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Dmitry_Shevchenko [17]
3 years ago
5

An alpha particle(the nucleus of a helium atom) has a mass of 6.64*10^-27kg and a charge of +2e. What are the magnitude and dire

ction of the electric field that will balance the gravitational force on it? Mp= 1.67*10^-27kg
Physics
2 answers:
emmasim [6.3K]3 years ago
8 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Weight = electric force 
<span>mg = qE </span>

<span>6.64x10^-27 x 9.81 = (2 x 1.60x10^-19) E 
</span>qE =mg, 
<span>E = mg/q = 6.64•10^-27•9.8/2•1.6•10^-19 =2.03•10^-7 V/m</span>
Rainbow [258]3 years ago
3 0

Answer:

Magnitude of electric field E= 2.03×10^-7NC^-1

The charge is positive,the electric force is in the same direction as the electric field according to F= qE

Therefore,the direction of the electric field is upward.

Explanation:

Fg=Fe

Where Fe is the force of the electric field

Fg is force of the gravitational field.

Mg= we

Mg= 2eE

E=mg/2e

E= (6.64×10^-27)(9.8)/2(1.6×10^-19)

E=( 6.50×10^-26)/(3.2×10^-19)

E= 2.03×10^-7NC^-1

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I'm pretty sure what you are trying to ask for is radiative energy, light energy, and electronic energy.
Radiative since the microwave is releasing radiation,
Light since there is light inside the microwave,
Electronic since it is plugged in and uses electricity.
You can also use sound, but I don't think every microwave makes sound. 
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3 years ago
Which is a sub-atomic particle?
Natalka [10]
A particle that is smaller than an atom or a cluster of particles.
4 0
3 years ago
You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation
Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, \omega = 25.13 rad/s

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = \frac{1}{T}

f = \frac{1}{0.25}

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

\omega = 2\pi \times f

\omega = 2\pi \times 4

\omega = 25.13 rad/s

5 0
3 years ago
As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.33 m l
deff fn [24]

Answer: 62 μT

Explanation:

Given

Length of rod, l = 1.33 m

Velocity of rod, v = 3.19 m/s

Induced emf, e = 0.263*10^-3 V

Using Faraday's law, the induced emf of a rod can be gotten by the formula

e = blv where,

e = induced emf of the rod

b = magnetic field of the rod

l = length of the rod

v = velocity of the rod. On substituting, we have

0.263*10^-3 = b * 1.33 * 3.19

0.263*10^-3 = b * 4.2427

b = 0.263*10^-3 / 4.2427

b = 0.0000620 T

b = 62 μT

Thus, the strength of the magnetic field is 62 μT

8 0
3 years ago
Read 2 more answers
How far can a sound wave travel in 90 seconds when the ambient air temperature is 10 C?
Ksju [112]

Answer:

s = 30330.7 m = 30.33 km

Explanation:

First we need to calculate the speed of sound at the given temperature. For this purpose we use the following formula:

v = v₀√[T/273 k]

where,

v = speed of sound at given temperature = ?

v₀ = speed of sound at 0°C = 331 m/s

T = Given Temperature = 10°C + 273 = 283 k

Therefore,

v = (331 m/s)√[283 k/273 k]

v = 337 m/s

Now, we use the following formula to calculate the distance traveled  by sound:

s = vt

where,

s = distance traveled = ?

t = time taken = 90 s

Therefore,

s = (337 m/s)(90 s)

<u>s = 30330.7 m = 30.33 km</u>

6 0
3 years ago
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