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Setler79 [48]
2 years ago
10

the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic f

ield for that wave at P
Physics
1 answer:
vesna_86 [32]2 years ago
7 0

Answer:

6.63\times 10^8\ N/C

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

\dfrac{E}{B}=c\\\\E=Bc

Put all the values,

E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C

So, the magnitude of the electric field is equal to 6.63\times 10^8\ N/C.

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A piece of wood has a mass of 25g and a volume of <br> 10cm3 What is its density?
Tasya [4]

Answer:

<h3>The answer is 2.5 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question we have

density =  \frac{25}{10}  =  \frac{5}{2}  \\

We have the final answer as

<h3>2.5 g/cm³</h3>

Hope this helps you

3 0
3 years ago
A building is 512 m high. What must be the minimum water pressure in a pipe at ground level in order to get water out of a tap i
vivado [14]

Answer:

Pressure =  5 x 10⁶ Pa

Explanation:

Given:

Height of building = 512 m

Find:

Pressure

Computation:

P2 = P1+dgh

P2 = 1 + (1000)(9.8)(512)

P2 = 51.2 atm

Pressure =  5 x 10⁶ Pa

6 0
3 years ago
Guys No one's answering my question so sad! Once again I'm asking the same question –Here
nexus9112 [7]

For the front glass of the car to get wet, V_c \geq 10 \ m/s.

The given parameters:

  • <em>Speed of the car, = Vc</em>
  • <em>Speed of the rain, = 10 m/s</em>

The relative velocity of the car with respect to the falling rain is calculated as;

V_{C/R} = V_C- V_R

  • If the speed of the car equals the speed of the rain, the rain will fall behind the car.
  • If the speed of the rain is greater than speed of the car, the rain will fall far in front of the car.
  • If the speed of the car is greater than speed of the rain, the rain will fall on the car.

Thus, for the front glass of the car to get wet, V_c \geq 10 \ m/s.

Learn more about relative velocity here: brainly.com/question/17228388

8 0
2 years ago
A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the
liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0
3 years ago
A copper sphere was moving at 25 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f
JulsSmile [24]
Increased temperature is 0.81 c
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