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3 years ago

The Tesla Roadster electric car is rated at 16.8kW how much energy does it use during a one hour car drive

Physics

1 answer:

Damm [24]3 years ago

7 0

Answer: $6.05\cdot 10^7 J$

Explanation:

Energy is related to power by:

$E=Pt$

where P is the power and t is the time. In this problem, we have:

$P=16.8 kW=16800 W$

$t=1 h=3600 s$

So, the energy used in one hour is

$E=(16800 W)(3600 s)=6.05\cdot 10^7 J$

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Svetllana [295]

Answer:

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Explanation:

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2 years ago

Inertia ____.

OlgaM077 [116]

The correct answer is (c.) resists a change in motion of an object. Inertia basically resists any physical changes in an object in terms of its state. As stated in the Newton's first law of motion or also known as law of inertia, an object that is at rest will stay at rest and an object that is in motion will stay in motion. In short, an object will keep doing what it is already doing UNLESS net force is acted upon it,

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3 years ago

If a wave has amplitude of 2 meters, a wavelength of 5 meters, and a frequency of 60 Hz, at what speed is it moving?

klemol [59]

The relationship between speed of a wave w, frequency f and wavelength $\lambda$ is
$v = \lambda f$
For the wave of our problem, $\lambda = 5 m$ and f=60 Hz, so its speed is
$v=(5 m)(60 Hz)=300 m/s$

5 0

3 years ago

Read 2 more answers

A gas is compressed in a cylinder from a volume of 20.0 l to 2.0 l by a constant pressure of 10.0 atm. calculate the amount of w

jok3333 [9.3K]

First, we need to convert the pressure in SI units. Keeping in mind that $1 atm = 1.01 \cdot 10^5 Pa$ :
$p=10.0 atm =1.01 \cdot 10^6 Pa$

The initial and final volume of the gas are (keeping in mind that $1.0 L = 0.001 m^3$ ):
$V_i = 20.0 L=0.020 m^3$
$V_f = 2.0 L=0.002 m^3$

So, the work done on the gas by the surrounding is
$W= -p \Delta V=-p(V_f-V_i)=-(1.01 \cdot 10^6 Pa)(0.002 m^3-0.020 m^3)=18180 J$
And the final positive sign means that this work corresponds to an increase in internal energy of the gas.

8 0

2 years ago

A camera lens focuses on an object 75.0 cm from the lensThe image forms 3.50 cm behind the lens. What is the magnification of th

N76 [4]

Answer:

7/150

Explanation:

The following data were obtained from the question:

Object distance (u) = 75cm

Image distance (v) = 3.5cm

Magnification (M) =..?

Magnification is simply defined as:

Magnification (M) = Image distance (v)/ object distance (u)

M = v /u

With the above formula, we can obtain the magnification of the image as follow:

M = v/u

M = 3.5/75

M = 7/150

Therefore, the magnification of the image is 7/150.

6 0

3 years ago