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2 years ago

9

. A 79 g sample of water at 21oC is heated until it becomes steam with a temperature of 143oC. Find the change in heat content o

f the system.

1 answer:

mote1985 [20]2 years ago

4 0

Answer:

40479.6 J

Explanation:

Applying,

q = cm(t₂-t₁).................... Equation 1

Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.

From the question,

Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C

Constant: c = 4200 J/kg.°C

Substitute these values into equation 1

q = 4200(0.079)(143-21)

q = 331.8(122)

q = 40479.6 J

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If the distance between two objects decreased, what would happen to the force of gravity between them?

evablogger [386]

Answer: A- It would increase

Explanation:

According to the law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

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As we can see, the gravity force is directly proportional to the mass of the bodies or objects and inversely proportional to the square of the distance that separates them.

In other words:

<h2>If we decrease the distance between both objects, the gravitational force between them will increase. </h2>

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3 years ago

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Match the following vocabulary words with their definitions:

Ronch [10]

Correct matching:

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2. speed --> the rate at which an object changes position when traveling in a certain direction

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2 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

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3 years ago

A pair of slits separated by 1 mm, are illuminated with monochromatic light of wavelength 411 nm. The light falls on a screen 1.

Ilya [14]

Answer:

$t = 0.192 \mu m$

Explanation:

Path difference due to a transparent slab is given as

$\Delta x = (\mu - 1) t$

here we know that

$\mu = 1.79$

now total shift in the bright fringe is given as

$Shift = \frac{D(\mu - 1)t}{d}$

Also we know that the fringe width of maximum intensity is given as

$\delta x = \frac{\lambda D}{d}$

now we have

$\frac{D}{d} = \frac{\delta x}{\lambda}$

now the shift is given as

$Shift = \frac{(\mu - 1) t \delta x}{\lambda}$

given that the shift is

$Shift = 0.37 \delta x$

here we have

$0.37 \delta x = \frac{(\mu - 1) t \delta x}{\lambda}$

now plug in all values in it

$0.37 = \frac{(1.79 - 1) t}{411 \times 10^{-9}}$

$t = 0.192 \times 10^{-6} m$

$t = 0.192 \mu m$

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3 years ago

ZanzabumX [31]

Answer: mass x height x gravitational field strength (g)

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Explanation:

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3 years ago