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Norma-Jean [14]
2 years ago
12

What is the rotational speed of the second hand on a clock that measures the seconds?

Physics
1 answer:
pickupchik [31]2 years ago
5 0

Explanation:

Below is an attachment containing the solution

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What is the radius of a nucleus of an atom?
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A boy whirls a stone in a horizontal circle of radius 1.4 m and at height 1.5 m above ground level. The string breaks, and the s
balandron [24]

Answer:233.23 m/s^2

Explanation:

Given

radius of circle=1.4 m

Height of stone above ground=1.5 m

Horizontal distance(R)=10 m

It is given at the time of break stone flies horizontally thus stone to cover a height of 1.5 m in time t before reaching ground

1.5=0+\frac{gt^2}{2}

t=0.55 s

Initial horizontal velocity at the time of break is given by u

R=u\times t

10=u\times 0.55

u=18.07 m/s

Therefore magnitude of centripetal acceleration is given by

a_c=\frac{u^2}{r}=\frac{18.07^2}{1.4}=233.23 m/s^2

6 0
3 years ago
Nitrogen at 100 kPa and 25oC in a rigid vessel is heated until its pressure is 300 kPa. Calculate (a) the work done and (b) the
nignag [31]

Answer:

A. The work done during the process is W = 0

B. The value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

Explanation:

Given Data

Initial pressure P_{1} = 100 k pa

Initial temperature T_{1} = 25 degree Celsius = 298 Kelvin

Final pressure P_{2} = 300 k pa

Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.

⇒ V_{1} = V_{2} ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

⇒ P ∝ T

⇒ \frac{P_{2} }{P_{1}} = \frac{T_{2} }{T_{1}}

⇒ Put all the values in the above formula we get the final temperature

⇒ T_{2} = \frac{300}{100} × 298

⇒ T_{2} = 894 Kelvin

(A). Work done during the process is given by W = P × (V_{2} -V _{1})

From equation (1), V_{1} = V_{2} so work done W = P × 0 = 0

⇒ W = 0

Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m C_{v} ( T_{2} -T_{1} )

Where m = mass of the gas = 1 kg

C_{v} = specific heat at constant volume of nitrogen = 0.743 \frac{KJ}{kg k}

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83 \frac{KJ}{kg}

Therefore the value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

6 0
3 years ago
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