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tamaranim1 [39]
3 years ago
11

A 535 kg roller coaster car began at rest at the top of a 93.0 m hill. Now it is at the top of the first loop-de-loop. This roll

er coaster’s track is nearly frictionless, so resistance can be ignored. Using g = 9.8 m/s2, what best describes the roller coaster car when it is at the top of the loop-de-loop?
Physics
1 answer:
Nikitich [7]3 years ago
3 0
The correct answer is <span>The car has both potential and kinetic energy, and it is moving at 24.6 m/s.</span>
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A 5.0 kg bucket of water is raised from a well by a rope. If the upward acceleration of the bucket is 3.0 m/s^2, find the force
Harman [31]
We know that m = 5 kg 
                      a  =  3.0 m/s^2
                      g = 9.8 (since not written, lets assume it's an international standard) 

T - mg =  ma

T = mg + ma

T = (5 . 9.8)  +  (5 . 3 )   

T = 49 + 15
  
T  = 64 N

Hope this helps

3 0
3 years ago
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Describe what happens when electrical impulses cross muscle fibers
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Answer:

Sometimes may cause involuntary responses like twitching

Explanation:

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3 years ago
How fast is a car going in m/5 if speedometer reads 45 miles?
fredd [130]

Answer:

i think you would have to divide to 2 numbers to get the sum

Explanation:

4 0
3 years ago
In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21
Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

z = 20km

And the velocity can be written as,

V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V = 980.55m/s

From the properties of standard atmosphere at altitude z = 20km temperature is

T = 216.66K

k = 1.4

R = 287 J/kg

Velocity of sound at this altitude is

a = \sqrt{kRT}

a = \sqrt{(1.4)(287)(216.66)}

a = 295.049m/s

Then the Mach number

Ma = \frac{V}{a}

Ma = \frac{980.55}{296.049}

Ma = 3.312

So front stagnation temperature

T_0 = T(1+\frac{k-1}{2}Ma^2)

T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)

T_0 = 689.87K

Therefore the temperature at its front stagnation point is 689.87K

6 0
4 years ago
If the rate of change of the magnetic field applied to a loop of wire is doubled, what happens to the induced emf in that loop a
slava [35]

Answer:

It is doubled.

Explanation:

The EMF( Electromotive force) is usually gotten from an energy source. In this case it is from the magnetic field. However when the other parameters are constant then the main focus is between the EMF and the magnetic field. They have a direct proportion relationship which is why when the magnetic field is doubled the EMF is doubled too.

3 0
4 years ago
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