All categories

3 years ago

Describe an instance when the motion energy of an object changed, causing some other energy change as well.

Physics

1 answer:

bonufazy [111]3 years ago

5 0

Answer:

For instance, an immobile object located at some height above the ground has potential energy. If the object starts to fall, its potential energy is converted to motion or kinetic energy. The heavier and the higher the object is, the more potential energy it will have. Transfer of thermal energy via conduction is aided by vigorously vibrating particles in a medium that has gained energy through heating.

Explanation:

You might be interested in

A person with mass mp = 76 kg stands on a spinning platform disk with a radius of R = 1.98 m and mass md = 191 kg. The disk is i

nalin [4]

<span>1.7 rad/s The key thing here is conservation of angular momentum. The system as a whole will retain the same angular momentum. The initial velocity is 1.7 rad/s. As the person walks closer to the center of the spinning disk, the speed will increase. But I'm not going to bother calculating by how much. Just remember the speed will increase. And then as the person walks back out to the rim to the same distance that the person originally started, the speed will decrease. But during the entire walk, the total angular momentum remained constant. And since the initial mass distribution matches the final mass distribution, the final angular speed will match the initial angular speed.</span>

3 0

4 years ago

A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two

AfilCa [17]

Answer:

please the answer below

Explanation:

(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

$V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\$

k=8.89*10^9

For both cases we have

$k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}$

(b) by replacing this values of r in the expression for V we obtain

$k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}$

hope this helps!!

3 0

3 years ago

Read 2 more answers

Explain why the extrapolated temperature is used to determine the maximun temperature of the mixture rather than the highest rec

Oksi-84 [34.3K]

The extrapolated temperature is used to define the maximum temperature of the mixture relatively than the highest recorded temperature in which the conclusion will effect in a higher specific heat value. Heat is bound to escape from whatever apparatus is using, therefore it is needed to account for the loss of the heat that does not go into increasing the temperature of the mixture.

6 0

3 years ago

Read 2 more answers

Please. Physics is so difficult.

Softa [21]

Answer:

0.010 m

Explanation:

So the equation for a pendulum period is: $y=2\pi\sqrt{\frac{L}{g}}$ where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:

$1.99 s = 2(3.14)\sqrt{\frac{L}{9.8 m\backslash s^2}}\\$

Evaluate the multiplication in front

$1.99 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}$

Divide both sides by 6.28

$0.317 s= \sqrt{\frac{L}{9.8 m\backslash s^2}}$

Square both sides

$0.100 s^2= \frac{L}{9.8 m\backslash s^2}$

Multiply both sides by m/s^2 (the s^2 will cancel out)

$0.984 m = L$

Now now let's find the length when it's two seconds

$2.00 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}}$

Divide both sides by 6.28

$0.318 s = \sqrt{\frac{L}{9.8 m\backslash s^2}$

Square both sides

$0.101 s^2 = \frac{L}{9.8 m\backslash s^2}$

Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)

$0.994 m = L$

So to find the difference you simply subtract

0.984 - 0.994 = 0.010 m

4 0

2 years ago

Read 2 more answers

Find the length of the third side of the triangle.

AlladinOne [14]

Answer:

Image result for What is the length of the hypotenuse of the triangle below

One way to solve this is to use Pythagorean theorem: the square of one leg of triangle plus square of other leg of the triangle equals c the hypotenuse (longest side of triangle). You might see this as the formula a^2 + b^2 = c^2, where a and b are the legs and c is the hypotenuse.Nov 23, 2016

Explanation:

4 0

3 years ago