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kumpel [21]
3 years ago
13

A loader sack of total mass

Physics
1 answer:
vampirchik [111]3 years ago
8 0

Question: A loader sack of total mass

is l000 grams falls down from

the floor of a lorry 200 cm high

Calculate the workdone by the

gravity of the load.​

Answer:

19.6 Joules

Explanation:

Applying

W = mgh........................ Equation 1

Where W = Workdone by gravity on the load, m = mass of the loader sack, h = height, g = acceleration due to gravity

From the question,

Given: m = 1000 grams = (1000/1000) kilogram = 1 kg, h = 200 cm = 2 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

W = (1×2×9.8)

W = 19.6 Joules

Hence the work done by gravity on the load is 19.6 Joules

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A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o
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The thickness of the oil slick is 1.95\times10^{-7}\ m

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Given that,

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2nt= m\lambda

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Put the value into the formula

2\times1.28 \times t=1\times\times500\times10^{-9}

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t=1.95\times10^{-7}\ m

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3 years ago
19. After a snowstorm, you put on your frictionless skis and tie a rope to the back of your friend’s truck. Your total mass is 7
Kisachek [45]

Explanation:

It is given that,

Total mass is 70 kg

The truck exerts a constant force of 20 N.

Then the net force is given by :

F = ma

a is acceleration of rider

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Initial velocity of rider is 0. So, using equation of kinematics to find the final velocity as :

v=u+at\\\\v=at\\\\v=\dfrac{2}{7}\times 15\\\\v=4.28\ m/s

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A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 20.0 cm long
noname [10]

Answer:

Part a)

\phi = 2.76 \times 10^{-7} T m^2

Part B)

M = 5.52 \times 10^{-5} H

Part C)

EMF = 0.1 V/s

Explanation:

Part a)

Magnetic field due to a long ideal solenoid is given by

B = \mu_0 n i

n = number of turns per unit length

n = \frac{N}{L}

n = \frac{350}{0.20}

n = 1750 turn/m

now we know that magnetic field due to solenoid is

B = (4\pi \times 10^{-7})(1750)(0.100)

B = 2.2 \times 10^{-4} T

Now magnetic flux due to this magnetic field is given by

\phi = B.A

\phi = (2.2 \times 10^{-4})(\pi r^2)

\phi = (2.2 \times 10^{-4})(\pi(0.02)^2)

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Part B)

Now for mutual inductance we know that

\phi_{total} = M i

\phi_{total} = N\phi

\phi_{total} = 20(2.76 \times 10^{-4})

\phi_{total} = 5.52 \times 10^{-6}

now we have

M = \frac{5.52 \times 10^{-6}}{0.100}

M = 5.52 \times 10^{-5} H

Part C)

As we know that induced EMF is given as

EMF = M \frac{di}{dt}

EMF = 5.52 \times 10^{-5} (1800)

EMF = 0.1 V/s

3 0
4 years ago
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