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maksim [4K]
3 years ago
15

What happens to the kinetic energy of a roller coaster as it approaches its lowest point and loses its potential energy?

Physics
1 answer:
Fiesta28 [93]3 years ago
3 0

Answer:

At the top of the roller coaster, there is a lot of potential energy. When it comes to the bottom, the roller coaster loses its potential energy and gains kinetic energy as it is going very fast here.

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Two forces and are applied to an object whose mass is 13.3 kg. The larger force is . When both forces point due east, the object
ANEK [815]

Answer:

Explanation:

First, It's important to remember F = ma, and in this problem m = 13.3 kg

This can be reduced to a simple system of equations problem.  Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them.  So let's call them F1 and F2, with F1 arger than F2.  Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.  

Can you solve this system of equations seeing them like this, or do you need more help?

6 0
2 years ago
1) A 1,600 kilogram car is also traveling in a straight line. Its momentum is 32,500 kg*m/s. What is the
BaLLatris [955]

Answer:

v = 20.31 m/s

Explanation:

p = mv -> v = p/m = 32,500 kg*m/s / 1,600 kg = 20.31 m/s

4 0
2 years ago
A truck was carrying a substance in a tank. The molecules of that substance were moving away from each other. The truck parked o
DochEvi [55]

Answer:

In the morning the molecules were moving away from each other with a smaller speed than when the truck was carrying the substance.

Explanation:

5 0
2 years ago
a radio antenna broadcasts a 1.0 MHz radio wav e with 21 kW of power. Assume that the radiation is emitted uniformly in all dire
My name is Ann [436]

Answer:

I=2.67\times 10^{-6}\ W/m

Explanation:

Given that,

Frequency of a radio antenna is 1 MHz

Power, P = 21 kW

We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :

I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m

So, the wave intensity 25 km from the antenna is 2.67\times 10^{-6}\ W/m.

5 0
3 years ago
Find the voltage across the 15 Q resistor.<br> [?] V<br><br> No links please
Margaret [11]

Answer:

Explanation:

same idea as before Liam,  first, find the parallel resistance in 35 || 20

(35*20)  / (35+20)  = 700 / 55  = 12.727272 ohms

now add the  12.727272 + 15 = 27.727272 ohms total resistance

V = IR

10 = I *    27.727272

10 / 27.727272 = I

0.360655 = I

V = IR  (again, but across the 15 ohm resistor)

V =     0.360655 * 15      

V = 5.4098                

4 0
2 years ago
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