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3 years ago

Three diffrent examples of accelerated motion

Physics

1 answer:

LekaFEV [45]3 years ago

3 0

Answer:

The three different examples of the accelerated motion are Falling/dropping of ball, Standing in circular rotating space, moving around the circle.

Explanation:

Acceleration is the change in velocity, which is related to the speed and direction in which the object is travelling. Hence, speeding up, slowing down and turning are few types . A simple example would be dropping a ball: as it falls its speed increases, which is a type of acceleration. A more complicated example would be standing in a circular, rotating space station. A point on the station moves in a circle, meaning that as it travels it must be turning (to remain in circular motion) making this another example of acceleration

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Which characteristic is used to measure the amount of light radiated by a star?

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The answer is c
Luminosity

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3 years ago

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Which two forms of energy does a hairdryer convert electric energy into? A.)chemical energy

Ne4ueva [31]

D, because the hairdryer uses a fan
E, because it produces heat

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4 years ago

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A two-turn circular wire loop of radius 0.63 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.219 T. I

Basile [38]

Answer:

The magnitude of the average induced emf in the wire during this time is 9.533 V.

Explanation:

Given that,

Radius r= 0.63 m

Magnetic field B= 0.219 T

Time t= 0.0572 s

We need to calculate the average induce emf in the wire during this time

Using formula of induce emf

$E=-\dfrac{d\phi}{dt}$

$E=-B\dfrac{dA}{dt}$

$E=-B\dfrac{A_{2}-A_{1}}{dt}$

$E=B\dfrac{A_{1}-A_{2}}{dt}$ .....(I)

In reshaping of wire, circumstance must remain same.

We calculate the length when wire is in two loops

$l=2\times 2\pi\times r_{1}$

$l=2\times 2\pi\times 0.63$

$l=7.916\ m$

The length when wire is in one loop

$l=2\pi\times r_{2}$

$7.916=2\times \pi\times r_{2}$

$r_{2}=\dfrac{7.916}{2\times \pi}$

$r_{2}=1.259\ m$

We need to calculate the initial area

$A_{1}=N\times\pi\times r_{1}^2$

Put the value into the formula

$A_{1}=2\times3.14\times(0.63)^2$

$A_{1}=2.49\ m^2$

The final area is

$A_{2}=N\times\pi\times r_{2}^2$

$A_{2}=1\times\pi\times(1.259)^2$

$A_{2}=4.98\ m^2$

Put the value of initial area and final area in the equation (I)

$E=0.219\dfrac{2.49-4.98}{0.0572}$

$E=-9.533\ V$

Negative sign shows the direction of induced emf.

Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.

6 0

3 years ago

A boy pushes a stationary box of mass 20 kg with a force of 50 N.

pav-90 [236]

Answer:

2.5m/s^2

Explanation:

Since there is no friction, the only force acting on the box is the applied force

We can now use the equation Fnet = ma

and we can rearrange for a

a = Fnet/m

now input the values

a = 50N/20kg

a = 2.5 m/s^2

6 0

3 years ago

Circuit A in a house has a voltage of 208 V and is limited by a 40.0-A circuit breaker. Circuit B is at 120.0 V and has a 20.0-A

Setler79 [48]

Given Information:

Voltage of circuit A = Va = 208 Volts

Current of circuit A = Ia = 40 Amps

Voltage of circuit B = Vb = 120 Volts

Current of circuit B = Ib = 20 Amps

Required Information:

Ratio of power = Pa/Pb = ?

Answer:

Ratio of power = Pa/Pb = 52/15

Explanation:

Power can be calculated using Ohm's law

P = VI

Where V is the voltage and I is the current flowing in the circuit.

The power delivered by circuit A is

Pa = Va*Ia

Pa = 208*40

Pa = 8320 Watts

The power delivered by circuit B is

Pb = Vb*Ib

Pb = 120*20

Pb = 2400 Watts

Therefore, the ratio of the maximum power delivered by circuit A to that delivered by circuit B is

Pa/Pb = 8320/2400

Pa/Pb = 52/15

4 0

3 years ago