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GalinKa [24]
3 years ago
7

A particle moves in a circular path of radius 0.10 m with a constant angular speed of 5 rev/s. The acceleration of the particle

is: A. 0.10p m/s2 B. 0.50 m/s2 C. 500p m/s2 D. 1000p2 m/s2 E. 10p2 m/s2
Physics
1 answer:
Darya [45]3 years ago
7 0

Answer:

2.5 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s²

For circular motion, the expression for acceleration is given as,

a = ω²r ................ Equation 1

Where a = acceleration of the particle, ω = angular speed of the particle, r = radius of the circular path.

Given: ω = 5 rev/s = 31.42 rad/s, r = 0.10 m.

Substitute into equation 1

a = 5²(0.10)

a = 25(0.10)

a = 2.5 m/s²

Hence the acceleration of the particle = 2.5 m/s²

Hence, none of the option  is correct

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sweet-ann [11.9K]
Output can not be greater than input because the conversion of energy can not be greater than 100%.
4 0
3 years ago
Electronic flash unit cameras contain a capacitor for storing the energy used to produce the flash. In one such unit the flash l
Stolb23 [73]

Answer:

(A) 421 J energy stored in the capacitor for one flash.

(B) The value of capacitance is 0.0537 F

Explanation:

Given :

(A)

Time t = \frac{1}{675}

Average power P = 2.7 \times 10^{5} W

From power equation,

   P= \frac{E}{t}

So energy in one light is given by,

   E = Pt

  E = 2.7 \times 10^{5} \times \frac{1}{675}  = 400 J

Since efficiency is 95 % so we can write, energy stored in one flash,

E_{tot} = \frac{400}{0.95} = 421 J

(B)

From the formula of energy stored in capacitor,

 E = \frac{1}{2}C V^{2}

Where E = E_{tot} and V = 125 V

 C = \frac{2E}{V^{2} }

 C = \frac{2 \times 421}{15625}

 C = 0.0537 F

8 0
3 years ago
A certain dielectric with a dielectric constant = 24 can withstand an electric field of 4 107 V/m. Suppose we want to use this d
mr_godi [17]

Answer:

Explanation:

Electric field E = 4 x 10⁷ V / m

Dielectric constant k = 24

capacitance of capacitor

C = kε₀ A / d

d = plate separation

A =  plate area

C = .89 x 10⁻⁶

V / d = electric field

for minimum d , electric field will be maximum

V / d  = 4 x 10⁷

1930 / d = 4 x 10⁷

d = 1930 / 4 x 10⁷

d = 482.5 x 10⁻⁷ m

= 48.25 x 10⁻⁶ m

C = kε₀ A / d

.89 x 10⁻⁶ = 24 ε₀ A / d

A = .89 x 10⁻⁶  X d /  24 ε₀

A = .89 x 10⁻⁶  X 48.25 x 10⁻⁶  /  24  x 8.85 x 10⁻¹²

= 42.9 / 212.4

= .2019 m²

8 0
3 years ago
Read 2 more answers
Can someone explain how they got their answer or how I get the change in number? :(
enyata [817]

Answer:

See Below

Explanation:

Okay, I thinkkk what it is asking by what you summarzied for me issss:

They split the total time into four quarters. They then took (for the first quarter) the start time. Then when the first quarter ends and the second quarter starts is the "end" time.

They then subtract the start time of the second quarter from the end time of the first quarter.

I hope this helps, good luck! :D

5 0
2 years ago
An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to
mafiozo [28]

Answer:

    wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

         w_f = w₀ - a t

         w₀ = w_f + a t

         w₀ = 109 + 4.7 1.87

         w₀ = 117.8 rad / s

let's reduce to revolutions / s

         w₀ = 117.8 rad / s (1 rev / 2pi rad)

         w₀ = 18.75 rev / s

8 0
3 years ago
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