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motikmotik
4 years ago
11

A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86

times 10^6 m/s.
a. Find the magnitude of the maximum magnetic force that could be exerted on the proton.
b. What is the magnitude of the maximum acceleration of the proton?
c. Would the field exert the same magnetic force on an electron moving through the field with the same speed? (Assume that the electron is moving in the direction as the proton.)

1. Yes
2. No
Physics
1 answer:
Vladimir [108]4 years ago
3 0

.Answer;

Using Fmax=qVB

F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)

ANS=1.29*10^-12 N

2. Using Amax=Fmax/ m

Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)

ANS=1.93*10^15 m/s^2*

3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton

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Answer:

1. 1800 W

2. $ 17.3

Explanation:

From the question given above, the following data were obtained:

Current (I) = 15 A

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Duration = 31 days

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1. Determination of the power.

Current (I) = 15 A

Voltage (V) = 120 V

Power (P) =?

P = IV

P = 15 × 120

P = 1800 W

Thus, 1800 W of power is required.

2. Determination of the cost per month (31 days).

We'll begin by converting 1800 W to KW.

1000 W = 1 KW

Therefore,

1800 W = 1800 W × 1 KW / 1000 W

1800 W = 1.8 KW

Next, we shall determine the energy consumption for 31 days. This can be obtained as follow:

Power (P) = 1.8 KW

Time (t) = 2 h per day

Time (t) for 31 days = 2 × 31 = 62 h

Energy (E) =?

E = Pt

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Finally, we shall determine the cost of consumption. This can be obtained as follow:

1 KWh = 15.5 cents

Therefore,

111.6 KWh = 111.6 KWh × 15.5 cents / 1 KWh

111.6 KWh = 1729.8 cents

Converting 1729.8 cents to dollar, we have:

100 cents = $ 1

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1729.8 cents = $ 17.3

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4 0
3 years ago
The pilot directs the aircraft to fly due north at 600km/h. A side-wind blows at
loris [4]

Answer:

678.2 km/h and 80.54° north of east

Explanation:

From the question,

Using pythagoras theorem,

a² = b²+c²..................... Equation 1

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Substitute these values into equation 1

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R² = 360000+10000

R² = 460000

R  = √460000

R = 678.2 km/h.

And the direction is

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tanθ = 6

tanθ = 6

θ = tan⁻¹(6)

θ = 80.54°.

Hence the resultant velocity of the aircraft is 678.2 km/h and 80.54° north of east

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Answer:

there it is fella

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