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Jet001 [13]
2 years ago
9

3) 4 electrons are placed - one electron per corner - at the corners of a square of side 1 meter. One fixed proton is placed in

the middle of the square.The 4 electrons are held in place by some mechanism. The 4 electrons are released by the mechanism at the same time. They move and reach the corners of a square of side 0.8 meters, and keep on moving . Find the velocity of each electron at the corners of the square of side 0.8 meters.
Physics
1 answer:
Eduardwww [97]2 years ago
4 0

Explanation:

3

i believe that they are all going at 3.2 meters each, I did 4 times 0.8

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How far could a rabbit run if it ran 36km/h for 5.0min?
Korolek [52]

3 kilometers, it is just 5/60 or 1/12 multiplied by 36.

4 0
2 years ago
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Waves that move the particles of the medium parallel to the direction in which the waves are traveling are called
Nikolay [14]

1. a. longitudinal waves.

There are two types of waves:

- Transverse waves: in transverse waves, the oscillations of the wave occur in a direction perpendicular to the direction of propagation of the wave

- Longitudinal waves: in longitudinal waves, the oscillations of the waves occur parallel to the direction in which the waves are travelling.

So, these types of waves are called longitudinal waves.


2. d. a medium

There are two types of waves:

- Electromagnetic waves: these waves are produced by the oscillations of electric and magnetic field, and they can travel both in a medium and also in a vacuum (they do not need a medium to propagate)

- Mechanical waves: these waves are produced by the oscillations of the particles in a medium, so they need a medium to propagate - therefore, the correct choice is d. a medium


3. a. AM/FM radio

Analogue signals consist of continuous signals, which vary in a continuous range of values. On the contrary, digital signals consist of discrete signals, which can assume only some discrete values. For AM and FM radios, signals are transmitted by using analogue signals.

5 0
2 years ago
A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

5 0
2 years ago
Read 2 more answers
If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k
Paraphin [41]
Efficiency η of a Carnot engine is defined to be: 
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>

<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>

<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
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6 0
2 years ago
High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 43.7 m/s just before it strikes a 45
Helga [31]

Answer:

41.27m/s

Explanation:

According to law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the velocity after impact

Given

m1 = 0.2kg

u1 = 43.7m/s

m2 = 45.9g = 0.0459kg

u2 = 30.7m/s

Required

Velocity after impact v

Substitute the given parameters into the formula

0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v

8.74+1.409 = 0.2459v

10.149 = 0.2459v

v = 10.149/0.2459

v = 41.27m/s

Hence the speed of the golf ball immediately after impact is 41.27m/s

8 0
2 years ago
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