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bonufazy [111]
3 years ago
12

if a cat is running at a constant speed of 10km/h for 5 s, what is its average speed and what is its instantaneous speed at 4 s?

Physics
1 answer:
iris [78.8K]3 years ago
8 0

Here a cat is running at constant speed which is given as 10 km/h for  5s

So here the average speed is defined as total distance moved in total time interval

so here it is given by

v_{avg} = \frac{distance}{time}

since

here speed of cat is constant so it will remain the same

And hence the average speed and instantaneous speed at any instant for this duration will remain the same

so here answer would be

<em>average speed = 10 km/h</em>

<em>instantaneous speed = 10 km/h</em>

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Standards for all managers ethical responsibilities are covered in a company's?
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Answer:

C. Code of ethics

Explanation:

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3 years ago
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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

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3 years ago
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Alex777 [14]

Answer:

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totally internally reflected

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3 years ago
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A 0.5-kg ball moving at 5 m/s strikes a wall and rebounds in the opposite direction with a speed of 2 m/s. If the impulse occurs
mart [117]

<em>Given that:</em>

                       mass of the ball (m) = 0.5 Kg ,

                    ball strikes the wall (v₁) = 5 m/s ,

rebounds in opposite direction (v₂) = 2 m/s,

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We know that from Newton's II law,

                                <em>F = m. a</em>  Newtons  

                                  (velocity acting in opposite direction, so <em>a = ( (v₁ + v₂)/t</em>

                                   = m × (v₁ + v₂)/t

                                   = 0.5 × (5 + 2)/0.01

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<em>The force acting up on the ball is 350 N</em>

                                     

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