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Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.30 m/s in 0.812 s. (a) What is the magnitude of the linear impulse experienced by a 62.0-kg passenger in the car during the time the car accelerates? kg · m/s (b) What is the magnitude of the average total force experienced by a 62.0-kg passenger in the car during the time the car accelerates? N

Answers

Answer 1

Answer:

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

$I=\Delta p = m\Delta v$

where

m = 62.0 kg is the mass of the passenger

$\Delta v$ is the change in velocity of the car (and the passenger), which is

$\Delta v = 5.30 m/s - 0 = 5.30 m/s$

So, the linear impulse experienced by the passenger is

$I=(62.0 kg)(5.30 m/s)=328.6 kg m/s$

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

$I=F \Delta t$

where in this case

$I=328.6 kg m/s$ is the linear impulse

$\Delta t = 0.812 s$ is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

$F=(I)/(\Delta t)=(328.6 kg m/s)/(0.812 s)=404.7 N$

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Two identical metal spheres a and b are in contact. both are initially neutral. 1.0×1012 electrons are added to sphere a, then the two spheres are separated. you may want to review ( pages 639 - 641) . part a afterward, what is the charge of sphere a?

Answers

The charge on the sphere A and sphere B after they are separated is $\boxed{ - 80\,{\text{nC}}}$ each.

Further Explanation:

Given:

The number of electrons transferred to sphere $A$ is $1.0 * {10^(12)}$ .

Concept:

The amount of charge carried by the electrons when reaches the spheres kept in contact with each other is first distributed equally on each sphere. Later as the spheres are moved away from one another, the charge on each sphere remains the same as it was when they were in contact.

The amount of charge on one electron is $- 1.6 * {10^( - 19)}\,{\text{C}}$ .

So, the amount of charge carried by the $1.0 * {10^(12)}$ electrons is given as.

$\begin{aligned}Q&= \left( {1.0 * {{10}^(12)}} \right)\left( { - 1.6 * {{10}^( - 19)}} \right)\n&= - 1.6 * {10^( - 7)}\,{\text{C}}\n\end{aligned}$

Since the charge is disturbed equally on the two sphere, so the amount of charge carried by each sphere s half of the total charge.

$\begin{aligned}{Q_A}&= \frac{{ - 1.6 * {{10}^( - 7)}}}{2}\n&= - 8 * {10^( - 8)}\,{\text{C}}\n &= - 8{\text{0}}\,{\text{nC}}\n\end{aligned}$

Thus, the amount of the charge carried by each sphere after separating from each other is $\boxed{ - 80\,{\text{nC}}}$ .

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Answer Details:

Grade: College

Chapter: Electrostatics

Subject: Physics

Keywords: Metal spheres, two identical, in contact, neutral, charged, electrons, charge on electron, charge on metallic sphere, charge of sphere A.

The charge on the sphere A and sphere B after they are separated is $-80\mu C$ each

What is Charge?

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field.

The amount of charge carried by the electrons when reaches the spheres kept in contact with each other is first distributed equally on each sphere. Later as the spheres are moved away from one another, the charge on each sphere remains the same as it was when they were in contact.

The amount of charge on one electron is $-1.6* 10^(-19) \ C$

So, the amount of charge carried by the electrons is given as.

$Q=(1* 10^(12))(-1.6* 10^(-19))$

$Q=-1.6* 10^(-7)\ C$

Since the charge is disturbed equally on the two sphere, so the amount of charge carried by each sphere s half of the total charge.

$Q_A=(-1.6* 10^(-7))/(2)$

$Q_A=-8* 10^(-8)\ C$

$Q_A=-80\ \mu C$

Thus, the amount of the charge carried by each sphere after separating from each other is $-80\mu C$

To know more about Charge follow

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he magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 1.9 AA , how many turns of wire would you need? Express your answer using two significant figures.

Answers

Answer: 3400

Explanation:

Given

Magnetic field, B = 0.1 T

Diameter of magnet, d = 2 cm = 0.02 m

Length of magnet, l = 8 cm = 0.08 m

Current of the magnet, I = 1.9 A

Number of turns needed, N = ?

To solve this problem, we would use the formula,

N = (LB) / (μI), where

μ = 1.257*10^-6 Tm/A, so that

N = (0.08 * 0.1) / (1.257*10^-6 * 1.9)

N = 0.008 / 2.388*10^-6

N = 3350

N ~ 3400

Therefore, the number of turns of wire needed is 3400

A car is initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at a constant acceleration.(a) how large was the acceleration in m/s ^2
(b)how large was the acceleration, in units go g= 9.80 m/s ^2

Answers

acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
= 37.5 m/s
velocity initial = 35 km/hr x 1hr /3600 s x 1000 m/1 km
= 9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units = (2.646m/s^2)/(9.8m/s^2)
= 0.27 units

An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and inductor, or a capacitor. At time t = 0 the voltage is zero. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -Imax, where Imax is the current amplitude. What is the unknown element?

Answers

Answer:

This satisfy the above given condition so we can say that this capacitor.

Explanation:

Let's take one by one option and check whether is wrong or right

For inductor:

$I=I_osin(wt-(\pi )/(2))$

Given that at t=T/4 ,I=0 and we know that

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

So at T/4

$I=I_osin((2\pi )/(T)* (T)/(4)-(\pi )/(2))$

I=0 A

At t=T/2

$I=I_osin((2\pi )/(T)* (T)/(2)-(\pi )/(2))$

$I=I_o$

It means that this not a indutor.

For capacitor:

$I=I_osin(wt+(\pi )/(2))$

At T/4, I=0

At t=T/2

$I=I_osin((2\pi )/(T)* (T)/(2)+(\pi )/(2))$

$I= -I_o$

This satisfy the above given condition so we can say that this capacitor.

Final answer:

The nature of the unknown ideal element in the given AC circuit can be determined based on the phase difference between the current and voltage. In this case, since the current is zero at T/4 and a maximum at T/2, it suggests the current is lagging the voltage, indicating that the element in the circuit is a capacitor.

Explanation:

The question relates to an alternating current (AC) circuit connected to an unknown ideal element, and from the given conditions, it appears this element is a capacitor. Let us understand why.

Firstly, in an AC circuit, we can determine the nature of the circuit elements—resistor, inductor, or capacitor—based on the phase difference between the current and voltage. In a resistor, the current and voltage are in phase. In an inductor, the current lags behind the voltage by 90 degrees (or π/2 radians), whereas in a capacitor, the current leads the voltage by 90 degrees (or π/2 radians).

Based on the given problem, at time t = T/4, the current is zero. Considering that in one period of the AC voltage, it starts from zero, reaches a maximum, comes back to zero (at T/2), goes to a minimum (at 3T/4), and returns back to zero (at T), the current would reach its peak either at T/4 (if it's leading, a capacitor) or at 3T/4 (if it's lagging, an inductor). Here, since the current is zero at T/4 and it is a maximum at T/2 (albeit negative), it suggests the current is lagging the voltage, and hence, it suggests the element in the circuit is a capacitor.

Learn more about AC Circuits here:

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A wagon is pulled at a speed of 0.40 m/s by a horse exerting 1800 Newtons of horizontal Force. how much work was done by the horse

Answers

The amount of work done per second by the horse exerting a force of 1800 N on a wagon moving with a speed of 0.4 m/s is 720 J/s.

What is power?

Power is the workdone by a body in one second.

To calculate the work done by the horse in one seconds, we use the formula below

Formula:

P = Fv................ Equation 1

Where:

P = work done on the horse in one second
F = Force of the horse
v = Velocity of the wagon

From the question,

Given:

F = 1800 N
v = 0.4 m/s

Substitute these values into equation 1

P = 1800×0.4
P = 720 J/s

Hence, the amount of work done per second by the horse is 720 J/s.

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Complete question: A wagon is pulled at a speed of 0.40 m/s by a horse exerting 1800 Newtons of horizontal Force. how much work was done by the horse per second.

Four +2 µC point charges are at the corners of a square of side 2 m. Find the potential at the center of the square (relative to zero potential at infinity) for each of the following conditions.(a) All the charges are positive(b) Three of the charges are positive and one is negative(c) Two are positive and two are negative

Answers

Answer:

(a) 51428.59 J/C

(b) 25714.29 J/C

(c) 0 J/C

Explanation:

Parameters given:

Q1 = 2 * 10^-6 C

Q2 = 2 * 10^-6 C

Q3 = 2 * 10^-6 C

Q4 = 2 * 10^-6 C

=> Q1 = Q2 = Q3 = Q4 = Q

Side of the square = 2m

The center of the square is the midpoint of the diagonals, i.e. Using Pythagoras theorem:

BD² = 2² + 2²

BD² = 8

BD = √(8) = 2.8m

OD = 1.4m

(The attached diagram explains better)

Hence, the distance between the center and each point charge, r, is 1.4m.

Electric Potential, V = kQ/r

k = Coulombs constant

(a) If all charges are positive:

V(Total) = V1 + V2 + V3 + V4

V1 = Potential due to Q1

V2 = Potential due to Q2

V3 = Potential due to Q3

V4 =Potential due to Q4

Since Q1 = Q2 = Q3 = Q4 = Q

=> V1 = V2 = V3 = V4

=> V(Total) = 4V1

V = (4 * 9 * 10^9 * 2 * 10^-6)/1.4

V = 51428.59J/C

(b) If 3 charges are positive and 1 is negative:

Since Q1 = Q2 = Q3 = Q

and Q4 = -Q

The total potential becomes:

V(Total) = V1 + V2 + V3 - V4

Since V1, V2, V3 and V4 have the same value,

V(Total) = V1 + V2

V(Total) = 2V1

V(Total) = (2 * 9 * 10^9 * 2 * 10^-6)/1.4

V(Total) = 25174.29 J/C

Since Q1 = Q2 = Q

and Q3 = Q4 = -Q

The total potential becomes:

V(Total) = V1 + V2 - V3 - V4

Since V1, V2, V3 and V4 have the same value,

V(Total) = 0 J/C

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