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adoni [48]
3 years ago
11

Someone answer asapp

Physics
2 answers:
garik1379 [7]3 years ago
6 0

Potential energy is highest at the top of the loop, and kinetic energy is highest at the bottom of the loop.

ohaa [14]3 years ago
5 0

Answer:

<h2>See below</h2>

Explanation:

Let us first give a formal definition of the terms.

1. Potential Energy: Energy that has the ability to do work, but is stationary at the moment.

2. Kinetic Energy: The energy of an object in motion.

Now that we know the definitions of these terms, we can answer the question. Remember that the area of greatest potential energy is that spot where there is potential to do work, but that energy is not causing any movement at the current time. The spot where there is the greatest amount of potential energy in the image is at the top of the roller coaster. There is little movement at this juncture, but there is plenty of potential energy. As the coaster goes down, the kinetic energy goes up, since the coaster is now in motion. The spot where there is the greatest amount of kinetic energy is near the bottom of the roller coaster.

<em>Hope this helps</em>

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What is occurring when a light wave goes through a pane of glass in a window?
ycow [4]

Answer:

transmission: the passing of a wave through an object

Explanation:

5 0
2 years ago
I'm completely lost in physics, the Kinematics and dynamics units.. My exam is coming up.. If a biker travelling at 6.4m/s sees
Alexeev081 [22]
D=rt
when biker A catches biker B, the time they've been riding is the same, so 
t=t, or d/r=d/r
the rates are 6.4 and 4.7, so
d/6.4=d/4.7
biker B is 34m ahead, so 
(d+34)/6.4=d/4.7
multiply both sides by 6.4*4.7:
4.7(d+34)=6.4d
4.7d+=6.4d+159.8
1.7d=159.8
d=94 meters

Another way to think of it is that biker A gains 1.7 meters on B every second (6.4-4.7=1.5), so the time it'll take for him to gain 34 meters is 34/1.7=20 seconds. In that time, biker B travels 4.7*20=94 meters
8 0
3 years ago
Help me please, need more assistance
Dmitrij [34]

Explanation:

12) q = mCΔT

125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)

T = 82.0°C

13) Solving for ΔT:

ΔT = q / (mC)

a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C

b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C

c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C

d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C

e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C

f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C

14) q = mCΔT

q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)

q = 502,000 J

20) q = mCΔT

q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)

q = 742,000 J

24) q = mCΔT

q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)

q = -0.091 J

6 0
3 years ago
How much acceleration does a car need if it is to go a distance of 620 ft. In a time of 20 sec while it is speeding up
yan [13]
31ft/second . 620/20
3 0
3 years ago
A man pushes a box along a flat, frictionless surface using a force of 500 N. The box was moved a distance of 2.5 m. The actual
andrey2020 [161]

Answer:

Workdone = 1250Nm

Explanation:

<u>Given the following data;</u>

Force, F = 500N

Distance, d = 2.5m

Workdone is given by the formula;

Workdone = force * distance

Substituting into the equation, we have

Workdone = 500 * 2.5  

Workdone = 1250Nm

Therefore, the actual work done by the worker is 1250 Newton-meter.

4 0
3 years ago
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