Physics High School

a potential difference of 0.020 v is developed across the 10-cm -long wire of (figure 1) as it moves through a magnetic field perpendicular to the plane of the figure. figure1 of 1 a horizontal 10 centimeter long wire segment has positive charges on the left end and negative charges on the right end. the segment moves vertically upward with a velocity of 5.0 meters per second. part a what is the strength of the magnetic field?

Answers

Answer 1

If the segment moves vertically upward with a velocity of 5.0 meters per second, the strength of the magnetic field is 0.040 T.

To solve for the strength of the magnetic field, we need to use the equation:

EMF = B*L*V

where EMF is the potential difference developed across the wire, B is the strength of the magnetic field, L is the length of the wire, and V is the velocity of the wire.

Substituting the given values, we get:

0.020 V = B*(10 cm)*(5.0 m/s)

First, we need to convert the length of the wire from centimeters to meters:

L = 10 cm = 0.1 m

Substituting this value, we get:

0.020 V = B*(0.1 m)*(5.0 m/s)

Simplifying, we get:

B = 0.020 V / (0.1 m * 5.0 m/s)

B = 0.040 T

Therefore, the strength of the magnetic field is 0.040 T.

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Related Questions

An object is placed in front of a concave mirror, between the center of
curvature of the mirror and its focal point, as shown in the diagram below.
Three light rays are traced, along with their corresponding reflected rays.
Which statement below best describes the image formed?

Answers

The image formed by the concave mirror is enlarged or magnified.

optionC.

What type of image is formed?

When an object is placed in front of a concave mirror, between the center of curvature of the mirror and its focal point, the image formed by the concave mirror has the following characteristics;

the image formed is beyond the center of curvature. the image formed is realthe image formed is invertedthe image formed is magnified

So based on the given options, we can that the option that falls in the characteristics given above is "the image is enlarged or magnified.

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in the following equation, a is acceleration, m is mass, v is velocity, r is radius, t is time, is an angle, and c is a constant. a=c mv2sin0/rtif this equation is valid, which of the following could be the units of c?a.s/kgb.m/s2c.m2/sd.kg/me.kg m/s2

Answers

The units of c are: [c] = m²/s³. The answer is b.

The given equation is a = cmv²sinθ/rt, where a is acceleration, m is mass, v is velocity, r is radius, t is time, θ is an angle, and c is a constant.

To determine the units of c, we can analyze the units of each term in the equation and then determine the units of c such that the units of the equation are consistent.

Units of each term in the equation are:

a: m/s²

m: kg

v: m/s

r: m

t: s

sinθ: dimensionless

Substituting these units in the given equation, we get:

[m/s²] = [c] x [kg] x [m/s]² x [dimensionless] / [m] x [s]

Simplifying the above equation, we get:

[c] = [m/s²] x [m] x [s] / [kg] x [m/s]² x [dimensionless]

Therefore, the units of c are:

[c] = m²/s³

Hence, option (b) m²/s³ could be the units of c.

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g what is the angular velocity (in rad/s) of a 62.0 cm diameter tire on an automobile traveling at 93.5 km/h? (enter the magnitude.)

Answers

The angular velocity of the tire is 84.02 rad/s

To find the angular velocity of the tire, we need to convert the linear velocity of the automobile into angular velocity of the tire using the formula:v = ωrwhere v is the linear velocity, ω is the angular velocity, and r is the radius of the tire.First, we need to convert the speed of the car from km/h to m/s:93.5 km/h = 26.0 m/sThe radius of the tire is half the diameter:r = 0.5(62.0 cm) = 0.31 mSubstituting these values into the formula, we get:26.0 m/s = ω(0.31 m)Solving for ω, we get:ω = 84.02 rad/sTherefore, the angular velocity of the tire is 84.02 rad/s.In physics, the rotational velocity or angular velocity ( or ), also known as the angular frequency vector, is a pseudovector representation of how quickly an object spins or revolves in relation to a point or axis.

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