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allsm [11]
4 years ago
8

The amplitude of the electric field for a certain type of electromagnetic wave is 570 N/C. What is the amplitude of the magnetic

field for that wave? (c = 3.00 × 108 m/s)
Physics
1 answer:
vesna_86 [32]4 years ago
8 0

Given Information:  

Amplitude of the electric field = E =  570 N/C

Speed of light = c = 3.0x10⁸ m/s

Required Information:  

Amplitude of the magnetic field = ?

Answer:

Amplitude of the magnetic field =  1.90x10⁻⁶ T

Explanation:

The relation between the amplitude of the electric field and magnetic field is given by

B = E/c

Where c is the speed of light

B = 570/3.0x10⁸

B = 1.90x10⁻⁶ T

Therefore, the amplitude of magnetic field is 1.90x10⁻⁶ T

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Radiation heat transfer can be due to what?
riadik2000 [5.3K]

Answer: Heat transfer due to emission of electromagnetic waves is known as thermal radiation.

Explanation: Heat transfer through radiation takes place in form of electromagnetic waves mainly in the infrared region. Radiation emitted by a body is a consequence of thermal agitation of its composing molecules.

4 0
3 years ago
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A rocket lifts off the pad at cape canaveral. according to newton's law of gravitation, the force of gravity on the rocket is gi
Alenkinab [10]
The equation is the Law of Universal Gravitation. The gravitational constant G is equal to 6.67×10⁻¹¹ Nm²/kg². The mass of the Earth is <span>5.972 ×10</span>²⁴ kg. Compared to the mass of the Earth, the mass of the rocket is negligible. So, we don't need to know the mass of the rocket. Substituting the values:

F = (6.67×10⁻¹¹ Nm²/kg²)(5.972 ×10²⁴ kg)/(4000 miles*(1.609 km/1mile))²
F = 9616423.08 N

The work is equal to
W = Fd
W = (9616423.08 N)(2000 miles*1.609 km/mile)
W = 9.095×10¹⁰ Joules
8 0
4 years ago
In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a
SOVA2 [1]

Answer:0.084 m/s^2

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance d_1=8.13 km

time taken=25 min =1500 s

initial speed v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s

after 8.13 km mark steve started to accelerate

speed after 60 s

v_2=v_1+at

v_2=5.6+a\times 60

distance traveled in 60 sec

d_2=v_1\times 60+\frac{a60^2}{2}

d_2=336+1800 a

time taken in last part of journey

t_3=1663.6-1560=103.6 s

distance traveled in this time

d_3=v_2\times t_3

d_3=\left ( 5.6+a\times 60\right )103.6

and total distance=d_1+d_2+d_3

10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6

1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6

a=0.084 m/s^2

5 0
4 years ago
A car (mass = 1090 kg) is traveling at 30.4 m/s when it collides head-on with a sport utility vehicle (mass = 2880 kg) traveling
Thepotemich [5.8K]

Answer:

The sport utility vehicle was traveling at V2= 11.5 m/s.

Explanation:

m1= 1090 kg

V1= 30.4 m/s

m2= 2880 kg

V2= ?

m1*V1 = m2*V2

V2= (m1*V1)/m2

V2= 11.5 m/s

7 0
4 years ago
Find<br>the equevalent unit<br>for Ohm using<br>the<br>geren<br>formula.<br>늘​
Brut [27]

Answer:

V/A

Explanation:

Resistance equals voltage divided by current. You already know resistance's unit is the Ohm. Voltage's unit is the Volt and current's is ths Ampere. Now you can do something like this:

1Ω=\frac{1V}{1A} ⇔

Ω=V/A

Hope this helped :)

7 0
3 years ago
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