Answer:

**(a) 7.18**

The electric field within a parallel plate capacitor with dielectric is given by:

(1)

where

is the surface charge density

k is the dielectric constant

is the vacuum permittivity

The area of the plates in this capacitor is

while the charge is

So the surface charge density is

The electric field is

So we can re-arrange eq.(1) to find k:

**(b) **

The surface charge density induced on each dielectric surface is given by

where

is the initial charge density

k = 7.18 is the dielectric constant

Substituting,

And by multiplying by the area, we find the charge induced on each surface:

Find a glass jar with a screw-top metal lid. Close the lid snugly and put the jar into the refrigerator. Leave it there for about 10 minutes and then take the jar out and try to open the lid. (a) Did the lid become tighter or looser? Explain your observation.

The graph to the right shows the change in Canada‘s harvest of Atlantic cod from 1950-2004 what year shows the clearest evidence of a collapse of fishing stocks?A.1965B.1985C.1995D.2005

*PLEASE HELP*A baseball is pitched with a horizontal velocity of 25.21 m/s. Mike Trout hits the ball, sending it in the opposite direction (back toward the pitcher) at a speed of -50.67 m/s. The ball is in contact with the bat for 0.0014 seconds. What is theacceleration of the ball?

Alex throws a 0.15-kg rubber ball down onto the floor. The ball’s speed just before impact is 6.5 m/s, and just after is 3.5 m/s. If the ball is in contact with the floor for 0.025 s, what is the magnitude of the average force applied by the floor on the ball?

A car with tires of radius 0.25 m come to a stop from 28.78 m/s (100 km/hr) in 50.0 m without any slipping of tires. Find: (a) the angular acceleration of the wheels; (b) number of revolutions made while coming to rest.

The graph to the right shows the change in Canada‘s harvest of Atlantic cod from 1950-2004 what year shows the clearest evidence of a collapse of fishing stocks?A.1965B.1985C.1995D.2005

*PLEASE HELP*A baseball is pitched with a horizontal velocity of 25.21 m/s. Mike Trout hits the ball, sending it in the opposite direction (back toward the pitcher) at a speed of -50.67 m/s. The ball is in contact with the bat for 0.0014 seconds. What is theacceleration of the ball?

Alex throws a 0.15-kg rubber ball down onto the floor. The ball’s speed just before impact is 6.5 m/s, and just after is 3.5 m/s. If the ball is in contact with the floor for 0.025 s, what is the magnitude of the average force applied by the floor on the ball?

A car with tires of radius 0.25 m come to a stop from 28.78 m/s (100 km/hr) in 50.0 m without any slipping of tires. Find: (a) the angular acceleration of the wheels; (b) number of revolutions made while coming to rest.

**(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)**

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

where is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

**(b) 4159 N**

The buoyant force is given by

where is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

is the air density

is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

**(c) 1524 N**

The mass of the helium gas inside the balloon is

where is the helium density; so we the total mass of the balloon+helium gas inside is

So now we can find the weight of the balloon:

And so, the net force on the balloon is

**(d) The balloon will rise**

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

**(e) 155 kg**

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

where m' is the additional mass. Re-arranging the equation for m', we find

**(f) The balloon and its load will accelerate upward.**

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

**(g) The decrease in air density as the altitude increases**

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

The physics involved in the functioning of helium balloons is based on buoyancy and Archimedes' Principle. The forces at play include the force due to gravity, the buoyant force and the net force, which determines the motion of the balloon. The balloon's height limit is determined by the decrease in air density with altitude.

The several parts of this question are related to the principles of buoyancy and Archimedes' Principle. First, regarding the force diagram for the balloon (part a), it would show two primary forces. The force due to gravity (**Fg**) acting downwards and the buoyant force (**Fb**) acting upwards, which is a result of the displacement of air by the balloon. The net force mentioned in part (c) is calculated as the difference between these two forces.

Calculating the buoyant force (part b) involves multiplying the volume of the balloon by the density of the air and the acceleration due to gravity (Fb = V * ρ_air * g). For the net force on the balloon (part c), this is calculated by subtracting the weight of the balloon from the buoyant force (F_net = Fb - Fg). If the net force is positive, the balloon will rise, if it's negative, the balloon will fall, and if it is zero, the balloon will remain stationary.

The maximum additional mass the balloon can support in equilibrium (part d) is calculated using the net force divided by gravity. If the mass of the load is less than this value (part e), the balloon and its load will accelerate upward.

Lastly, the limit to the height to which the balloon can rise (part f) is determined by the decreasing density of the air as the balloon ascends. The buoyant force reduces as the balloon rises because the air density is lower at higher altitudes.

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**Answer: the answer i for c is yes 0& 10 **

**Explanation:**

B. Ultraviolet light waves

C. Infrared Waves

D. Microwaves

The heat emitted from anything is carried in the form of infrared waves. (C)

**Answer:**

31.29 m/sec

**Explanation:**

We have given density of substance

We have convert this into

We know that 1 lb = 0.4535 kg. so 0.14 lb = 0.14×0.4535 = 0.06349 kg

We know that 1 kg = 1000 g ( 1000 gram )

So 0.06349 kg = 63.49 gram

And we know that 1 gram = 1000 milligram

So 63.49 gram

We know that

So

So =\frac{63.49\times 10^3}{0.2249\times 10^{-5}}=276.74\times 10^8lb/m^3[/tex]

In second part we have to convert 70 mi/hr to m/sec

We know that 1 mi = 1609.34 meter

So 70 mi = 70×1609.34 = 112653.8 meter

1 hour = 3600 sec

So 70 mi/hr

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:

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 **charge****distribution** remains the same, only the length changes.

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

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

let's find the **ratio** :

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

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

B. 3.75 × 10–7 N toward C

C. 2.00 × 10–7 N toward D

D. 1.15 × 10–7 N toward D

The magnitude and **direction **of the net gravitationalforce on mass A due to the other **masses **is 1.15 × 10⁻⁷ N toward D.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the **distance **between them. In this case, all of the masses are equal to 3 kg, and the distance between mass A and mass D is 3 m.

The **gravitational **force between mass A and mass D is therefore:

F = G * m_A * m_D / r²

= 6.674 × 10⁻¹¹ N m² / kg² * 3 kg * 3 kg / 3 m²

= 1.15 × 10⁻⁷ N

The direction of the gravitational **force **is towards mass D.

Therefore, the net gravitational force on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

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**Answer:**

THE ANSER IS B

**Explanation:**