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weqwewe [10]
2 years ago
9

An amusement park ride moves a rider at a constant speed of 14 meters per second in a horizontal circular path of radius 10. met

ers. What is the rider's centripetal acceleration in terms of g, the acceleration due to gravity?
A) 1g
B) 2g
C) 3g
D) 0g
Physics
1 answer:
ASHA 777 [7]2 years ago
4 0

Answer:

B) 2g

Explanation:

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

Velocity, v = 14m/s

Radius, r = 10m

To find the centripetal acceleration;

Acceleration, a = \frac {v^{2}}{r}

Substituting into the equation, we have;

Acceleration, a = \frac {14^{2}}{10}

Acceleration, a = \frac {196}{10}

Acceleration, a = 19.6m/s²

In terms of acceleration due to gravity, g = 9.8m/s²

We would divide by g;

Acceleration, a = 19.6/9.8 = 2

Hence, centripetal acceleration = 2g

Therefore, the rider's centripetal acceleration in terms of g, the acceleration due to gravity is 2g.

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Explanation:

it is given that, the linear charge density of a charge, \lambda=\dfrac{\lambda_ox_o}{x}

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

dE=\dfrac{k\ dq}{x^2}..........(1)

The linear charge density is given by :

\lambda=\dfrac{dq}{dx}

dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx

Integrating equation (1) from x = x₀ to x = infinity

E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx

E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}

E=\dfrac{k\lambda_o}{2x_o}

Hence, this is the required solution.

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3 years ago
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Which tricks do you use to reduce pressure in everyday activities? Which tricks do you use to reduce pressure in everyday activi
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Answer:

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8 0
3 years ago
Your roommate leaves a 120W fan running in your apartment.Over the course of an hour,how much thermal energy does the fan add to
zubka84 [21]

Answer:

4.32\cdot 10^5 J

Explanation:

Power is related to energy by the following relationship:

P=\frac{E}{t}

where

P is the power used

E is the energy used

t is the time elapsed

In this problem, we know that

- the power of the fan is P = 120 W

- the fan has been running for one hour, which corresponds to a time of

t = 1 h \cdot (60 min/h)(60 s/min)=3600 s

So we can re-arrange the previous equation to find E, the energy (in the form of thermal energy) released by the fan:

E=Pt=(120 W)(3600 s)=4.32\cdot 10^5 J

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Explanation:

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A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 12 A, while that
andreev551 [17]

Answer:

a) 1450watts

b) 564watts

c) 1.11

Explanation:

Power consumed = IV

I is the current rating

V is the operating voltage

If a blow-dryer and a vacuum cleaner each operate with a voltage of 120 V and the current rating of the blow-dryer is 12 A, while that of the vacuum cleaner is 4.7 A then their individual power rating is calculated thus;

a) For blow-dryer

Operating voltage = 120V

Its current rating = 12A

Power consumed = IV

= 120×12

= 1440watts

b) For vacuum cleaner:

Operating voltage is the same as that of blow dryer = 120V

Its current rating = 4.7A

Power consumed = IV

= 120×4.7

= 564watts

c) Energy used = Power consumed × time taken

Energy used = Power × time

Energy used by blow dryer = 1440×20×60

= 1,728,000Joules

Energy used up by vacuum cleaner = 564×46×60

= 564×2760

= 1,556,640Joules

Ratio of the energy used by the blow-dryer in 20 minutes to the energy used by the vacuum cleaner in 46 minutes will be 1,728,000/1,556,640 = 1.11

4 0
3 years ago
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