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STatiana [176]
3 years ago
9

An electric car uses a 45-kW (160-hp) motor. If the battery pack is designed for 340V, what current would the motor need to draw

from the battery? Neglect any energy losses in getting energy from the battery to the motor.
Physics
1 answer:
alukav5142 [94]3 years ago
5 0

Answer:

Current = 132.35 A

The motor needs to draw 132.35 Amperes current from the battery.

Explanation:

The formula of electric power is given as follows:

Power = (Voltage)(Current)

Current = Power/Voltage

In this question, we have:

Power = 45 KW = 45000 W

Voltage of Battery Pack = 340 V

Current needed to be drawn = ?

Therefore,

Current = 45000 W/340 V

<u>Current = 132.35 A</u>

<u>The motor needs to draw 132.35 Amperes current from the battery.</u>

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50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t
r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

PE=mgh

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

KE=\frac{1}{2}mv^2

where v is the speed of the object

When the ball is sitting on the top of the building, we have

  • h=40 m, therefore the potential energy is not zero
  • v=0, since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

h=20 m

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

E=PE+KE=const.

At the top of the building,

E=PE_{top}

While halfway through the fall,

PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}

And the mechanical energy is

E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}

which means

KE_{half}=\frac{E}{2}

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

  • The height of the ball relative to the ground is now zero: h=0. This means that the potential energy of the ball is zero: PE=0
  • The kinetic  energy, instead, is not zero: in fact, the ball has gained speed during the fall, so v\neq 0, and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

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

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

PE=(2)(9.8)(40)=784 J

5)

The potential energy of the ball as it is halfway through the fall is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

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

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

PE=(2)(9.8)(20)=392 J

6)

The kinetic energy of the ball halfway through the fall is given by

KE=\frac{1}{2}mv^2

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the  fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

KE=\frac{1}{2}(2)(19.8)^2=392 J

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

KE=\frac{1}{2}mv^2

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

KE=\frac{1}{2}(2)(28)^2=784 J

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0
3 years ago
A truck traveled 400 meters north in 80 seconds and then it traveled 300 meters east in 70 seconds. The magnitude of the average
Varvara68 [4.7K]

Answer:

3.3m/s

Explanation:

You first get the total time (80 + 70 = 150s).

Then you would find the displacement of the truck. To do that you would do component method (vector addition), so since its a right triangle (North and East), displacement is 400^2 + 300^2 = d^2.

d= 500m.

So now that you have displacement and time, you can find the velocity:

v=d/t

v=500/150

v=3.3

5 0
3 years ago
The same amount of substance was added to four beakers of water. The treatments were placed in the chart.
vfiekz [6]

Answer: Solution W and Y solution have more solubility than X and Z

Solutions are homogeneous mixtures of two or more components. By uniform mix we mean that its structure and properties are the same in the whole mix. Generally, the component which is present in the largest quantity is known as solvent. Solvent determines the physical condition in which the solution exists. In addition to the solvent, one or more component present in the solution is called solutes. In this unit we will only consider binary solutions (i.e., with two components)

The structure of the solution can be described by expressing its concentration. The latter can either be expressed qualitatively or quantitatively. For example, in qualitatively we can say that the solution is diluted (i.e., relatively small amounts of solubility) or it is concentrated (i.e., relatively rarely sighs). But in real life such details may be very confusing and thus require a quantitative description of the solution. There are several ways that we can quantitatively describe the concentration of solutions. (i) Mass Percentage (W / W): The mass percentage of a component of the solution is defined as: mass of the component = mass of the component in the solution = 100 Total mass of the solution .For example, if by mass A solution is described by 10% glucose in water, it means that 10 grams of glucose dissolved in 90 grams of water, resulting in 100 grams of solution. The concentration described by a large percentage of the population is usually used in industrial chemical applications. For example, the commercial bleaching solution contains 3.62 mass percentages of sodium hypochlorite in water. (ii) Volume Percentage (V / V): Volume Percentage is defined as: Total Volume of Component Volume 100 (component) Volume% of Component  

Explanation:

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D. 65.1 is the answer
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What happens when the mass of an object is greter
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The acceleration goes up.
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