The unit of electric potential or electromotive force is the

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Answer 1

Answer: That would be called VOLT

Related Questions

1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of θ = 37◦ with the horizontal as shown in figure-1. The block is sliding at a constant velocity over a frictionless floor.(a) Find the value of the normal force on the block.(b) Find the magnitude of force F~2 that is acting on the block(c) Find the magnitude of force F~ 2 if the block accelerates with a magnitude of a = 2.5 m/s2 along the direction of F~ 2 .
You have devised an experiment to measure the kinetic coefficient of friction between a ramp and block. You place the block on the ramp at an angle high enough that it starts sliding. You measure the time it takes to fall down a known distance. The time it takes to fall down the ramp starting from a standstill is 0.5 sec, ???? = 1 kg, θ = 45o, and the distance it falls, L, is 0.5 m. What is µk? (8 pts)
A car travels 13 km in a southeast direction and then 16 km 40 degrees north of east. What is the car's resultant direction?
"A proton is placed in a uniform electric field of 2750 N/C. You may want to review (Page) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Electron in a uniform field. Calculate the magnitude of the electric force felt by the proton. Express your answer in newtons.(F = ? )Calculate the proton's acceleration.( a= ? m/s2 )Calculate the proton's speed after 1.40 {\rm \mu s} in the field, assuming it starts from rest.( V= ? m/s )"
For a certain transverse wave, the distance between two successive crests is 1.20 m, and eight crests pass a given point along the direction of travel every 12.0 s. calculate the wave speed.

While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: yellow light from a sodium street lamp, radio waves from an AM radio station, radio waves from an FM radio station, and microwaves from an antenna of a communications system. Rank these types of waves in terms of increasing photon energy, lowest first.

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Answer:

Explanation:

1. radio waves from am

2. radio waves from fm

3.yellow light from a sodium street lamp

4. microwaves from an antenna of a communications system.

What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingrod of radius R1 inside of a thin-walled conducting shell of radius 2(both are infinite length). Suppose the inner rod hasradiusR1= 1.3 mm and outer shell has radiusR2= 10R1Ifthe net charge density on the center rod isq1= 3.4×10−12C/mand the outer shell isq2=−2q1,a.)What is the magnitude and direction of the electric field atradial distancer= 5R1from the center rod

Answers

Answer:

E = 9.4 10⁶ N / C, The field goes from the inner cylinder to the outside

Explanation:

The best way to work this problem is with Gauss's law

Ф = E. dA = qint / ε₀

We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.

The flow on the faces is zero, since the field goes in the radial direction of the cylinders.

The area of the cylinder is the length of the circle along the length of the cable

dA = 2π dr L

A = 2π r L

They indicate that the distance at which we must calculate the field is

r = 5 R₁

r = 5 1.3

r = 6.5 mm

The radius of the outer shell is

r₂ = 10 R₁

r₂ = 10 1.3

r₂ = 13 mm

r₂ > r

When comparing these two values we see that the field must be calculated between the two housings.

Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is

λ = q / L

Qint = λ L

Let's replace

E 2π r L = λ L /ε₀

E = 1 / 2piε₀ λ / r

Let's calculate

E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3

E = 9.4 10⁶ N / C

The field goes from the inner cylinder to the outside

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1400 kg, which is required to travel upward 37 m in 3.6 min, starting and ending at rest. The elevator's counterweight has a mass of only 930 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable?

Answers

Answer:

789.8 W

Explanation:

mass of the cab = 1400 kg, the counter weight of the elevator = 930 kg

weight of the cab = 1400 × 9.81 where weight = mg and m is mass and g is acceleration due to gravity.

weight of the cab = 13734 N

counter weight of the elevator = 930 × 9.81 = 9123.3 N

the exerted force of the elevator = weight of the cab - counter weight of the elevator = 13734 - 9123.3 = 4610.7 N

Average power by the motor P = F × v = F × distance / time

where v is speed in m/s, and time is in seconds

P = 4610.7 × 37 / ( 3.6 × 60) = 789.80 W

where (3.6 × 60 ) is the time in seconds

A thin flashlight beam traveling in air strikes a glass plate at an angle of 52° with the plane of the surface of the plate. If the index of refraction of the glass is 1.4, what angle will the beam make with the normal in the glass?

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To solve this problem it is necessary to apply Snell's law and thus be able to calculate the angle of refraction.

From Snell's law we know that

$n_1sin\theta_1 = n_2 sin\theta_2$

Where,

n_i = Refractive indices of each material

$\theta_1$ = Angle of incidence

$\theta_2$ = Refraction angle

Our values are given as,

$\theta_1 = 38\°$

$n_1 = 1$

$n_2 = 1.4$

Replacing

$1*sin38 = 1.4*sin\theta_2$

Re-arrange to find $\theta_2$

$\theta_2 = sin^(-1) (sin38)/(1.4)$

$\theta_2 = 26.088°$

Therefore the angle will the beam make with the normal in the glass is 26°

You say goodbye to your friend at the intersection of two perpendicular roads. At time t=0 you drive off North at a (constant) speed v and your friend drives West at a (constant) speed ????. You badly want to know: how fast is the distance between you and your friend increasing at time t?

Answers

Answer:

$d'=√(v^2+w^2)$

Explanation:

Rate of Change

When an object moves at constant speed v, the distance traveled at time t is

$x=v.t$

We know at time t=0 two friends are at the intersection of two perpendicular roads. One of them goes north at speed v and the other goes west at constant speed w (assumed). Since both directions are perpendicular, the distances make a right triangle. The vertical distance is

$y=v.t$

and the horizontal distance is

$x=w.t$

The distance between both friends is computed as the hypotenuse of the triangle

$d^2=x^2+y^2$

We need to find d', the rate of change of the distance between both friends.

Plugging in the above relations

$d^2=(v.t)^2+(w.t)^2$

$d^2=v^2.t^2+w^2.t^2=(v^2+w^2)t^2$

Solving for d

$d=√((v^2+w^2)t^2)$

$d=√((v^2+w^2)).t$

Differentiating with respect to t

$\boxed{d'=√(v^2+w^2)}$

Final answer:

The problem is solved using Pythagoras' Theorem, representing the two travel paths forming a right triangle. The rate at which the distance increases between two points moving perpendicularly can be found by differentiating the resulting equation, which yields the expression sqrt[(v^2)+(u^2)].

Explanation:

The question is about the rate at which the distance between you and your friend is increasing at time t. It's a typical problem in kinematics. Because the roads are perpendicular to each other, we can solve the problem using Pythagoras' Theorem which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Let's denote the distance you've traveled as D1 = v*t (because distance = speed * time) and the distance your friend has travelled D2 = u*t. The distance between you can be computed using Pythagoras' Theorem as D = sqrt(D1^2 + D2^2). Hence, D = sqrt[(v*t)^2 + (u*t)^2]. Differentiating D with respect to t using the chain rule will give us the rate at which the distance between you is increasing, which is sqrt[(v^2)+(u^2)].

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Which of the following describes the net force acting on an object?The sum of all forces acting on an object
The gravitational force minus any contact forces acting on an object
The difference between the normal force and the gravitational force acting on an object
The sum of all the forces acting on an object in the same direction

Answers

The sum of all forces acting on an object in the same direction is described for the net force acting on an object.

What is a Net force?

When the forces are acting in the same direction of movement of the object it can be said as sum of the two individual forces will be equal to the "Net Force" .

The net force is the combined force of all individual forces acting on an object.
If the object with the forces in the opposite direction, then the net force will not be equal to the sum of the forces.

Example : If two forces (2 kids pushing in the same direction to move the object big box) act on an object (big box) in the same direction, then the net force is equal to the sum of the two forces. If the kids pushed in the opposite direction, the net force will not occur.

Hence, Option D is the correct answer.

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Answer:

The sum of all the forces acting on an object in the same direction.

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