In pan C of the experiment, an electrophorus will be used to get more charge than we can from rubbing rods. An electrophorus is a metal disk (grey in the diagram below) with an insulating handle (black in the diagram below). Suppose that you have a plastic sheet that has already been positively charged by being rubbed with a cloth. The rubbed side of the sheet is placed face-down. What will happen to Actually, the glass flask that encases an electroscope becomes charged after some time. If enough charge appears on the glass flask, then the electroscope doesn't respond as we expect. Use your second electroscope in these cases. In the meantime, the casing of the first electroscope will lose its charge and again behave as you expect. charge in the electrophorus if it is put on the charged plastic sheet? Explain in words what will happen. Copy the following diagram in your pre-lab and indicate the charged regions. |3| Let's think about a different situation before returning to the electrophorus. 'Ground' is just what you think: The Earth. Suppose that a (small) charged conductor Is connected to ground. Explain why the conductor becomes uncharged. |2| [It might help for you to replace the word 'ground' with the phrase 'a really big nail'. In electrostatics I, you thought about what is happening to charge inside a nail when a charged object approaches it.] Let's return to the electrophorus. Typically, the charged plastic sheet is on a bench and the electrophorus is put on top of it. The top surface of the METAL disk of the electrophorus is briefly connected to ground and then disconnected from it. What is the charge-state of the electrophorus after the electrophorus is momentarily connected to ground and then disconnected from it? [1] Explain, assuming that the positive charge is free to move. [2] Imagine the charge-state of the electrophorus is as in the previous paragraph before any connection to ground is made. Now let's change our perspective. Your previous explanation assumed that positive charge was free to move. Now assume that positive charge is stationary and that negative charge moves. Using this assumption to explain the charge-state of the electrophorus after the electrophorus is momentarily connected to ground and then disconnected from it. |2|

The magnitude of the initial acceleration of the object is 4.2 m/s².

The tension in the string once the object starts moving is 13.65 N.

The magnitude of the initial acceleration of the object is calculated by applying Newton's second law of motion as follows;

F(net) = ma

m₂g - μm₁g cosθ = a(m₁ + m₂)

where;

(4.75 x 9.8) - (0.535 x 3.25 x 9.8 x cos40) = a(3.25 + 4.75)

33.5 = 8a

a = 33.5/8

a = 4.2 m/s²

The tension in the string once the object starts moving is calculated as;

T = m₁a

T = 3.25 x 4.2

T = 13.65 N

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Answer:

the resultant wave is the algebraic sum of all the waves reaching that particular point at a given time.

Explanation:

imagine two or three waves reaching a particular particle x at the same time. The particle will vibrate those waves and give out or transmit a resultant wave which is the algebraic sum of the incoming two waves. If both the waves have the same amplitude and phase, the resultant wave will be amplified. However if the waves have the same amplitude and equal but opposite phase then the resultant wave will be a straight line

Answer:

basketball

Explanation: cause it is

Answer:

It honestly depends on which one you're better at.

Explanation: I mean that's what I think.

We have that for the Question "a)what is the resistivity in (ohm-cm) of the material? b) what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material"

Answer:

From the question we are told

This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.

A) Resistivity is given as,

\(p = \frac{RA}{l}\)

where,

\(R = \frac{V}{I}\)

Therefore,

\(p = \frac{VA}{Il}\\\\p = \frac{1*(0.1*0.5)}{0.019*1.5}\\\\p = 1.754 ohm-cm\)

B) Conductivity is given as,

\(U = \frac{\rho}{pe}\\\\U = \frac{10^{17}}{10^{10}*1.6*10^{-19}}\\\\U = 6.25*10^{25} cm^3/V-s\)

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Total weight in tons

\(\\ \sf\longmapsto 2.3\times 10^6\times 2.4\)

\(\\ \sf\longmapsto 5.52\times 10^6\)

\(\\ \sf\longmapsto 5520000ton\)

\(\\ \sf\longmapsto 5520000\times 1000\)

\(\\ \sf\longmapsto 5520000000kg\)

\(\\ \sf\longmapsto 5.52\times 10^9Kg\)

Newton's second law tells us that force is directly proportional to mass and acceleration, such that:

Where force is measured in newtons (N).

The three fundamental laws of Newton are:

Newton's laws are fundamental to understanding the interaction between bodies. Today they are still valid, although at the atomic level it has some exceptions that quantum mechanics studies.

Data:

m = 7.5 kg

F = ?

a = 3.2 m/s²

The formula to calculate the force is: F = m * a

We clear, but in this case it is not necessary, since we only want to calculate the force.

The force you need to accelerate down the alley is 24 Newtons.

\(\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}\)

Actually Welcome to the concept of resistors,

The equivalent circuit is given as,

1/Rp = 1/3 + 1/6 + 1/9

1/Rp = 6+3+2/18

1/Rp = 11/18

Rp = 18/11 ==> 1.63 ==> 1.6 ohms

hence the equivalent resistance of the circuit is 1.6 ohms

The equivalent resistance of the circuit is 1.6Ω. The correct option is B.

Ohm’s law state that ” At constant temperature, the current (I) flowing through a conductor is directly proportional to the potential difference(V) across the two endpoints of the given conductor.”

I.e V ∝ I

V=IR

Where V= potential difference across the conductor.

I = current flowing through the conductor.

R= constant pf proportionality i.e resistance which unit is ohm(Ω).

There are two ways we can connect the resistors.

(i) series connection

If a number of resistors are said to be connected in series When the same current (I) flows through them.

Let R1, R2, and R3 be the resistors connected in series.

Then the R equivalent is

Req=R1+R2+R3

(ii) parallel connection

A number of resistors are said to be connected in parallel when the same potential difference(V) exists across each of them.

Let R1, R2, and R3 be the resistors connected in parallel.

Then the R equivalent is

1/Req=1/R1+1/R2+1/R3

In this question,

The three resistance connected in parallel by applying the above formula we get,

1/Req=1/R1+1/R2+1/R3

1/Req = 1/3 + 1/6 + 1/9      .................  (∵R1=3Ω,R2=6Ω AND R3=9Ω)

1/Req =11/18

Req=18/11

Req=1.6363Ω

Req≈1.6Ω

Therefore, The equivalent resistance of the circuit is 1.6Ω.

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The resistance of working potential difference of the 6.2 V, 4.5 W lamp is 8.54 ohm. When terminal pd across the battery is 6.2 V, the emf of the battery is 8.01 V.

The opposition that a substance provides to the flow of electric current is referred to as resistance. The uppercase letter R represents it. The ohm is the standard unit of resistance, which is sometimes written as a word and sometimes represented by the uppercase Greek letter omega. Resistance is a measure of an electrical circuit's resistance to current flow. Resistance is measured in ohms, which is represented by the Greek letter omega ().

Here,

r=2.5 ohm

V= 6.2 volts

P=4.5 watt

P=V²/R

R=V²/P

R=6.2*6.2/4.5

R=8.54 ohm that is about 9 ohm.

I=V/R

I=0.726 A

EMF=V+Ir

=6.2+0.726*2.5

=6.20+1.81

=8.01 volts

The resistance of the 6.2 V, 4.5 W lamp at its working potential difference is 8.54 ohm. The emf of the battery when terminal pd across the battery is 6.2 V is 8.01 V.

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Answer:

The driver does not do it. The acceleration must be greater than 9.0m/s² in order to beat the time of 4.5s. The acceleration must be calculated using the equation a = (d/t)², where a is the acceleration, d is the distance, and t is the time. In this case, a = (100/4.5)², which is equal to approximately 27.78m/s². Therefore, the acceleration must be greater than 27.78m/s² in order to beat the time of 4.5s.

Explanation:

Answer:

50 W

Explanation:

Case 1

Power = V * I

100 = 220 * I

I = \(\frac{100}{220}\) A

Case 2

P = V * I

P = 110 * \(\frac{100}{220}\)

P = 50 W

I think the answer is 50 W

Hope it helps

Answer:

After the head-on collision, both identical billiard balls will bounce back at the same speed they had before the collision. This is because of the conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces act on it. In this case, the two billiard balls are a closed system, and their total momentum before the collision is equal and opposite to their total momentum after the collision. Therefore, they will both bounce back with the same speed they had before the collision.

Explanation:

Answer:

kilograms

Explanation:

The part of a road vehicle which must be tested to ensure that there is sufficient friction to stop the vehicle in an emergency is the tyre.

This is referred to as a force that resists the motion of one object against another when they roll or slide against each other.

When dealing with braking, the main factor is to have sufficient friction between the road surface and tyre to bring the vehicle to a standstill. If the tyres are wornout there won't be enough friction to make the vehicle stop during emergencies which is therefore the reason why it was chosen as the correct choice.

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Answer:

e

Explanation:

Answer:

The blocks c

Explanation:

Answer:

a = 1.152s

b = 0.817 m

c = 7.29m/s

Explanation: let the following

From the first equation of linear motion

V = u+at..........1

parameters be represented as :

t = Time taken

v = Final velocity

a = Acceleration due to gravity = 9.8m/s²

u = Initial velocity = 4 m/s

s = Displacement

V = 0

Substitute the values into equation 1

0 = 4-9.8(t)

-4 = -9.8t

t = 4/9.8

t = 0.408s

From : s = ut+1/2at^2.........2

S = 4×0.408+0.5(-9.8)×0.408^2

S= 1.632-4.9(0.166)

S = 1.632-0.815

S = 0.817m

Her highest height above the board is 0.817 m

Total height she would fall is 0.817+1.90 = 2.717 m

From equation 2

s = ut+1/2at^2

2.717 m = 0t+0.5(9.8)t^2

2.717 m = 0+4.9t^2

2.717 m = 4.9t^2

2.717/4.9 = t^2

0.554 =t^2

t =√0.554

t = 0.744s

Hence, her feet were in the air for 0.744+0.408seconds

= 1.152s

Also recall from equation 1

V= u+at

V = 0+9.8(0.744)

V = 7.29m/s

Hence, the velocity when she hits the water is 7.29m/s

Finally,

a = 1.152s

b = 0.817 m

c = 7.29m/s

When the liquid cools down, it loses heat energy.

As the liquid cools, it loses heat energy. As a result, its particles slow down in movement and come closer to one another. Attractive forces begin to hold particles and the crystals of a solid form.

If water is cooled, it can change into ice. If ice is warmed, it can change into a liquid state. Heating a substance makes the molecules move very fast whereas cooling a substance makes the molecules move very slowly.

Heating a liquid increases the speed of the molecules present in it. An increase in the molecule's speed competes with the attraction between molecules and results in the molecules moving apart whereas Cooling a liquid decreases the movement of the molecules.

So we can conclude that the liquid cools down when it loses heat energy.

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Beats can be defined as the periodic fluctuations in the frequency of sound waves. That is option D

Sound waves are those waves that are produced by the vibration of an object whose energy is usually propagated through a medium.

When two sound waves of different frequencies meets, a periodic variation that occurs is called beats.

Therefore, Beats can be defined as the periodic fluctuations in the frequency of sound waves.

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The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².

To calculate the average acceleration of the motorcycle, we can use the formula:

Average acceleration = (final velocity - initial velocity) / time

First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.

Converting the final velocity:

Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s

Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:

Average acceleration = (21.67 m/s - 0 m/s) / time

To find the time taken to reach this velocity, we need to use the formula for average speed:

Average speed = total distance/time

Rearranging the formula:

time = total distance / average speed

Plugging in the values:

time = 50 m / 21.67 m/s ≈ 2.31 seconds

Now we can calculate the average acceleration:

Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²

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The motorcycle had covered a distance of 16 meters in half the total time.

a) To calculate the acceleration, we can use the formula:

a = (v - u) / t

where a is the acceleration, v is the final velocity, u is the initial velocity (which is 0 since the motorcycle starts from rest), and t is the time.

Given:

u = 0 m/s (initial velocity)

v = ? (final velocity)

t = 4 s (time)

s = 64 m (distance)

Using the equation of motion:

s = ut + 1/2at^2

We can rearrange the equation to solve for acceleration:

a = 2s / t^2

a = 2(64) / (4)^2

a = 128 / 16

a = 8 m/s^2

Therefore, the acceleration of the motorcycle is 8 m/s^2.

b) To find the final velocity, we can use the formula:

v = u + at

v = 0 + (8)(4)

v = 32 m/s

Therefore, the final velocity of the motorcycle is 32 m/s.

c) To determine the time at which the motorcycle had covered half the total distance, we divide the total distance by 2 and use the formula:

s = ut + 1/2at^2

32 = 0 + 1/2(8)t^2

16 = 4t^2

t^2 = 4

t = 2 s

Therefore, the motorcycle had covered half the total distance at 2 seconds.

d) To calculate the distance covered in half the total time, we use the formula:

s = ut + 1/2at^2

s = 0 + 1/2(8)(2)^2

s = 0 + 1/2(8)(4)

s = 0 + 16

s = 16 m

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Answer:

\(c=0.45\ J/g^{\circ} C\)

Explanation:

Given that,

A 25 g sample of iron (initially at 800.00°C) is dropped into 200 g of water (initially at  30.00°C). The final temperature of the system is 40.22°C.

We need to find the specific heat of iron.

It can be calculated as:

Cooler water gains = hot metal loses

mc∆T = - mc∆T

Put all the values,

\(200g(4.184\ J/g^{\circ} C)(T_f-T_i) = -25g(c)(T_f-T_i) \\\\200g(4.184 )( 40.22-30.00) = -25\times (c)\times (40.22-800.00)\\\\8552.096 = 18994.5c\\\\c=\dfrac{8552.096 }{18994.5}\\\\c=0.45\ J/g^{\circ} C\)

So, the specific heat of iron is \(0.45\ J/g^{\circ} C\)

Answer:

Explanation:

Let the separation required be d .

Force between rod = 10⁻⁷ x  2 I₁ I₂ L / d

where I₁ and I₂ are current in them , d is distance of separation and L is length of wire .

Force between rod = 10⁻⁷ x  2 x 1200 x  1200 x .69  / d

= .1987 /d

Restoring Force by spring = k x where k is force constant and x is compression .

= 130 x .03

= 3.9 N

For balancing

Restoring Force by spring = Force between rod

.1987 /d = 3.9

d = .1987 /3.9

= .0509 m

= 5.09 cm .

To determine the thrust applied by the engines, we can use Newton's second law of motion, which states that force (thrust) is equal to mass times acceleration. In this case, we need to calculate the force required to decelerate the plane from 30 m/s to 25 m/s in 12 seconds.

First, we calculate the change in velocity (∆v):

\(\displaystyle\sf \Delta v=25\,m/s-30\,m/s=-5\,m/s\)

Next, we calculate the acceleration (∆a) using the formula:

\(\displaystyle\sf \Delta a=\frac{\Delta v}{\Delta t}\)

where ∆t is the change in time, which is 12 seconds in this case.

\(\displaystyle\sf \Delta a=\frac{-5\,m/s}{12\,s}\)

Now, we can determine the force (thrust) applied by the engines using Newton's second law:

\(\displaystyle\sf F=m\cdot a\)

where m is the mass of the airplane, which is 480000 kg.

\(\displaystyle\sf F=480000\,kg\cdot \left(\frac{-5\,m/s}{12\,s}\right)\)

Calculating the result:

\(\displaystyle\sf F=-200000\,N\)

Therefore, the engines applied a thrust of -200000 Newtons (N) to decelerate the plane. The negative sign indicates that the thrust is in the opposite direction of the motion.

Answer:

The brachistochrone curve is an idealized curve that provides the fastest descent possible.

Explanation:

There is actually an analytical solution to this case or, with some derivation work, we can use the PDE functionality of COMSOL Multiphysics to solve the problem.

Applying newton's second law:

Now, we can answer the questions.

1.

Answer:

50N

2.

Answer:

Right

3.

Answer:

Accelerate

4.

Answer:

Unbalanced force

Answer:

violet

Explanation:

(a) To solve for the mass of the rod, we can use the formula for rotational kinetic energy:

KE = (1/2) * I * w^2

where KE is the rotational kinetic energy, I is the moment of inertia, and w is the angular velocity.

At the beginning, when the bug is at the axis of rotation, the rotational kinetic energy of the rod is:

KE1 = (1/2) * I * w^2 = (1/2) * (3.10×10−3 kg⋅m^2) * (0.37 rad/s)^2 = 0.036 J

When the bug reaches the other end of the rod, the rotational kinetic energy is:

KE2 = (1/2) * I * w^2 = (1/2) * (3.10×10−3 kg⋅m^2) * (0.132 m/s)^2 / (0.480 m)^2 = 3.62×10^-5 J

The change in kinetic energy is due to the work done by the bug as it crawls along the rod. The work done by the bug can be calculated as the product of the force it exerts on the rod and the distance it crawls:

W = F * d

The force the bug exerts on the rod can be calculated using Newton's second law for rotational motion:

τ = I * α

where τ is the torque, α is the angular acceleration, and I is the moment of inertia.

Since the rod is rotating with a constant angular velocity, its angular acceleration is zero, so the net torque on the rod must be zero. This means that the torque exerted by the bug on the rod must be equal and opposite to the torque due to the angular momentum of the rod:

τ_bug = τ_rod

F * d = I * w

F = I * w / d

Substituting the values given, we get:

F = (3.10×10−3 kg⋅m^2) * (0.37 rad/s) / (0.480 m) = 0.0237 N

Now we can use the work-energy principle to find the mass of the rod:

W = KE2 - KE1 = F * d = (1/2) * m * v^2

where m is the mass of the rod and v is the tangential velocity of the bug at the end of the rod.

Substituting the values given, we get:

(1/2) * m * (0.132 m/s)^2 = 3.62×10^-5 J - 0.036 J

Simplifying and solving for m, we get:

m = 0.221 kg

Therefore, the mass of the rod is 0.221 kg.

(b) To find the mass of the bug, we can use the same equation we used to calculate the force it exerted on the rod:

F = I * α / d

where α is the angular acceleration of the rod caused by the bug crawling along it.

Since the rod is a thin uniform rod, its moment of inertia can be calculated as:

I = (1/3) * m * L^2

where L is the length of the rod.

Substituting the values given, we get:

I = (1/3) * (0.221 kg) * (0.480 m)^2 = 0.0202 kg⋅m^2

Now we can calculate the angular acceleration caused by the bug crawling along the rod. Since the rod is rotating with a constant angular velocity, its angular acceleration is given by:

α

Fade

finish

α = (w_f^2 - w_i^2) / (2 * θ)

where w_i is the initial angular velocity of the rod, w_f is the final angular velocity of the rod after the bug has crawled to the end, and θ is the angle through which the bug crawls, which is equal to the length of the rod:

θ = L = 0.480 m

Substituting the values given, we get:

α = (0.132 m/s)^2 / (2 * 0.480 m) = 0.072 rad/s^2

Now we can calculate the force exerted by the bug on the rod:

F = I * α / d = (0.0202 kg⋅m^2) * (0.072 rad/s^2) / (0.480 m) = 0.00303 N

Finally, we can use Newton's second law to find the mass of the bug:

F = m * a

where a is the tangential acceleration of the bug, given by:

a = r * α

where r is the distance from the bug to the axis of rotation, which is equal to the length of the rod minus the distance the bug has crawled, or:

r = L - d = 0.480 m - 0.480 m = 0 m

Therefore, the tangential acceleration of the bug is zero, and its mass is:

m = F / a = 0.00303 N / 0 m/s^2 = undefined

This means that the force exerted by the bug on the rod is not enough to cause any tangential acceleration, and therefore the mass of the bug is negligible compared to the mass of the rod. We can assume that the bug has zero mass for practical purposes.