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padilas [110]
3 years ago
9

A charged particle is observed traveling in a circular path in a uniform magnetic field. If the particle had been traveling four

times as fast, the radius of the circular path would be:
Physics
1 answer:
8090 [49]3 years ago
8 0

Answer:

The new radius of the trajectory of the particle is four times the previous radius

Explanation:

In order to know what is the radius of the trajectory of the charged particle, if its speed is four times as fast, you take into account the following formula, which describes the radius of a charged particle in a magnetic field:

r=\frac{mv}{qB}          (1)

If the speed of the particle is for time as fast, that is, v' = 4v, you obtain, in the equation (1):

r'=\frac{mv'}{qB}=\frac{m(4v)}{qB}=4\frac{mv}{qB}=4r

The new radius of the trajectory of the particle is four times the previous radius

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1. one-Half

2. Apogee

3. Any object that revolves around another object

4. Venus's gravitation pull

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The earth has a radius of 6.38 × 106 m and turns on its axis once every 23.9 h.
Vladimir [108]

Answer:

a) V = 465.9 m/s

b) θ = 70.529°

Explanation:

Let's first calculate angular velocity of earth:

\omega=\frac{2\pi}{23.9h}*1h/3600s

Velocity of a person on Ecuador will be:

V_E = \omega*R

V_E = 465.9 m/s

For part b, since angular velocity is the same:

\frac{\omega*R}{3}=\omega*(R*cos\theta )

Solving for θ:

\theta=acos(1/3)

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8 0
3 years ago
An object possessing an excess of 6.0x10^6 electrons has a net charge of
guajiro [1.7K]
A single electron has a charge of
e=-1.6 \cdot 10^{19}C
Therefore, if we have an excess of N=6.0\cdot10^6 electrons, the total net charge will be the product between the charge of a single electron and the total number of electrons in excess:
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.. A 15.0-kg fish swimming at 1.10 m>s suddenly gobbles up a 4.50-kg fish that is initially stationary. Ignore any drag effec
stira [4]

Answer:

(a) 0.846 m/s

(b) 2.097J

Explanation:

Parameters given:

Mass of big fish, M = 15 kg

Mass of small fish, m = 4.5 kg

Initial speed of big fish, U = 1.1 m/s

Initial speed of small fish, u = 0 m/s (it is stationary)

(a) We apply the principle of conservation of momentum:

Total initial momentum = Total final momentum

Since both fish have the same final speed, V, (the small fish is in the mouth of the big fish), we have:

MU + mu = (M + m)*V

(15 * 1.1) + (4.5 * 0) = ( 15 + 4.5) * V

16.5 = 19.5V

=> V = 16.5/19.5

V = 0.846 m/s

The speed of the large fish after the meal is 0.846 m/s.

(b) We need to find the change in Kinetic energy of the entire system to find the total mechanical energy dissipated.

Initial Kinetic energy:

KEini = (½ * M * U²) + (½ * m * u²)

KEini = (½ * 15 * 1.1²) + (½ * 4.5 * 0²)

KEini = 9.075 J

Final Kinetic Energy:

KEfin = (½ * M * V²) + (½ * m * V²)

KEfin = (½ * 15 * 0.846²) + (½ * 4.5 * 0.846²)

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Change in kinetic energy will be:

KEfin - KEini = 9.075 - 6.978

ΔKE = 2.097 J

The energy dissipated in eating the meal is 2.097 J

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The answer is B) air.
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