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1 year ago

Part A

Physics

1 answer:

Basile [38]1 year ago

6 0

The relationships to determine the number of calories to change 0.50 kg of 0°C ice to 0°C ice water is 1,080,000 cal.

<h3>How does heating ice that is at C affect it?</h3>

Ice melts and becomes liquid water at 0 degrees Celsius. Once all of the ice has been entirely transformed into liquid water, the temperature of the remaining ice begins to increase once more (in °C), continuing to rise until it reaches 100 °C, where it then stabilizes.

The water turns into steam when it reaches a temperature of 100 °C (D).

Water has a fusion latent heat of fusion of 80 cal/g.

Water has a 1 cal/g-C specific heat.

Water has a 540 cal/g latent heat of vaporization.

In light of this, the total amount of heat needed is 1500 g [(80 cal/g) + (1 cal/g-C)(100 - 0)C + (540 cal/g)] = 1500 g [(720 cal/g)] = 1,080,000 cal.

To learn more about Vaporization refer to:

brainly.com/question/26306578

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Impulse = Force * time
Impulse = 500N *0.5 s =250 N*s

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Mr. Patel is looking for his cat in the garden at night. It is completely dark outside. Mr Patel shines his torch at the tree. W

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Answer:

the tree's light reflects into his eyes.

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2 years ago

Salmon often jump waterfalls to reach their

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The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

height of the waterfall, h = 0.432 m
distance of the Salmon from the waterfall, s = 3.17 m
angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

$h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s$

The minimum velocity of the Salmon jumping at the given angle is calculated as;

$X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s$

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

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2 years ago

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

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Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

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3 years ago

Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to

Debora [2.8K]

Here we want to study how the linear charge density changes as we change the measures of our body.

We will find that we need to add 9*Q of charge to keep the linear charge density unchanged.

<em>I will take two assumptions:</em>

The charge is homogeneous, so the density is constant all along the wire.

As we work with a linear charge density we work in one dimension, so the wire "has no radius"

Originally, the wire has a charge Q and a length L.

The linear charge density will be given by:

λ = Q/L

Now the length of the wire is stretched to 10 times the original length, so we have:

L' = 10*L

We want to find the value of Q' such that λ' (the <u>linear density of the stretched wire</u>) is still equal to λ.

Then we will have:

λ' = Q'/L' = Q'/(10*L) = λ = Q/L

Q'/(10*L) = Q/L

Q'/10 = Q

Q' = 10*Q

So the new <u>charge must be 10 times the original charge</u>, this means that we need to add 9*Q of charge to keep the linear charge density unchanged.

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3 years ago