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bearhunter [10]
3 years ago
12

HELP PLEASE !! how are waves, engery, and matter related??

Physics
1 answer:
Makovka662 [10]3 years ago
7 0

Waves transport energy along a medium without transporting matter. The amount of energy carried by a wave is related to the amplitude of the wave. Thus, the higher the wave is from the resting line, the more energy is put in and vice-versa.

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A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±
gladu [14]

Answer:

E(final)/E(initial)=2

Explanation:

Applying the law of gauss to two parallel plates with  charge density equal σ:

E=\sigma/\epsilon_{o}=Q/(L^{2}*\epsilon_{o})\\

So, if the charge is doubled the Electric field is doubled too

E(final)/E(initial)=2

3 0
2 years ago
Two boxers are fighting. Boxer 1 throws his 5 kg fist at boxer 2 with a speed of 9 m/s.
Sladkaya [172]

Answer:

0.001 s

Explanation:

The force applied on an object is equal to the rate of change of momentum of the object:

F=\frac{\Delta p}{\Delta t}

where

F is the force applied

\Delta p is the change in momentum

\Delta t is the time interval

The change in momentum can be written as

\Delta p=m(v-u)

where

m is the mass

v is the final velocity

u is the initial velocity

So the original equation can be written as

F=\frac{m(v-u)}{\Delta t}

In this problem:

m = 5 kg is the mass of the fist

u = 9 m/s is the initial velocity

v = 0 is the final velocity

F = -45,000 N is the force applied (negative because its direction is opposite to the motion)

Therefore, we can re-arrange the equation to solve for the time:

\Delta t=\frac{m(v-u)}{F}=\frac{(5)(0-9)}{-45,000}=0.001 s

4 0
3 years ago
What three steps can you take to ensure your measurments are both accurate and precise?
Ivan

First, use a high-quality measurement tool. Next, measure carefully. Finally, repeat the measurement a few times. Hope it helps!

8 0
2 years ago
Phosphate never enters the<br> O atmosphere<br> O ground<br> O water<br> O ocean
gizmo_the_mogwai [7]
I think you can google this because I really don’t know the answer I’m so sorry
7 0
3 years ago
A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

KE=\frac{kQq}{r}\     \ \ \ \ \ \ ...i

k is Coulomb's constant.

#The electric field,E at radius r is expressed as:

E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii

From i and ii, we have:

KE=Eqr

r=(KE)/Eq

#Substitute actual values in our equation:

r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0
2 years ago
Read 2 more answers
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