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MrRa [10]
3 years ago
14

The friends take a few minutes to discuss their ideas. Which of Amy's comments with respect to a charged particle moving in a ma

gnetic field is false? "The magnitude FB of the magnetic force exerted on the particle is proportional to the magnitude of the charge q and to the speed v of the particle." "The magnetic force acting on a charged particle moving in a magnetic field is perpendicular to the particle's velocity." "If the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line." "The work done by the magnetic force on the particle is zero."
Physics
1 answer:
Marizza181 [45]3 years ago
7 0

Answer:

The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false.

Explanation:

The equation for the magnetic force on a charge q moving at velocity v on a magnetic field B is given by the (vectorial) Lorentz Force Law F=qv\times B

From it we can clearly see that the <em>magnitude of the magnetic force </em>exerted on the particle is <em>proportional to the magnitude of the charge q and to the speed v of the particle</em>, and that it is also <em>perpendicular to the particle's velocity</em>. This means that at each instant it moves perpendicularly to the force, so <em>the work done by the magnetic force on the particle is zero</em>.

The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false not only for this but for any force, a force always perpendicular to a velocity will curve the trajectory.

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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
A speeding car traveling at 24.8 m/s passes a police car that is at rest. The police car begins its pursuit at the instant the s
marusya05 [52]

Answer:

the acceleration required is 1.37m/s^2

Explanation:

The car is having a constant velocity movement, so if we calculate the time to reach 897m, we can use it to find the acceleration the policeman need to apply to reach the car.

x=v*t\\t=\frac{x}{v}\\t=\frac{897m}{24.8m/s}\\t=36.17s

the policeman is traveling with a constant acceleration starting from rest so:

x=\frac{1}{2}*a*t^2\\\\a=\frac{2*x}{t^2}\\\\a=1.37 m/s2

7 0
3 years ago
2. A solid plastic cube of side 0.2 m is submerged in a liquid of density 0.8 hgm calculate the
kotegsom [21]

Answer:

vpg = 0.064 N

Explanation:

Upthrust = Volume of fluid displaced

upthrust liquid on the cube g=10ms−2

vpg =0.2 x 0.2 x 0.2 x0.8 x 10= 0.064N

vpg = 0.064 N

hope it helps.

3 0
3 years ago
The free body diagram shows a box being pulled to the left up a 25-degree incline
Greeley [361]
The question is incomplete but still I answer to assume your thinking.
The picture is attached below!.
Here,
F is the force with which you pull up the incline.
N is the normal force.
w is the weight acting downward.
Axis are mentioned in the attached picture.
Concept:
You can see there is no movement of object in the y-direction that means acceleration is zero in y-direction, sum of all the forces in y-direction equal to zero.
According to newton second law,
<span>∑ F = ma
</span>As, acceleration is zero in y-direction, so right hand side is zero in the above equation.
<span>∑ F = 0</span>
N-wcosθ=0 
N= m*g*cos25°
N= m*(9.8)*(0.9063)
N= 8.8817*m
By putting the value of mass(m)(not given in the question) you will get the answer.
Hopefully, this is the answer of your question.


5 0
4 years ago
Read 2 more answers
Can someone please help me
Mars2501 [29]
The correct answer is letter b.

To find the answer follow the following steps.
1. 6524.96 x .25 = X

2. 1631.24 = X

This works for all of the given answers to find the correct answer.
5 0
3 years ago
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