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VMariaS [17]
2 years ago
6

What is the period and frequency of the second hand on a clock?

Physics
1 answer:
Tanzania [10]2 years ago
6 0

Answer:

Explanation:For second hand, period = 1 min = 3660 s; frequency = 0.0166 Hz}\\\mbox{For minute hand, period = 1 hr = 3600 s; frequency = 2.77}\times10^{-4} Hz\\\mbox{For hour hand, period = 12 hrs = 4300 s; frequency = 2.3}\times10^{-5} Hz.

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Describe the basis concept of the atwood’s machine
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Answer:The Atwood Machine is a device that demonstrates the basic principles of acceleration and dynamics. You'll mostly see Atwood machines in Physics laboratories and classrooms. It consists of two objects with different masses that hang vertically from a frictionless pulley that has a very small, negligible mass.

Explanation:

8 0
3 years ago
What does that mean?
Veseljchak [2.6K]

Answer:

acceleration

acceleration is the rate at which velocity change

i think

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3 years ago
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The voltage in the lines that carry electric power to homes is typically 2000 V. What is the required ratio of the loops in the
mariarad [96]

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yurrrrr 221n54

Explanation:

yurrrrrrrrrrrrrrrrrrrrrr

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3 years ago
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Two boxes on opposite ends of a massless board that is 3.0 m long. The board is supported in the middle by a fulcrum. The box on
rosijanka [135]

Answer:

b. 1.1 m

Explanation:

It is given that the total distance between the masses is equal to the length of the board, which is 3 m. Therefore,

s_{1} + s_{2} = 3\ m\\\\s_{2} = 3\ m - s_{1}\ --------- eqn(1)

where,

s₁ = distance of fulcrum from left mass

s₂ = distance of fulcrum from right mass

In order to achieve balance, the torque due to both masses must be equal:

T_{1} = T_{2}\\m_{1}s_{1} = m_{2}s_{2}\\(25\ kg)(s_{1}) = (15\ kg)(s_{2})\\\\\frac{15\ kg}{25\ kg}(s_{2}) = s_{1}\\\\using\ eqn(1):\\(0.6)(3\ m - s_{1}) = s_{1}\\1.8\ m = 1.6\ s_{1}\\s_{1} = \frac{1.8\ m}{1.6}

s₁ = 1.1 m

Hence, the correct option is:

<u>b. 1.1 m</u>

4 0
3 years ago
A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i = 1680 mA2 cos [πt/(0.
zhannawk [14.2K]

Answer:

(a) 1.58 V

(b) 0.0126 Wb

(c) 0.0493 V

Solution:

As per the question:

No. of turns in the coil, N = 400 turns

Self Inductance of the coil, L = 7.50 mH = 7.50\times 10^{- 3}\ H

Current in the coil, i = 1680cos[\frac{\pi t}{0.0250}] A

where

i_{max} = 1680\ mA = 1.680\ A

Now,

(a)  To calculate the maximum emf:

We know that maximum emf induced in the coil is given by:

e = \frac{Ldi}{dt}

e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]

e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]

For maximum emf, sin\theta should be maximum, i.e., 1

Now, the magnitude of the maximum emf is given by:

|e| = 7.50\times 10^{- 3}\times 1680\times 10^{- 3}\times \frac{\pi}{0.0250} = 1.58\ V

(b) To calculate the maximum average flux,we know that:

\phi_{m, avg} = L\times i_{max} = 7.50\times 10^{- 3}\times 1.680 = 0.0126\ Wb

(c) To calculate the magnitude of the induced emf at t = 0.0180 s:

e = e_{o}sin{\pi t}{0.0250}

e = 7.50\times 10^{- 3}\times sin{\pi \times 0.0180}{0.0250} = 2.96\times 10^{- 4} =0.0493\ V

7 0
3 years ago
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