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3 years ago

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Determine the heat energy required to vaporize 13.9 grams of liquid water at 100° C. O 2,006 cal O 47.8 cal O 7,506 cal O 24.9 c

al

1 answer:

Harlamova29_29 [7]3 years ago

6 0

Answer:

7506 cal

Explanation:

Given: Mass of water = m = 13.9 grams.

Heat energy required to determine the conversion of 13.9 grams of water to steam or vapor is m L,

where L = Latent heat of vaporization of water=540 cal/g.

Q = mL = (13.9)(540) = 7,506 cal.

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A circuit element consists of a resistor with value 20Ω and inductor with value 10mH connected in series. A voltage of LaTeX: v(

Flura [38]

Answer:

8.97 Watt

Explanation:

Resistance, R = 20 ohm

Inductance, L = 10 mH

V(t) = 20 Cos (1000 t + 45°)

Compare with the standard equation

V(t) = Vo Cos(ωt + Ф)

Ф = 45°

ω = 1000 rad/s

Vo = 20 V

Inductive reactance, XL = ωL = 1000 x 0.01 = 10 ohm

impedance is Z.

$Z = \sqrt{R^{2}+X_{L}^{2}}$

$Z = \sqrt{20^{2}+10^{2}}$

Z = 22.36 ohm

$V_{rms}=\frac{V_{0}}{\sqrt{2}}$

$V_{rms}=\frac{20}{\sqrt{2}} = 14.144 V$

$I_{rms}=\frac{V_{rms}}{Z}=\frac{14.144}{\sqrt{22.36}}=0.634 A$

Apparent power is given by

P = Vrms x Irms

P = 14.144 x 0.634

P = 8.97 Watt

6 0

3 years ago

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0

3 years ago

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit

Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7\times 10^{10} m$

$v_i=9.5\times 10^4 m/s$

$d_2=6\times 10^{12} m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

$\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f$

$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

4 0

3 years ago

During the execution of a play, a football player carries the ball for a distance of 33 m in the direction 58° north of east. To

podryga [215]

Answer:28 m

Explanation:

Given

Direction is $58^{\circ}$ North of east i.e. $58 ^{\circ}$ with x axis

Also ball moved by 33 m

therefore its east component is 33cos58=17.48 m

Northward component $=33sin58=27.98 m\approx 28 m$

8 0

3 years ago

What are the properties of a thermometer

lesya [120]

Answer:

The thermometer makes use of a physical property of a thermometric substance which changes continuously with temperature. The physical property is referred to as thermometric property.

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Thermometric Properties Used In Various Thermometers.

3 0

3 years ago