object come to test.

Kinetic energy of an object can be calculated by the equation,

K.E. = ½mv²

Where m = mass, v = velocity

So you get,

K.E. = ½×0.5×(120)² = 3600 J

8 0

3 years ago

You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.

dem82 [27]

Angle (θ) = 60°
Force (F) = 20 N
Distance (s) = 200 m
Therefore, work done
= Fs Cos θ
= (20 × 200 × Cos 60°) J
= (20 × 200 × 1/2) J
= (20 × 100) J
= 2000 J

Answer:

2000 J

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0

2 years ago

Read 2 more answers

Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear

Sidana [21]

Answer:

The amplitude of the eardrum's oscillation is 6.65×10^-13 m.

Explanation:

Given data:

The sound has a frequency of 262 Hz

The sound level is 84 dB

The air density is 1.21 kg/m^3

The speed of sound is 346 m/s

Solution:

As, Intensity of sound is given by,

I = Io×10^(s/10 db)

I = 2×π^2×ρ×v×f^2×Sm^2

Thus,

Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)

Now, put the values,

Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )

Sm = √(2.51×10^-4 / 5.66×10^8)

Sm = √0.443×10^-12

Sm = 6.65×10^-13 m.

8 0

3 years ago

Can some one please help!!!!!! ASAP pleaseeeeeeee❗️❗️❗️❗️❗️❗️❗️❗️

Thepotemich [5.8K]

a) The launch velocity of the rocket is 5.48 m/s

b) The maximum height is 1.53 m

Explanation:

a)

We can solve this part by applying the law of conservation of energy, by considering the kinetic energy and the elastic potential energy only, since there is no change in gravitational potential energy and no friction is involved.

The total energy when the spring is compressed is:

$E=KE_i + PE_{si}$

with

$KE_i = 0$ (initial kinetic energy is zero)

$PE_{si} = \frac{1}{2}kx^2$ is the elastic potential energy stored in the spring, with

k = 450 N/m (spring constant)

x = 0.10 m (compression of the spring)

The total energy when the spring is relased is:

$E=KE_f + PE_{sf}$

with

$KE_f = \frac{1}{2}mv^2$ (final kinetic energy), with

m = 0.15 kg (mass of the rocket)

v = velocity of launch of the rocket

$PE_{sf} = 0$ (elastic potential energy is zero when the spring is released)

Combining the two equations we get

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

And solving for v,

$v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(450)(0.10)^2}{0.15}}=5.48 m/s$

b)

In this part instead we consider only the kinetic energy and the gravitational potential energy, since the spring is at rest so its energy is now zero.

The total energy at the launch is:

$E=KE_i + PE_{gi}$

where

$KE_i = \frac{1}{2}mv^2$ (initial kinetic energy), with

m = 0.15 kg (mass of the rocket)

v = 5.48 m/s (velocity of launch of the rocket)

$PE_{gi}=0$ (initial gravitational potential energy is zero)

The total energy at the point of maximum height is:

$E=KE_f + PE_{gf}$

where

$KE_f = 0$ (kinetic energy is zero since speed is zero)

$PE_{gf}=mgh$ (final gravitational potential energy), with

m = 0.15 kg

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

h = ? (maximum height)

Combining the two equations we find

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

And solving for h,

$h=\frac{v^2}{2g}=\frac{(5.48)^2}{2(9.8)}=1.53 m$

Learn more about potential energy and kinetic energy:

brainly.com/question/1198647

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#LearnwithBrainly

3 0

3 years ago

A gold nucleus is 466 fm (1 fm = 10-15 m) from a proton, which initially is at rest. When the proton is released, it speeds away

Artist 52 [7]

Answer:

$v=6.8\times10^6m/s$

Explanation:

The sum of the kinetic and electric potential energies of the proton when initially released must be equal to their sum at infinity, so we have:

$K_i+U_i=K_f+U_f$

Which, since $K_i=0J$ because initially the proton is at rest, is:

$\frac{kqQ}{d}=\frac{mv^2}{2}+\frac{kqQ}{r_\infty}$

where $k=9\times10^9Nm^2/C^2$ is Coulomb's constant, $q=1.6\times10^{-19}C$ the charge of the proton, $Q=(79)(1.6\times10^{-19})C$ the charge of the gold nucleus, since it has 79 protons, $d=466\times10^{-15}m$ the initial separation between them, $m=1.67\times10^{-27}kg$ the mass of the proton and v its final velocity. $r_\infty$ is very far away, so the final electric potential will be 0J, and we have:

$v=\sqrt{\frac{2kqQ}{md}}=\sqrt{\frac{2(9\times10^9)(1.6\times10^{-19})^2(79)}{(1.67\times10^{-27})(466\times10^{-15})}}m/s=6839409m/s$

8 0

3 years ago