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just olya [345]

2 years ago

11

student uses a magnet to move a 0.025 kg metal ball magnet exerts a force of 5N which causes the ball to begin moving what is th

e accelaration of the ball when it begins to move

1 answer:

schepotkina [342]2 years ago

5 0

The acceleration of the ball when it begins to move is 200 m/s^2.

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A steel cable lifting a heavy box stretches by ΔL . In order for the cable to stretch by only half of ΔL , by about what factor

il63 [147K]

Answer:

2.0

Explanation:

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6 0

2 years ago

An individual white LED (light-emitting diode) has an efficiency of 20% and uses 1.0 W of electric power. a. How many LEDs must

Neporo4naja [7]

Answer:

8, 8 W

Explanation:

The useful power of 1 Light Emitting Diode is

$0.2\times 1=0.2\ W$

Total power required is 1.6 W

Number of Light Emitting Diodes would be

$n=\dfrac{1.6}{0.2}\\\Rightarrow n=8$

The number of Light Emitting Diodes is 8.

Power would be

$P=8\times 1=8\ W$

The power that is required to run the Light Emitting Diodes is 8 W

7 0

2 years ago

A flywheel with a very low friction bearing takes 1.6 h to stop after the motor power is turned off. The flywheel was originally

nadezda [96]

Answer: 5.76 rads/s

Explanation:

The initial rotation is 55 rpm

1 rev = 2π radians

55 revs = 55 × 2π/1 = 345.58 radians/min

345.58 rads/min = 345.58rads/60s = 5.76 rads/s

3 0

2 years ago

Consider two circular metal wire loops each carrying the same current I as shown below. In what r... Consider two circular metal

NeX [460]

Answer:

1) The magnetic field outside the loop is zero.

In region III the magnetic fields due to the two wire loops point in the opposite direction andhence cancel each other. Therefore the magnetic field is zero in region I, III and V

The diagram is attached

6 0

2 years ago

A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i

victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

$F= \frac{kqQ}{r^2}$

Where,

$k = 8.99 * 10^9 N m^2 / C^2.$

q = charges of the objects

r= distance/radius

Our values are previously given, so

$q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59$

Replacing,

$F=\frac{kqQ}{r^2}$

$F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}$

$F= 4.8423N$

The force acting on the block are given by,

$F-mgsin\theta = ma$

$a = \frac{F-mgsin\theta}{m}$

$a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}$ $a = 10.31m/s^2$

Therefore the box is accelerated upward.

3 0

2 years ago