If you double an object's velocity, its kinetic energy increases by a factor of four. True False

Answers

Answer 1

True. Doubling an object's velocity increases its kinetic energy by a factor of four.

The relationship between kinetic energy (KE) and velocity (v) is given by the equation [tex]KE=\frac{1}{2}*m * V^{2}[/tex]

where m is the mass of the object. According to this equation, kinetic energy is directly proportional to the square of the velocity. If we consider an initial velocity [tex]V_1[/tex], the initial kinetic energy would be:

[tex]KE_1=\frac{1}{2} * m * V_1^{2}[/tex].

Now, if we double the velocity to [tex]2V_1[/tex], the new kinetic energy would be [tex]KE_2=\frac{1}{2} * m * (2V_1)^2 = \frac{1}{2} * m * 4V_1^2[/tex].

Comparing the initial and new kinetic energies, we can see that [tex]KE_2[/tex] is four times larger than [tex]KE_1[/tex]. Therefore, doubling the velocity results in a fourfold increase in kinetic energy.

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Related Questions

An RLC series circuit has a 1.00 kΩ resistor, a 130 mH
inductor, and a 25.0 nF capacitor.
(a)
Find the circuit's impedance (in Ω) at 490 Hz.
Ω
(b)
Find the circuit's impedance (in Ω) at 7.50 k

Answers

An RLC series circuit has a 1.00 kΩ resistor, a 130 mH inductor, and a 25.0 nF capacitor.(a)The circuit's impedance at 490 Hz is approximately 1013.53 Ω.(b)The circuit's impedance at 7.50 kHz is approximately 6137.02 Ω.

(a) To find the circuit's impedance at 490 Hz, we can use the formula:

Z = √(R^2 + (XL - XC)^2)

where Z is the impedance, R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.

Given:

R = 1.00 kΩ = 1000 Ω

L = 130 mH = 0.130 H

C = 25.0 nF = 25.0 × 10^(-9) F

f = 490 Hz

First, we need to calculate the inductive reactance (XL) and capacitive reactance (XC):

XL = 2πfL

= 2π × 490 × 0.130

≈ 402.12 Ω

XC = 1 / (2πfC)

= 1 / (2π × 490 × 25.0 × 10^(-9))

≈ 129.01 Ω

Now we can calculate the impedance:

Z = √(R^2 + (XL - XC)^2)

= √((1000)^2 + (402.12 - 129.01)^2)

≈ √(1000000 + 27325.92)

≈ √1027325.92

≈ 1013.53 Ω

Therefore, the circuit's impedance at 490 Hz is approximately 1013.53 Ω.

(b) To find the circuit's impedance at 7.50 kHz, we can use the same formula as before:

Z = √(R^2 + (XL - XC)^2)

Given:

f = 7.50 kHz = 7500 Hz

First, we need to calculate the inductive reactance (XL) and capacitive reactance (XC) at this frequency:

XL = 2πfL

= 2π × 7500 × 0.130

≈ 6069.08 Ω

XC = 1 / (2πfC)

= 1 / (2π × 7500 × 25.0 × 10^(-9))

≈ 212.13 Ω

Now we can calculate the impedance:

Z = √(R^2 + (XL - XC)^2)

= √((1000)^2 + (6069.08 - 212.13)^2)

≈ √(1000000 + 36622867.96)

≈ √37622867.96

≈ 6137.02 Ω

Therefore, the circuit's impedance at 7.50 kHz is approximately 6137.02 Ω.

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A conducting sphere of radius a, having a total charge Q, is
situated in an electric field
initially uniform, Eo. Determine the potential at all points
outside the sphere.

Answers

The potential at all points outside the sphere is given by,V = Q / (4πε₀r) + Q / (4πε₀a)

We are given that a conducting sphere of radius a, having a total charge Q, is situated in an electric field initially uniform, Eo. We need to determine the potential at all points outside the sphere.Potential at any point due to a point charge Q at a distance of r from it is given by the equation,V = Q / (4πε₀r)

The conducting sphere will be at equipotential because the electric field is initially uniform. Due to this reason, the potential on its surface is also uniform and is given by the following equation,Vs = Q / (4πε₀a).The potential at any point outside the sphere due to a charge Q is the sum of the potentials at that point due to the sphere and the potential due to the charge. Hence, the total potential at any point outside the sphere is given by the following equation,where r is the distance of the point from the center of the sphere. Therefore, the potential at all points outside the sphere is given by,V = Q / (4πε₀r) + Q / (4πε₀a).

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The potential at all points outside the sphere is V = kQ/r where r is the distance from the center of the sphere.

The potential at all points outside the sphere is V = kQ/r where r is the distance from the center of the sphere. If we calculate the potential at a distance r from the center of the sphere, we can use the formula:

V = kQ/r where Q is the total charge and k is Coulomb’s constant which equals 9 x 10^9 N.m²/C².

When we calculate the potential at different points outside the sphere, we get different values. When the distance r is infinity, the potential is zero. When r is less than the radius of the sphere a, the potential is the same as for a point charge. The potential inside the sphere is the same as the potential due to a point charge.

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