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Phantasy [73]
1 year ago
11

A laboratory electromagnet produces a magnetic field of magnitude 1.50T . A proton moves through this field with a speed of 6.00

×10⁶ m/s . (a) Find the magnitude of the maximum magnetic force that could be exerted on the proton. Explain.
Physics
1 answer:
hram777 [196]1 year ago
3 0

The magnitude of maximum magnetic force that could be exerted on the proton is 1.44 x 10^-12 N.

The magnitude of the force on a charged particle moving in a magnetic field is given by the formula,

F= qvB

Here q is the charge on proton = 1.6 x 10^-19 C.

v is the velocity with which the particle is moving = 6.00 x 10^6 m/s

And B is the value of the magnetic field = 1.5 T

Putting the given values in the above equation,

F = 1.6 x 10^-19 x 6 x 10^6 x 1.5 = 1.44 x 10^-12 N.

Hence, the magnitude of maximum magnetic force that could be exerted on the proton is 1.44 x 10^-12 N.

To know more about "magnetic force", refer to the link given below:

brainly.com/question/13791875?referrer=searchResults

#SPJ4

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How do you think the amount of a material affects its tendency to sink or float?
Viktor [21]

Answer:

Things float when they are positively buoyant, or less dense than the fluid in which they are sitting. This does not mean that an object has to be lighter than the fluid, as in the case of a boat; objects just need to have a greater ratio of empty space to mass than the fluid.

Explanation:

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6 0
3 years ago
Need answer fast plz
vazorg [7]

Answer:

a = 1.20m\s^{2}

Explanation:

225 x 9.8

= 441N

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710 - 441

= 269N

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\frac{269}{225} = 1.19

a = 1.20

8 0
2 years ago
At takeoff, an aircraft travels at 62 m/s, so that the air speed relative to the bottom of the wing is 62 m/s. Given the sea lev
olganol [36]

Answer:

the aircraft must travel at a speed of <em>73.4 m/s</em> in order to create the ideal lift.

Explanation:

We will use Bernoulli's theorem in order to determine the pressure lift:

ΔP = 1/2 (ρ)(v₂² - v₁²)

the generated pressure lift is ΔP = 1000 N/m²

Therefore,

1000 = 1/2(ρ)(v₂² - v₁²)

v₂² - v₁² = 2000 / ρ

v₂² = (2000 N/m² / 1.29 kg/m³) + (62 m/s)²

v₂ = √[ (2000 N/m² / 1.29 kg/m³) + (62 m/s)² ]

<em>v₂ = 73.4 m/s </em>

<em></em>

Therefore, the aircraft must travel at a speed of <em>73.4 m/s</em> in order to create the ideal lift.

5 0
3 years ago
A __________ provides a visual summary of the improvements that a data mining project provides on a binary classification proble
babymother [125]

Answer:

Option (B)

Explanation:

A lift chart usually refers to a graphical representation that is mainly used in order to improve the drawbacks of a mining model by making a comparison with any random guess, and also helps in determining the changes that occur in terms of lift scores.

It describes the binary classification of the problems associated with the mining activity. This type of chart is commonly used to differentiate the lift scores for a variety of models, and picking the best one out of all.

Thus, the correct answer is option (B).

5 0
4 years ago
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar
defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency \zeta = 3%

where;

\zeta = \dfrac{W_{out}}{Q_{supplied }}

Q_{supplied } = \dfrac{2}{0.03} \ MW

Q_{supplied } = 66.66 \ MW

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}

Also;

\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K

\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K

LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}

LMTD = \dfrac{8}{In (5)}

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

Q_H = UA (LMTD)

where;

U = overall heat coefficient given as 1200 W/m².K

66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\  A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\  \mathbf{A = 11178.236 \ m^2}

The mass flow rate:

Q_{H} = mC_p(T_{in} -T_{out} )  \\ \\  66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{  66.667 \times 10^6}{4.18 \times 8} \\ \\  \mathbf{m = 1993630.383 \ kg/s}

3 0
3 years ago
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