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makkiz [27]
3 years ago
5

At a location near the equator, the earth’s magnetic field is horizontal and points north. An electron is moving vertically upwa

rd from the ground. What is the direction of the magnetic force that acts on the electron?1.North
2.West
3.The magnetic force is zero.
4.East
5.South
Physics
1 answer:
Elan Coil [88]3 years ago
5 0

Answer:

2. west

Explanation:

Given an electron is moving vertically upward from ground.

Now Fleming right hand rule state that: make L shape with thumb and index finger then point middle finger perpendicular to index and thumb.Then index finger points in the direction of moving charge , middle finger points in the direction of the magnetic field and thumb points in the direction of the magnetic force.

According to Fleming right hand rule the direction of the magnetic that acts on the electron is west.

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What are beats
N76 [4]

Beats can be defined as the periodic fluctuations in the frequency of sound waves. That is option D

<h3>What are sound waves?</h3>

Sound waves are those waves that are produced by the vibration of an object whose energy is usually propagated through a medium.

When two sound waves of different frequencies meets, a periodic variation that occurs is called beats.

Therefore, Beats can be defined as the periodic fluctuations in the frequency of sound waves.

Learn more about waves here:

brainly.com/question/15663649

#SPJ1

4 0
2 years ago
The density of air on the Earth decreases almost linearly with height from 1.22kgm–3 at sea level
Paraphin [41]
P = 10^5 - ρ (average)gh

where ρ (average) = average air density between 0 and 5000 m = (1.22+0.74)/2 = 0.98

we know g & h

I get P = 52470 Pa
3 0
3 years ago
A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle res
garri49 [273]

Answer:

ω = 22.36 Hz

f = 3.56 Hz

T = 0.28 s.      

Explanation:

a) The angular frequency (ω), can be calculated using the following equation:

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{F}{x*m}}  

<u>Where:</u>

<em>k: is the spring constant = F/x</em>

<em>m: is the mass of the particle = 0.500 kg</em>

<em>F: is the force applied = 7.50 N       </em>

<em>x: is the displacement = 3.00 cm = 0.03 m </em>            

\omega = \sqrt{\frac{7.50 N}{0.03 m*0.500 kg}} = 22.36 s^{-1} = 22.36 Hz    

Therefore, the angular frequency of the motion is 22.36 Hz.

b) To find the frequency (f) we can use the next equation:

f = \frac{\omega}{2 \pi} = \frac{22.36 Hz}{2 \pi} = 3.56 Hz

Hence, the frequency of the motion is 3.56 Hz.

c) The period (T) is equal to:

T = \frac{1}{f} = \frac{1}{3.56 Hz} = 0.28 s

Therefore, the period of the motion is 0.28 s.

I hope it helps you!

4 0
3 years ago
Which of these would be a good model to demonstrate the fact that the galaxies are moving apart as the universe is expanding?
Sergio [31]

Answer: Option A: The spots on the balloon move away as the balloon is inflated.

one of the scientific models describes the galaxies are moving apart as the universe is expanding. This expansion theory came from the observation of red-shifted spectrum from all the directions indicating that the galaxies are moving away.  This can be understood from the inflated balloon. Initially spots can be marked using a colored pen on the balloon.On inflating the balloon, it would be noticed that the spots on the balloon move away. Actually the position of the spots won't change, but the distance between the spots would expand. This is a good model to explain the expanding universe. The galaxies are neither moving in any random direction nor moving forward. So, rest of the options are not good models to explain the theory.

3 0
3 years ago
Three children are riding on the edge of a merry‑go‑round that has a mass of 105 kg and a radius of 1.40 m . The merry‑go‑round
Irina18 [472]

Answer:

new angular velocity  is 25.20 rpm

Explanation:

Given data

mass = 105 kg

radius =  1.40 m

spinning = 20.0 rpm

masses = 22.0 , 28.0 and 33.0 kg

to find out

new angular velocity

solution

we apply here conservation momentum

L initial  = L final   .........1

we know Ip = 1/2 mr² = 1/2 × 105 ×1.4² = 102.9 kg/m³

and for children initial I(i) = ( 33 + 28 + 22 )=   83 × 1.4² = 162.68 kg/m²

and  I(f) children final = 33+ 22 = 55 × 1.4² = 107.80 kg/m²

so as that from equation 1

Ip  + I(i) × 20 = Ip ×  I(f)  × ω

(102.9  + 162.68 ) × 20 = (102.9 +  107.80 ) × ω

ω =  25.209302

so  new angular velocity  is 25.20 rpm

3 0
3 years ago
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