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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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Which of the following represent units of capacitance? You may choose more than one correct answer:
inessss [21]

Answer:

Capacitance is a derived physical quantity measured in farad

4 0
2 years ago
Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec
AysviL [449]
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
E= \frac{\sigma}{2\epsilon_0}
where
\sigma is the charge density
\epsilon_0 is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
E=30 N/C
7 0
3 years ago
2. A portion of the body receives 0.15 mGy from radiation with a quality factor Q = 6 and 0.22 mGy from radiation with Q = 10. (
DiKsa [7]

Answer with Explanation:

We are given that

D_1=0.15mGy

D_2=0.22mGy

Q_1=6

Q_2=10

a.We have to find the total dose

Total dose=D=D_1+D_2

Using the formula then, we get

D=0.15+0.22

D=0.37mGy

b.We have to find the total dose equivalent

Total dose equivalent=H=D_1Q_1+D_2Q_2

Using the formula

H=0.15(6)+0.22(10)

H=3.1mSv

7 0
3 years ago
A dust particle with mass of 5.4×10−2 g and a charge of 2.3×10−6 C is in a region of space where the potential is given by V(x)=
user100 [1]

Answer:

1.6 m/s^2

Explanation:

Hello!

To calculate the acceleration we must know the electric field. The electric field and the potential are related by:

E = -\frac{dV}{dx} =- 2.6(\frac{V}{m^{2}})x + 8.1(\frac{V}{m^{3}})x^{2}

If the particle starts at 2.3m, the electric field is:

E = 36.869 V/m = 36.869 N/C

So, the force on the particle is:

F = q E =  2.3×10^−6 C * 36.869 N/C = 8.48 x 10^-5 N

And its acceleration is :

a = F/m =  8.48 x 10^-5 N / 5.4×10−5 kg = 1.57 m/s^2

Rounded to two significant figures:

1.6 m/s^2

6 0
3 years ago
A train takes 2h to reach station B from station A and 3h to return back to A.The distance between the station is 200km, then it
serg [7]

Answer:

80km/hr

Explanation:

Total distance: 400km

Total time 5 hours

Average speed= distance/time

400/5= 80km/hr

8 0
3 years ago
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