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2 years ago

if you have a kinetic energy of 1470 J, and you are 60kg mass and 0 m above the ground, what is you velocity?

Physics

2 answers:

laiz [17]2 years ago

8 0

Answer:

The 39.

Explanation:

Anni [7]2 years ago

6 0

KE = 1/2 · m · v²

1470 = (1/2 · 60) · (v²)

v² = 1470 / 30

v² = 49

v = √49

v = 7 m/s

(Note: Kinetic energy has nothing to do with height above ground.)

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Did the bigbang violate the law of conservation of energy and matter?

Pie

Under general relativity, there is no 'before the Big Bang'. The problem is that time is itself a part of the universe and is affected by matter and energy. Because of the huge densities just after the Big Bang, time itself is warped in such a way that it cannot go back before that event. It is somewhat like asking what is north of the north pole.

The conservation of matter and energy states that the total amount of mass and energy at one time is the same at any other time. Notice how time is a crucial part of this statement. To even talk about conservation laws, you have to have time.

The upshot is that the Big Bang did not break the conservation laws because time itself is part of the universe and started at the Big Bang and because the conservation laws need to have time in their statements.

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3 years ago

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noname [10]

Answer:

Power is the rate at which work is done or energy is transferred in a unit of time. Power is increased if work is done faster or energy is transferred in less time.

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3 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

LC Circuit

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

3 years ago

A trunk is dragged 3.0 meters across an attic floor by a force of 2.0N how much positive work is done on the trunk

Novay_Z [31]

The formula for work is

F*d

Therefore work=2.0N*3.0=6N*m

8 0

3 years ago

What type of material is wrapped around an object could cause the object to lose the most heat

Dmitrij [34]

A sheet of copper could cause the object to lose the most amount of heat. Copper is an essential element and a good conductor of heat. Heat can transfer from one end of a piece of copper to the other end.

7 0

3 years ago