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3 years ago

14

A proton travels with a speed of 4.2×106 m/s at an angle of 30◦ west of north. A magnetic field of 2.5 T points to the north. Fi

nd the magnitude of the magnetic force on the proton. (The magnetic force experienced by the proton in the magnetic field is proportional to the component of the proton’s velocity that is perpendicular to the magnetic field.)

1 answer:

Arada [10]3 years ago

6 0

Answer:

$8.4\cdot 10^{-13} N$

Explanation:

The magnitude of the magnetic force on the proton is given by:

$F=qvB sin \theta$

where:

$q=1.6\cdot 10^{-19} C$ is the proton charge

$v=4.2\cdot 10^6 m/s$ is the proton velocity

$B=2.5 T$ is the magnetic field

$\theta=30^{\circ}$ is the angle between the direction of v and B

Substituting into the formula, we find

$F=(1.6\cdot 10^{-19}C)(4.2\cdot 10^6 m/s)(2.5 T) sin 30^{\circ}=8.4\cdot 10^{-13} N$

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Which has high frequency? i never figure this out

sashaice [31]

Answer: D

Explanation:

It would be answer D, youre correct. Frequency is how many "cycles" or "things" per second. The reason D looks so different is because the amplitude is not the same.

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3 years ago

Someone please help me with finding the resistance of these circuits! I've been asking for an hour now. I will give brainliest i

vivado [14]

Answer:

1. 59 Ω

2. 3 Ω

3. 0.625 kΩ

Explanation:

1. The total resistance in a series circuit is equal to the sum of the resistance.

$R_T=R_1+R_2+R_3...\\R_T=20+19+20\\R_T=59$

Therefore, the total resistance in the first circuit is 59 Ω.

2. The total resistance in a parallel circuit is equal to the sum of the reciprocals of the resistance.

$\frac{1}{R_T} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} ...\\\frac{1}{R_T} = \frac{1}{6.0} +\frac{1}{12} +\frac{1}{36}+\frac{1}{18} \\\frac{1}{R_T} = \frac{1}{3} \\R_T=3$

Therefore, the total resistance in the second circuit is 3 Ω.

3. This is another parallel circuit, so we use the same equation from above:

$\frac{1}{R_T} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} ...\\\frac{1}{R_T} = \frac{1}{10} +\frac{1}{2} +\frac{1}{1} ...\\\frac{1}{R_T} =1.6\\R_T=\frac{1}{1.6}$

Therefore, the total resistance in the third circuit is $\frac{1}{1.6}$ kΩ, or 0.625 kΩ.

I hope this helps!

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3 years ago

If the depth of water in a well is 10m, what is the pressure exerted by it the bottom of the well ? ( Use g = 10 m/s2)

Tasya [4]

Answer:

The precise answer depends on the density and therefore the temperature of the water, but we can obtain a reasonable approximation by assuming that the density of the water is 1000 kilograms per cubic meter (kg/m³).

Since the depth of the water in the well is 10 m, the volume of water directly above an area A of a square meters (m²) at the bottom of the well is 10×a m³.

Since the density of the water is 1,000 kg/m³, the mass of water directly above area A is (1,000 kg/m³) × (10×a m³) = (1000×10×a kg) = 10,000×a kg.

Since g = 9.8 m/s², the force of gravity acting on the water directly above area A is (9.8 m/s²) × (10,000×a kg) = 9.8×10,000×a N (newtons) = 98,000×a N.

So the pressure of water acting on area A is (98,000×a N)/(a m²) = (98,000×a)/a N/m² = 98,000 pascals (pa). And since A could be any given area at the bottom of the well, this is the pressure at any point at the bottom of the well.

So the pressure at the bottom of the well is 98,000 pascals (or 98,000/101,325 standard atmospheres = 560/579 atmospheres ~ 0.967 standard atmospheres).

Please comment below if you have any questions.

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3 years ago

A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t

tresset_1 [31]

Answer:

Part a)

$Reading = 2.00 kg$

Part b)

$Reading = 2.00 kg$

Part c)

$Reading = 4.04 kg$

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part b)

When elevator is decending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

$F_{net} = ma$

$T - mg = ma$

$T = mg + ma$

Reading is given as

$Reading = \frac{mg + ma}{g}$

$Reading = 2.00\frac{9.81 + 10}{9.81}$

$Reading = 4.04 kg$

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0

3 years ago

If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of

kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

<em>F</em> ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0

3 years ago