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3 years ago

Change in Energy Quick Check

Physics

1 answer:

nika2105 [10]3 years ago

4 0

Answer:

The force of gravity on the object decreases. (FALSE)

The potential energy of the object decreases. (TRUE)

The acceleration due to gravity decreases. (FALSE)

The kinetic energy of the object decrease (FALSE)

Explanation:

FORCE OF GRAVITY:

Force of gravity on the object is the weight of object, which depends upon the mass and value of acceleration due to gravity (W = mg). Since, the value mass is constant on Earth and acceleration due to gravity is also constant on earth.

Therefore, force of gravity remains same on the object. The statement is false.

POTENTIAL ENERGY:

Potential energy of object depends upon the mass, value of acceleration due to gravity, and change in height of object (P.E = mgΔh). Since, the value mass is constant on Earth and acceleration due to gravity increases as the object moves towards the surface of earth. But, the height of object is decreasing.

Therefore, potential energy of object decreases. The statement is true.

ACCELERATION DUE TO GRAVITY:

Acceleration due to gravity depends upon the altitude (gh = g[1 - 2h/Re]). Since, the height of object is decreasing.

Therefore, acceleration due to gravity increases. The statement is false.

But, this change is not significant.

KINETIC ENERGY:

Kinetic Energy of object, which depends upon the mass and velocity of the object (K.E = mv²/2). Since, the value mass is constant on Earth and velocity increases as the object moves towards the surface of earth.

Therefore, kinetic energy of the object also increases. The statement is false.

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Answer:

The time is 0.5 sec.

Explanation:

Given that,

Voltage V= 12.00 V

Inductance L= 1.20 H

Current = 3.00 A

Increases rate = 8.00 A

We need to calculate change in current

$\Delta A = 8.00-3.00= 5.00\ A$

We need to calculate the time interval

Using formula of inductor

$V=L\dfrac{\Delta A}{\Delta t}$

$\Delta t =\dfrac{L\Delta A}{V}$

Where, $\Delta A$ = change in current

V = voltage

L = inductance

Put the value into the formula

$\Delta t=\dfrac{1.20\times5.00}{12.00}$

$\Delta t=0.5\ sec$

Hence, The time is 0.5 sec.

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3 years ago

If light has a speed of 122,000 mps in a transparent medium, what is the index of refraction of the medium? A. n = 1.52

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Answer:

$A.\hspace{3}n=1.52$

Explanation:

The refractive index of a medium is a measure to know how much the speed of light within the medium is reduced. It can be calculated with the next equation:

$n=\frac{c}{v}$ (1)

Where:

$c=Speed\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}vacuum\\n=Refractive\hspace{3}index\\$

$v=Velocity\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}the\hspace{3}medium$

The speed of light in the vacuum is approximately 300,000 km/s. In order to work with the same units let's do the proper conversion with the velocity of the medium:

$122,000\frac{mi}{s} *\frac{1.60934km}{1mi}=196339.48\frac{km}{s}$

Finally, replacing the data in (1):

$n=\frac{300,000}{196339.48} =1.527965746\approx1.52$

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3 years ago

Explain what the core muscles do and why it's important to have a strong set of core muscles.

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Answer:

Since your spine runs by your core, having a strong core will help protect and stabilize your spine, and you're also are able to control movements such as walking or standing better with a strong core.

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4 years ago

Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a

lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

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