You are studying circular motion by placing pennies on a turntable and then turning it on so that it will spin. You keep increasing the speed until one of the pennies slips off. You repeat this procedure and observe that the penny close to the outer edge always slip off first. What is the best inference?

1.The penny near the edge has a greater tangential velocity than the one in the center, so it experiences more air resistance. It’s the effect of the air “blowing” it off.

2.The centripetal force required to keep the pennies in place increases with the distance from the center. Eventually, as the turntable spins faster, the friction force between the turntable and the penny near the edge is not enough to supply the required centripetal force.

3.The centrifugal force acting on the pennies is stronger on the one near the edge than the one near the center.

Answer:

It should be geothermal energy

Answer:

d. kT ln 3

Explanation:

Given;

k as Boltzmann constant

Let initial initial microstate, = Фi

Let the final microstate, = Фf = 3Фi

at constant temperature, T

The energy input to gas as heat is given by;

Q = TΔS = Tk(lnФf - lnФi)

[tex]Q= kT*\frac{ln \phi _f}{ln \phi_i} \\\\Q= kT*\frac{3ln \phi _i}{ln \phi_i}\\\\Q= kT ln3[/tex]

Therefore, the energy input to gas as heat is kT ln 3

Answer: (a) B = 2 x 10⁻⁵T

               (b) B = 1.94 x 10⁻⁵T

               (c) B = 1.8 x 10⁻⁴T

Explanation: A magnetic field due to a current passing through a straight wire is calculated using the Biot-Savart Law:

[tex]dB=\frac{\mu_{0}}{4.\pi} \frac{IdLXR}{r^{3}}[/tex]

where

dL is current length element

[tex]\mu_{0}[/tex] is permeability of free space ([tex]4.\pi.10^{-7}[/tex]T.m/A)

(a) For a infinite straight wire:

[tex]B=\frac{\mu_{0}I}{2.\pi.R}[/tex]

[tex]B=\frac{4.\pi.10^{-7}.2}{2.\pi.2.10^{-2}}[/tex]

B = 2x10⁻⁵T

For an infinite, long and straight wire, magnetic field is 2x10⁻⁵T.

(b) For a finite wire:

[tex]B=\frac{\mu_{0}I}{2.\pi.R}\frac{L}{\sqrt{L^{2}+R^{2}} }[/tex]

[tex]B=\frac{4.\pi.10^{-7}.2}{2.\pi.2.10^{-2}} \frac{8.10^{-2}}{\sqrt{(8.10^{-2})^{2}+(2.10^{-2})^{2}} }[/tex]

B = 1.94x10⁻⁵T

The magnetic field for a finite wire in the same conditionsas infinite wire is 1.94x10⁻⁵T.

(c) For a finite wire at a point distant from the end of the wire:

[tex]B=\frac{\mu_{0}I}{4.\pi.L\sqrt{2} }[/tex]

[tex]B=\frac{4.\pi.10^{-7}.2}{4.\pi.8.10^{-2}\sqrt{2} }[/tex]

B = 0.18x10⁻⁵T

At a point at the end, magnetic field is 1.8x10⁻⁴T.

Answer:

D. 10 m/s

Explanation:

Answer:

34.3N

Explanation:

The normal force, denoted by F(N), is the force exerted on an object by a surface. The normal force in this case is acting in an opposite manner but equal magnitude to the force of gravity.

F(N) = Fg

Where;

F(N) = Normal force

Fg = gravitational force

Fg = mg

Where;

m = mass (kg)

g = acceleration due to gravity

F(N) = mg

F(N) = 3.5kg × 9.8m/s²

F(N) = 34.3N

Hence, the normal force acting on the object is 34.3N.

[tex]a = \frac{vf - vi}{t} [/tex]

here initial velocity vi=0 as ball release from rest

the final velocity is vf=4.0

time is t=6

so putting all these values in above equation

[tex]a = \frac{ 4.0- 0}{6} [/tex]

[tex]a = 0.6667m \s {}^{2} [/tex]

Answer:

substance and mixture

Answer:

1. s

2. m

3. start m/s

4. end m/s

5. m/s^2

Explanation:

1. t is the suffix for the time and it is related to seconds [s]

2. d is the suffix for distance and it is related to meters [m]

3. Vi  is the suffix for the initial velocity and it is related to meters per second [m/s]

4. Vf  is the suffix for the final velocity and it is related to meters per second [m/s]

5. a  is the suffix for the acceleration and it is related to meters per squared second [m/s^2]

Answer:

Assuming that 4000 J of heat transfer occurs from it the change in entropy will be=2.33j/k

Explanation:

Given data

Th=619 K

Tc= 455K

We are going to assume that 4000 J of heat transfer occurs from it, since it was not specified in the question.

we know that the change in entropy is given as

ΔStotal=ΔShot+ΔScold .

ΔShot=−Q/Th=−4000J/619K=−6.46J/K

For the cold reservoir,

ΔScold=Q/Tc=4000J/455K=8.79J/K

therefore  the total is

ΔStotal=ΔShot+ΔScold

=(−6.46+8.79)J/K

=2.33j/k

Answer:

simple machines such as ramps lessen the moment required to do work. if a triangle has a base of 5 and the height  is 7, a ramp would make the hypotenuse of this triangle lessoning the total distance. using a²+b²=c² 25+49=c² 74≈8.6 and it is obvious that 8.6 is less than 12 in every unit. other simple machines such as pulleys make it lighter making it simply easier for an object to be lifted.

Explanation:

Answer:

Simple machines are useful because they reduce effort or extend the ability of people to perform tasks beyond their normal capabilities. Simple machines that are widely used are the wheel and axle, pulley, inclined plane, screw, wedge and lever.

a train cuz its 120km in 3 hours

Answer:

(a) Negative

(b) Positive

Explanation:

Here is a question on sign.

(a) When the rock is going up, it is working against gravity. In this case, the acceleration due to gravity will take a negative value. This is because gravity works or acts downwards

(b) When the rock is coming down, it is actually working with gravity. So in this case, the sign will ne positive

Answer:

Explanation:

The mass of the object can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

where

f is the force

a is the acceleration

So we have

[tex]m = \frac{34}{4} = \frac{17}{2} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

The value of q is [tex]\dfrac{Q}{8}[/tex]

Explanation:

Given that,

Each charge = -Q

Distance between charges = L

Reduced force = [tex]\dfrac{F}{2}[/tex]

Suppose, Two particles each with a charge -Q are fixed a distance L apart as shown above. Each particle experiences a net electric force F. A particle with a charge +q is now fixed midway between the original two particles.

We know that,

The force on each end is

[tex]F=\dfrac{kQ^2}{L^2}[/tex]...(I)

If the charge q is placed at mid point then

The  force on each end charge is

[tex]\dfrac{F}{2}=F+F'[/tex]....(II)

We need to calculate the value of q

Using equation (II)

[tex]\dfrac{F}{2}=F+F'[/tex]

Put the value of F into the formula

[tex]\dfrac{\dfrac{kQ^2}{L^2}}{2}=k\dfrac{Q^2}{L^2}+k\dfrac{q\times(-Q)}{(\dfrac{L}{2})^2}[/tex]

[tex]\dfrac{kq(-Q)}{(\dfrac{L}{2})^2}=-\dfrac{kQ^2}{2L^2}[/tex]

[tex]\dfrac{q}{\dfrac{1}{4}}=\dfrac{Q}{2}[/tex]

[tex]q=\dfrac{Q}{8}[/tex]

Hence, The value of q is [tex]\dfrac{Q}{8}[/tex]

Goal posts at the end of football field are padded as a safety measure to players who might run into them,these thick paddings of the goal posts are done to prevent injuries to the players because the padding of the goal post increases the time of impact and this in-turn decreases the force on the the players body or the force it will have when a player's body hit against it and therefore decreasing the force applied to bring the player to a stop.

We can also say that it increases the time of collision between the player's body and the post.

Answer:

147.27N

Explanation:

Power = workdone/time

Power = Force*distance/time

Given

Power = 90Watts

Distance = 1.1m

Time = 1.8secs

Force = ?

Substitute the given parameters into the formula:

[tex]90 = \frac{1.1d}{1.8}\\cross \ multiply\\ 90 \times 1.8 = 1.1F\\162 = 1.1F\\1.1F = 162\\F = \frac{162}{1.1} \\F = 147.27N[/tex]

Hence the magnitude of the force that you exert on the handle is 147.27N

Answer:

b

Explanation:

Answer: 10,000 N

Explanation: USE , f = ma

Answer:

it would take 5 hours

Explanation:

plz vote this up i promise its correct

Complete question is;

Jason works for a moving company. A 75 kg wooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11°.

What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving UP the ramp?

Answer:

F = 501.5 N

Explanation:

We are given;

Mass of wooden crate; m = 75 kg

Angle of ramp; θ = 11°

Now, for the wooden crate to slide upwards, it means that the force of friction would be acting in an opposite to the slide along the inclined plane. Thus, the force will be given by;

F = mgsin θ + μmg cos θ

From online values, coefficient of friction between wooden surfaces is μ = 0.5

Thus;

F = (75 × 9.81 × sin 11) + (0.5 × 75 × 9.81 × cos 11)

F = 501.5 N

Answer:

Explanation:

According to energy conservation which states that the workdone is equal to change in the system

Workdone = change in kinetic energy + (frictional force * distance)

Workdone = ΔK + fd

Workdone = kf-Ki + fd

Workdone = = 1/2(m(v-u)^2) + fd

Given

Mass m = 495kg

final velocity v = 105m/s

initial velocity = 0m/s

Force f= 1400N

distance d = 395m

Substitute

Workdone = 1/2(495(105-0)^2) + 1400(395)

Workdone = 2,728,687.5+553000

Workdone = 3,281,687.5 Joules

Time = 8.2secs

Power output = Workdone/Time

Power output = 3,281,687.5/8.2

Power output = 885,766.768

Power output = 8.858 * 10^5 watts

Answer:

the correct answer is The English Bill of Rights

Explanation:

This is the answer because i have taken the test.

Answer:

Transition Metals

Explanation:

__________________________

Answer:Explanation:

A microscopic hair examination can also determine if a hair was forcibly removed, artificially treated or diseased. A comparison microscope can be used to compare a questioned hair to a known hair sample in order to determine if the hairs are similar and if they could have come from a common source.

Answer:

P=300 W

W=1800 J

P=W/t

t=W/P

t=6 s

Answer:

457.81 Hz

Explanation:

From the question, it is stated that it is a question under Doppler effect.

As a result, we use this form

fo = (c + vo) / (c - vs) × fs

fo = observed frequency by observer =?

c = speed of sound = 332 m/s

vo = velocity of observer relative to source = 45 m/s

vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).

fs = frequency of sound wave by source = 459 Hz

By substituting the the values to the equation, we have

fo = (332 + 45) / (332 - (-46)) × 459

fo = (377/ 332 + 46) × 459

fo = (377/ 378) × 459

fo = 0.9974 × 459

fo = 457.81 Hz