PHYSICS COLLEGE

In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its speed is decreased linearly from 60 mph to 30 mph in 10 seconds. Calculate the theoretical maximum energy in kWh that can be recovered during this interval. Ignore all losses.

Answers

Answer 1

Answer:

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

m = 4000 / 2.20462 = 1814.37 kg

Initial velocity $V_(i)$ = 60 mph = 26.8224 m/s

Final velocity $V_(f)$ = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = $(1)/(2)$ m( $V_(i)$ ² - $V_(f)$ ² )

we substitute

Δk = $(1)/(2)$ ×1814.37( (26.8224)² - (13.4112)² )

Δk = $(1)/(2)$ × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

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A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 150-mL mark with 34.5°C acetone. After the flask is filled, the acetone cools and the flask warms so that the combination of acetone and flask reaches a uniform temperature of 32.0°C. The combination is then cooled back to 20.0°C. (The average volume expansion coefficient of acetone is 1.50 10-4(°C)−1.) (a) What is the volume of the acetone when it cools to 20.0°C?

Answers

Answer:149.73 ml

Explanation:

Given

$\beta \ of\ acetone=1.50* 10^(-4) ^(\circ)C^(-1)$

change in volume is given by

$\Delta V=V_(final)-V_(initial)$

$\Delta V=\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]$

$V_(final)=\nu_(initial)+\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]$

$V_(final)=150+150* 1.50* 10^(-4)\left [ 20-32\right ]$

$V_(final)=149.73 ml$

Final answer:

The volume of the acetone when it cools to 20.0°C is approximately 142.39 mL.

Explanation:

In order to determine the volume of the acetone when it cools to 20.0°C, we can use the equation for the volume change caused by a temperature change at constant pressure, known as Charles's law. Charles's law states that the volume of a gas is directly proportional to its temperature in Kelvin. We can use the formula V2 = V1 * (T2 / T1) to calculate the volume of the acetone at the lower temperature.

Given that the initial volume of the acetone is 150 mL at a temperature of 34.5°C, we need to convert this temperature to Kelvin by adding 273.15. Therefore, T1 = 34.5°C + 273.15 = 307.65 K.

Since the final temperature is 20.0°C, the final temperature in Kelvin will be T2 = 20.0°C + 273.15 = 293.15 K. We can now plug these values into the equation to find the volume of the acetone at the lower temperature: V2 = 150 mL * (293.15 K / 307.65 K) = 142.39 mL.

Learn more about Volume calculation here:

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Captain Kiddo is trapped inside a submarine. In order to escape, she first needs to open a heavy revolving door. a. Would be easier for her to apply all her force close to the axle of the door or as far away as possible from it?
b. To finally get the hatch open she needs to grab hold of the wheel with her hands on opposite sides and rotate the wheel by exerting a force toward the top of the door on one side while exerting the same force towards the bottom of the door on the opposite side. Is there a net force applied on the wheel? Is there a net torque applied? Explain your answers

Answers

Answer:

a) It will be easier to apply all her force as far as possible from the axle

b) There is a net torque applied

Explanation:

a) Applying a larger force far away from the axle of the door produces a larger force on the torque than pushing it near the axle. This is because Increasing the lever arm between the axle and the point at which you push the door increases the torque on the door

b)The door is turning faster and faster because you are exerting a torque on it and its undergoing angular acceleration. There is a net torque which is the addition of the torque applied on the opposite sides of the door

A manufacturer claims that a carpet will not generate more than 5.8 kV of static electricity What magnitude of charge would have to be transferred between a carpet and a shoe for there to be a 5.8 kV potential distance d = 2.8 mm ? Approximate the area of a shoe as 30 cm x 8 cm. Express your answer using two significant figures.

Answers

Answer:

4.4×10⁻⁷ Coulomb

Explanation:

V = Voltage = 5.8 kV

d = Potential distance = 2.8 mm = 0.0028 m

A = Area = 0.3×0.08 = 0.024 m²

ε₀ = permittivity constant in a Vacuum= 8.85×10⁻¹² F/m

$(Q)/(V)=(A\epsilon_0)/(d)\n\Rightarrow \Q=V(A\epsilon_0)/(d)\n\Rightarrow Q=5.8* 10^3(0.024* 8.85* 10^(-12))/(0.0028)\n\Rightarrow Q=4.4* 10^(-7)\ C$

Magnitude of charge transferred between a carpet and a shoe is 4.4×10⁻⁷ Coulomb.

Charge q1 is placed a distance r0 from charge q2 . What happens to the magnitude of the force on q1 due to q2 if the distance between them is reduced to r0/4 ?What is the electrostatic force between and electron and a proton separated by 0.1 mm?

Answers

Answer:

The electrostatic force between and electron and a proton is $F=2.30* 10^(-20)\ N$

Explanation:

It is given that, charge $q_1$ is placed at a distance $r_o$ from charge $q_2$ . The force acting between charges is given by :

$F=(kq_1q_2)/(r_o^2)$

We need to find the force if the distance between them is reduced to $r_o/4$ . It is given by :

$F'=(kq_1q_2)/((r_o/4)^2)$

$F'=16* (kq_1q_2)/(r_o^2)$

$F'=16* F$

So, if the the distance between them is reduced to $r_o/4$ , the new force becomes 16 times of the previous force.

The electrostatic force between and electron and a proton separated by 0.1 mm or $10^(-4)\ m$ is :

$F=(kq_1q_2)/(r_o^2)$

$F=(9* 10^9* (1.6* 10^(-19))^2)/((10^(-4))^2)$

$F=2.30* 10^(-20)\ N$

So, the electrostatic force between and electron and a proton is $F=2.30* 10^(-20)\ N$ . Hence, this is the required solution.

A 110 kg ice hockey player skates at 3.0 m/s toward a railing at the edge of the ice and then stops himself by grasping the railing with his outstretched arms. During the stopping process, his center of mass moves 30 cm toward the railing. (a) What is the change in the kinetic energy of his center of mass during this process? (b) What average force must he exert on the railing?

Answers

Answer:

(a). The change in the kinetic energy of his center of mass during this process is -495 J.

(b). The average force is 1650 N.

Explanation:

Given that,

Mass = 110 kg

Speed = 3.0 m/s

Distance = 30 cm

(a). We need to calculate the change in the kinetic energy of his center of mass during this process

Using formula of kinetic energy

$\Delta K.E=K.E_(2)-K.E_(1)$

$\Delta K.E=(1)/(2)mv_(f)^2-(1)/(2)mv_(i)^2$

Put the value into the formula

$\Delta K.E=(1)/(2)*110*0^2-(1)/(2)*110*(3.0)^2$

$\Delta K.E=-495\ J$

(b). We need to calculate the average force must he exert on the railing

Using work energy theorem

$W=\Delta K.E$

$Fd=\Delta K.E$

$F=(\Delta K.E)/(d)$

Put the value into the formula

$F=(-495)/(30*10^(-2))$

$F=-1650\ N$

The average force is 1650 N.

Hence, (a). The change in the kinetic energy of his center of mass during this process is -495 J.

(b). The average force is 1650 N.

Answer

given,

mass of ice hockey player = 110 Kg

initial speed of the skate = 3 m/s

final speed of the skate = 0 m/s

distance of the center of mass, m = 30 cm = 0.3 m

a) Change in kinetic energy

$\Delta KE = (1)/(2)mv_f^2 - (1)/(2)mv_i^2$

$\Delta KE = (1)/(2)m(0)^2 - (1)/(2)* 110 * 3^2$

$\Delta KE = - 495\ J$

b) Average force must he exerted on the railing

using work energy theorem

W = Δ KE

F .d = -495

F x 0.3 = -495

F = -1650 N

the average force exerted on the railing is equal to 1650 N.

A 1.5v battery stores 4.5KJ of energy. How long can it light a flashlight bulb that draws 0.60A

Answers

Answer:

The 1.5V battery can power the flashlight bulb drawing 0.60A for 83.33 minutes before it is depleted.

Explanation:

To determine how long a 1.5V battery can power a flashlight bulb drawing 0.60A, you can use the formula for calculating the energy (in joules) consumed by an electrical device over time:

Energy (Joules) = Power (Watts) × Time (Seconds)

In this case, the power (P) is given by the product of the voltage (V) and current (I):

Power (Watts) = Voltage (Volts) × Current (Amperes)

So, first, calculate the power consumption of the flashlight bulb:

Power (Watts) = 1.5V × 0.60A = 0.90 Watts

Now, you want to find out how long the battery can power the bulb, so rearrange the energy formula to solve for time:

Time (Seconds) = Energy (Joules) / Power (Watts)

Given that the battery stores 4.5 kJ (kilojoules), which is equivalent to 4,500 joules, and the power consumption is 0.90 watts:

Time (Seconds) = 4,500 J / 0.90 W = 5,000 seconds

Now, to express the time in more practical units, convert seconds to minutes:

Time (Minutes) = 5,000 seconds / 60 seconds/minute ≈ 83.33 minutes

So, the 1.5V battery can power the flashlight bulb drawing 0.60A for approximately 83.33 minutes before it is depleted.

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