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bija089 [108]
3 years ago
13

The Biot-Savart Law describes the relationship between which of the following? Select all that apply.

Physics
1 answer:
faltersainse [42]3 years ago
8 0
The correct answer is magnetic field, electric field, and charges.


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By what factor would your weight increase if you could stand on the sun? (never mind that you can't.)
Fofino [41]

Gravity on the surface of sun is given as

g = \frac{GM}{R^2}

here we know that

M = 1.98 \times 10^{30} kg

R = 6.95 \times 10^8 m

now we will have

g = \frac{(6.67 \times 10^{-11})(1.98 \times 10^30)}{(6.95 \times 10^8)^2}

g = 273.4 m/s^2

now we need to find the ratio of weight on surface of sun and on surface of Earth

R = \frac{mg_{sun}}{mg_{earth}}

R = \frac{273.4}{9.8} = 28 times

so weight will increase by 28 times

6 0
3 years ago
Does potential energy increase with temperature?
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-- The potential energy of a 12-lb bowling ball up on the shelf
doesn't have anything to do with the temperature of the ball or
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-- The potential energy of a jar full of gas does depend on the
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to blow the jar to smithereens.
6 0
3 years ago
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An amusement ride consists of a car moving in a vertical circle on the end of a rigid boom. The radius of the circle is 10 m. Th
AveGali [126]

Answer:

3.7 kN (Up)

Explanation:

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Which device will allow you to reduce the voltage from 120 vac to vdc in order to charge a cell phone?
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It should be 4. An alternator
6 0
3 years ago
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
4 years ago
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