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1 year ago

calculate the kinetic energy of a 820 kg compact car moving at 23 m/s.

Physics

1 answer:

WARRIOR [948]1 year ago

8 0

Answer:

216,890 J

Explanation:

We first define the Kinetic Energy Formula:

K.E =

$\frac{1}{2} m {v}^{2}$

where m is the mass of the object in kg, and v is the velocity in m/s.

Given from the information from the question, we substitute the values in to find K.E.

K.E of car =

$\frac{1}{2} (820) {(23)}^{2} \\ = 216890 \: joules$

Note that the SI unit for Energy is Joules or J.

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The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}$

$g \approx 9.82\,\frac{m}{s^{2}}$

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

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3 years ago

A 7.2 magnitude earthquake, with an epicenter in Northern Mexico, was felt over 100 miles away in Southern California because wa

ryzh [129]

Answer:

because waves generated

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1 year ago

A train moving with a velocity of 42.9 km/hour North, increases its speed with a uniform acceleration of 0.250 m/s^2 North until

ankoles [38]

Answer:

3658.24m

Explanation:

Hello!

the first thing that we must be clear about is that the train moves with constant acceleration

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

$\frac {Vf^{2}-Vo^2}{2.a} =X$

Vf = final speed =160km/h=44.4m/s

Vo = Initial speed =42.9km/h=11.92m/s

A = acceleration =0.25m/s^2

X = displacement

solving

$\frac {44.4^{2}-11.92^2}{2.(0.25)} =X\\X=3658.24m$

the distance traveled by the train is 3658.24m

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3 years ago

If heat is transferred away from a gas, what could happen?

NeX [460]

<span>♦The gas may become a liquid as it loses energy♦</span>

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3 years ago

Consider a projectile launched with an initial velocity of v0 = 120 ft/s, inclined at an angle, θ with the horizontal. Let us as

Natali [406]

Answer:

How to find the maximum height of a projectile?

if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...

if α = 45°, then the equation may be written as: ...

if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.

Explanation:

4 0

3 years ago

￼calculate the kinetic energy of a 820 kg compact car moving at 23 m/s.

calculate the kinetic energy of a 820 kg compact car moving at 23 m/s.