PHYSICS COLLEGE

A mouse is running across a room with a speed of 2.2 m/s. The mass of the mouse is 1.4 kg. What is the Kinetic Energy of the mouse?

Answers

Answer 1

Answer:

Answer:

3.39 J

Explanation:

The kinetic energy of an object can be found by using the formula

$k = (1)/(2) m {v}^(2) \n$

m is the mass

v is the velocity

From the question we have

$k = (1)/(2) * 1.4 * {2.2}^(2) \n = 0.7 * 4.84 \n = 3.388 \: \: \: \: \: \: \: \: \:$

We have the final answer as

3.39 J

Hope this helps you

Answer 2

Answer:

Answer:

im not sure

Explanation:

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Answers

Answer: D i am pretty sure

Explanation:

Answer:

all

Explanation:

Which type of diffraction occurs when the point source and the screen are at finite distances from the obstacle forming the diffraction pattern?A. fraunhofer
B. fresnel
C. far-field
D. single slit

Answers

Fresnel diffraction

A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very far from the wire. What is the ratio Ff/Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?

Answers

The ratio of the electric force on the proton after the wire segment is shrunk to three times its original length to the force before the segment was shrunk is 3.

The electric force between a point charge and a segment of wire with a distributed charge is given by Coulomb's law.

The formula for the electric force on a point charge q due to a segment of wire with charge Q distributed along its length L is:

$F=(k.q.Q)/(L)$

where:

F is the electric force on the point charge,

k is Coulomb's constant ( 8.988 × 1 0⁹ Nm²/ C²),

q is the charge of the point charge,

Q is the charge distributed along the wire segment, and

L is the length of the wire segment.

When the wire segment is shrunk to one-third of its original length, the new length becomes 1/3 L.

The chargedistribution remains the same, only the length changes.

So, the new electric force $F_f$ on the proton after the segment is shrunk becomes:

$F_f=(k.q.Q)/((1)/(3)L)$

The original electric force $F_i$ on the proton before the segment was shrunk is:

$F_i = (k.q.Q)/(L)$

let's find the ratio $(F_f)/(F_i)$ :

$(F_f)/(F_i)=((k.q.Q)/((1)/(3)L))/((k.q.Q)/(L))$

$(F_f)/(F_i)=3$

Hence, the ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk is 3.

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Final answer:

The ratio of the electric force on the proton after the wire segment is shrunk is equal to the ratio of their charges.

Explanation:

The ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk can be found using Coulomb's law. Coulomb's law states that the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In this case, the charges involved are the charge of the wire segment and the charge of the proton. Since the wire segment contains 10 nC of charge, we can consider it as one of the charged objects. The proton is very far from the wire, so we can assume that the distance between them remains the same before and after the wire segment is shrunk. Therefore, the ratio of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk is equal to the ratio of their charges.

Let's assume that the initial force on the proton is Fi and the final force on the proton is Ff. Using the given information, we have:

Fi = k(q1 * q2) / r^2

where k is the electrostatic constant, q1 and q2 are the charges of the wire segment and the proton respectively, and r is the distance between them.

After the wire segment is shrunk to one-third of its original length, the charge of the wire segment remains the same and the distance between the wire segment and the proton also remains the same. Therefore, the ratio Ff/Fi can be calculated as:

Ff/Fi = (q1 * q2) / (q1 * q2) = 1

A 13.0-Ω resistor, 13.5-mH inductor, and 50.0-µF capacitor are connected in series to a 55.0-V (rms) source having variable frequency. If the operating frequency is twice the resonance frequency, find the energy delivered to the circuit during one period.

Answers

Answer:

E = 0.13 J

Explanation:

At resonance condition we have

$\omega = \sqrt{(1)/(LC)}$

$\omega = \sqrt{(1)/((13.5 * 10^(-3)){50* 10^(-6))}}$

$\omega = 1217.2 rad/s$

now if the frequency is double that of resonance condition then we have

$x_L = 2\omega L$

$x_L = 2(1217.2)(13.5* 10^(-3))$

$x_L = 32.86 ohm$

now we have

$x_c = (1)/(2(1217)(50* 10^(-6)))$

$x_c = 8.22 ohm$

now average power is given as

$P = i_(rms)V_(rms)* (R)/(z)$

$P = (55)/(√((32.86 - 8.22)^2 + 13^2)))(55)* (13)/(√((32.86 - 8.22)^2 + 13^2))$

$P_(avg) = 50.67 Watt$

Now time period is given as

$T = (2\pi)/(\omega)$

so total energy consumed is given as

$E_(avg) = 50.67((2\pi)/(2(1217.2)))$

$E = 0.13 J$

According to the World Flying Disk Federation, the world distance record for a flying disk throw in the men’s 85-years-and-older category is held by Jack Roddick of Pennsylvania, who on July 13, 2007, at the age of 86, threw a flying disk for a distance of 54.0 m. If the flying disk was thrown horizontally with a speed of 13.0 m/s, how long did the flying disk remain aloft? (Jack Roddick was also a physics teacher! Read more about him at

Answers

Answer:

t = 4.15 seconds

Explanation:

It is given that,

Distance traveled by a flying disk, d = 54 m

The speed at which it was thrown, v = 13 m/s

We need to find the time for which the flying disk remain aloft. Let the distance is d. We know that, speed is equal to the distance covered divided by time. So,

$t=(d)/(v)\n\nt=(54\ m)/(13\ m/s)\n\nt=4.15\ s$

Hence, for 4.15 seconds the flying disk remain aloft.

The mass of the Sun is 2 × 1030 kg, the mass of the Earth is 6 × 1024 kg, and their center-to-center distance is 1.5 × 1011 m. Suppose that at some instant the Sun's momentum is zero (it's at rest). Ignoring all effects but that of the Earth, what will the Sun's speed be after 3 days? (Very small changes in the velocity of a star can be detected using the "Doppler" effect, a change in the frequency of the starlight, which has made it possible to identify the presence of planets in orbit around a star.)