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makvit [3.9K]
3 years ago
6

Suppose you move along a wire at the same speed as the drift speed of the electrons in the wire. Do you now measure a magnetic f

ield of zero?
Physics
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

False. Field is non-zero

Explanation:

If you were moving along with the electrons, they would appear stationary to you. You would measure a current of zero. However, the fixed positive charges in the wire seem to move backwards relative to you, creating the equivalent current as if you weren't moving. You would measure the same field, but the field would be caused by the 'backward' motion of positive particles.

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The visible spectrum refers to the portion of the electromagnetic spectrum that we ________.
yKpoI14uk [10]
<h2>Answer: can see</h2>

Explanation:

The portion visible by the human eye of the electromagnetic spectrum is between 380 nm (violet-blue) and 780 nm (red) approximately.  Which  means this part of the spectrum is located between ultraviolet light and infrared light.  

Note the fact only part of the whole electromagnetic spectrum is visible to humans is because the receptors in our eyes are only sensitive to these wavelengths.

Therefore:

<h2>The visible spectrum refers to the portion of the electromagnetic spectrum that <u>we </u><u>can see</u></h2>
8 0
3 years ago
According to the law of mass, 8 grams of hydrogen and 32 grams of oxygen will combine to form how many grams of water?
seropon [69]

Answer:

36 grams of water

Explanation:

mass of H2O = 0.5 x 18 = 9g

7 0
2 years ago
Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject
Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}

where

\mu_{sun} is gravitational parameter of sun =  1.32712 x 10^20 m^3 s^2.t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}

t = 12,105.96 sec

6 0
3 years ago
A bicyclist travels 4.5 km west, then travels 6.7 km at an angle 27.0 degrees South of West. What is the magnitude of the bicycl
Dimas [21]

Answer:

<em>10.90km</em>

Explanation:

Magnitude of the total displacement is expressed using the equation

d = √dx²+dy²

dx is the horizontal component of the displacement

dy is the vertical component of the displacement

dy = -6.7sin27°

dy = -6.7(0.4539)

dy = -3.042

For the  horizontal component of the displacement

dx = -4.5 - 6.7cos27

dx = -4.5 -5.9697

dx = -10.4697

Get the magnitude of the bicyclist's total displacement

Recall that: d = √dx²+dy²

d = √(-3.042)²+(-10.4697)²

d = √9.2538+109.6146

d = √118.8684

<em>d = 10.90km</em>

<em>Hence the magnitude of the bicyclist's total displacement is 10.90km</em>

<em></em>

6 0
3 years ago
What is the magnitude of the gravitational force acting on the earth due to the sun?
expeople1 [14]

Answer: 3.524(10)^{22}N

Explanation:

According to Newton's law of Gravitation, the force F exerted between two bodies of masses m1 and m2  and separated by a distance r  is equal to the product of their masses and inversely proportional to the square of the distance:

F=G\frac{(m1)(m2)}{r^2}   (1)

Where:

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}  

m1=1.99(10)^{30}kg is the mass of the Sun

m2=5.972(10)^{24}kg is the mass of the Earth

r=1.50(10)^{11}m  is the distance between the Sun and the Earth

Substituting the values in (1):

F=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.99(10)^{30}kg)(5.972(10)^{24}kg)}{(1.50(10)^{11}m)^2}   (2)

Finally:

F=3.524(10)^{22}N   This is the gravitational force acting on the earth due to the sun

3 0
3 years ago
Read 2 more answers
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