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Paul [167]
3 years ago
12

A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals

during the time the flashlight is in operation, how long was the flashlight used?
Physics
1 answer:
slamgirl [31]3 years ago
5 0

Answer:

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI................ Equation 1

Where P = Power, V = Voltage, I = Current.

make I the subject of the equation

I = P/V................ Equation 2

From the question,

Given: P = 0.414 W, V = 1.50 V

Substitute into equation 2

I = 0.414/1.50

I = 0.276 A

Also,

Q = It............... Equation 3

Where Q = amount of charge, t = time

make t the subject of the equation

t = Q/I.................. Equation 4

From the question,

4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs

Q = 7.899×10⁻¹⁹ C

Substitute these value into equation 4

t =  7.899×10⁻¹⁹/0.276

t = 2.86×10⁻¹⁸ seconds

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3 years ago
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A cylindrical wire made of an unknown alloy hangs from a support in the ceiling. You measure the relaxed length of the wire to b
IceJOKER [234]

Answer: hello  some of your values are wrongly written hence I will resolve your question using the right values

answer:

stiffness =  1.09 * 10^-6 N/m

Explanation:

Given data:

Length ( l ) = 16 m

radius of wire ( r ) = 3.5 m

mass ( m ) = 5kg

<u>Distance stretched (  Δl ) = 4 * 10^-3 m </u> ( right value )

<u>average bond length ( between atoms ) = 2.3 * 10^-10 m </u>( right value)

first step : calculate the area

area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2

        γ          = MgL / A Δl

                    = [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]

                    = 784.8 / 0.165 = 4756.36 N/m^2

hence : stiffness =   γ  * bond length

                           =  4756.36 * 2.3 * 10^-10  = 1.09 * 10^-6 N/m

8 0
3 years ago
1
Darina [25.2K]

Answer:

7.5 x 10⁻⁸N

Explanation:

Given parameters:

Mass 1  = 60kg

Mass 2  = 75kg

Distance between the bodies  = 2m

Unknown:

Gravitational fore  = ?

Solution:

The gravitational force between the two bodies can be derived using;

  F  = \frac{G mass 1 x mass 2}{distance^{2} }  

    G is the universal gravitation constant  = 6.67 x 10⁻¹¹m³kg⁻¹s⁻²

Insert the parameters and solve;

  F  = \frac{6.67 x 10^{-11}  x 60 x 75}{2^{2} }   = 7.5 x 10⁻⁸N

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3 years ago
Kingda Ka has a maximum speed of 57.2 m/s. Determine the height of the hill on this roller coaster. Explain to get full credit,
kodGreya [7K]

This question involves the concepts of the law of conservation of energy, kinetic energy, and potential energy.

The height of the hill is "166.76 m".

<h3>LAW OF CONSERVATION OF ENERGY:</h3>

According to the law of conservation of energy at the highest point of the roller coaster ride, that is, the hill, the whole (maximum) kinetic energy of the roller coaster is converted into its potential energy:

Maximum\ Kinetic\ Energy\ Lost = Maximum\ Potential\ Energy\ Gain\\\\&#10;\frac{1}{2}mv_{max}^2=mgh\\\\&#10;h=\frac{v_{max}^2}{2g}

where,

  • h = height of the hill = ?
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Therefore,

h=\frac{(57.2\ m/s)^2}{2(9.81\ m/s^2)}\\\\&#10;

<u>h = 166.76 m</u>

Learn more about the law of conservation of energy here:

brainly.com/question/101125

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AnnZ [28]

Answer:

0.75

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Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).

So, μ = F/N

= 300 N/400 N

= 3/4

= 0.75

So, the  coefficient of static friction μ = 0.75

6 0
3 years ago
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