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3 years ago

When it goes through a chemical change, a substance changes what

Physics

1 answer:

Anestetic [448]3 years ago

8 0

Answer:

form

I'm pretty sure...

let me know if I'm right, if I am give brainliest

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A force of 16 N to the west is applied to each object below. Which object will

Montano1993 [528]

41kg object that is moving east at 5 m s

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3 years ago

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A nuclear reaction only involves valence electrons. True or False?

dalvyx [7]

I believe it would be false

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3 years ago

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Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the

anygoal [31]

Answer:

The value is the temperature of the air inside the tire $T_{2} =$ 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet $P_{1}$ = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet $P_{2}$ = 3.2 + 1 = 4.2 atm

Temperature at inlet $T_{1}$ = 300 K

(a) Volume of the system is constant so pressure is directly proportional to the temperature.

$\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }$

$\frac{T_{2} }{300} = \frac{4.2}{3.7}$

$T_{2} =$ 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

$m_{1} T_{1} = m_{2} T_{2}$

$\frac{m_{2} }{m_{1} } = \frac{T_{1} }{T_{2} }$

$\frac{m_{2} }{m_{1} } = \frac{300}{354.54}$

$\frac{m_{2} }{m_{1} } =0.00294$

value of the original mass of air in the tire should be released is $\frac{m_{2} - m_{1}}{m_{1}}$

$\frac{0.00294-1}{1}$

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

8 0

3 years ago

A jetliner, traveling northward, is landing with a speed of 71.9 m/s. Once the jet touches down, it has 675 m of runway in which

Leviafan [203]

Answer:

The value is $a = -3.7 \ m/s^2$

Explanation:

From the question we are told that

The landing speed is $u = 71.9 \ m/s$

The distance traveled is $d = 675 \ m$

The velocity it is reduced to is $v = 11.3 \ m/s$

Generally the average acceleration is mathematically represented as

$a = \frac{ v^2 - u^2 }{ 2 * d }$

=> $a = \frac{ 11.3^2 - 71.9^2 }{ 2 * 675 }$

=> $a = -3.7 \ m/s^2$

5 0

3 years ago

Make the curvature radius 0.6 m, the refractive index 1.5, and the diameter 0.6 m. place the lamp so that the source of light is

vodka [1.7K]

We are given
r = 0.6 m
n = 1.5
D = 0.6 m, R1 = 30 cm
R2 = 120 cm

We are asked to get the focal length and the distance of the focal plane from the lens

We use the formula
1 / f = ( n - 1) (1/R1 - 1/R2)
Substituting and solving for f
1/ f = (1.5 - 1) (1/30 - 1/120)
f = 80 cm

The focal length is 80 cm and the distance of the focal plane from the lesn is 80 cm - 30 cm = 50 cm.<span />

5 0

3 years ago