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3 years ago

A 6.7 kg object moves with a velocity of 8 m/s. what is its kinetic energy

Physics

2 answers:

blagie [28]3 years ago

3 0

The Correct answer to this question for Penn Foster Students is: 214.4 J

marta [7]3 years ago

3 0

Answer:

$E_k=214.4 J$

Explanation:

Kinetic energy is a form of energy, known as motion energy. The kinetic energy of an object is that produced by its movements, it depends on the mass and speed of the object. In formula form:

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

$m=Mass\hspace{3}of\hspace{3}the\hspace{3}object\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}object$

Replacing the data provided, the kinetic energy is:

$E_k=\frac{1}{2} (6.7)(8^{2} )=214.4J$

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A toy car is moving across a table with a velocity of 7.0 m/s and drives off the end. The table is 1.8 m tall. How long will the

KatRina [158]

Answer:

Option D - 0.2 s

Explanation:

We are given;

Initial velocity; u = 7 m/s

Height of table; h = 1.8m

Now,since we want to find the time the car spent in the air, we will simply use one of Newton's equation of motion.

Thus;

h = ut + ½gt²

Plugging in the relevant values, we have;

1.8 = 7t + ½(9.8)t²

4.9t² + 7t - 1.8 = 0

Using quadratic formula to find the roots of the equation gives us;

t = -1.65 or 0.22

We can't have negative t value, thus we will pick the positive one.

So, t = 0.22 s

This is approximately 0.2 s

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4 years ago

In the mixtures lab, you waited to see if the liquid would form lumpy or fluffy masses, which would indicate it was a _____.

Over [174]

B) colloid because i took the test and that the answer

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3 years ago

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Review 1: A plane is located x = 40 miles (horizontally) away from an airport at an altitude of h miles. Radar at the airport de

-Dominant- [34]

Explanation:

Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,

$s(t)^2={h^2+x^2}$ ...........(1)

Given, h = 4, x = 40 and s(t) = -20 mph

Differentiate equation (1) wrt t

$2s(t)s'(t)=2x(t)x'(t)$

$x'(t)=\dfrac{s(t)s'(t)}{x(t)}$

When x = 40, $s(t)=\sqrt{40^2+4^2}=40.19\ m$

$x'(t)=\dfrac{-240s(t)}{x(t)}$

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4 years ago

A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal in

erica [24]

Explanation:

Given that,

Mass of a freight car, $m_1=30,000-kg$

Speed of a freight car, $u_1=0.85\ m/s$

Mass of a scrap metal, $m_2=110,000\ kg$

(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :

$30000\times 0.85=(30000+110000)V\\\\V=\dfrac{30000\times 0.85}{30000+110000}\\\\=0.182\ m/s$

So, the final velocity of the loaded freight car is 0.182 m/s.

(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy

$\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J$

Lost in kinetic energy is 8518.82. Negative sign shows loss.

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3 years ago

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matrenka [14]

Answer:

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3 years ago

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