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3 years ago

Science please help!

Physics

2 answers:

Aleks [24]3 years ago

7 0

I do believe the answer you are looking for is Matter.

Romashka [77]3 years ago

6 0

Inertia is a property of matter
i hope this help

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gizmo_the_mogwai [7]

I think you can google this because I really don’t know the answer I’m so sorry

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3 years ago

How does a metamorphic rock become an igneous rock?

Oksana_A [137]

Answer:

If the newly formed metamorphic rock continues to heat, it can eventually melt and become molten (magma). When the molten rock cools it forms an igneous rock. Metamorphic rocks can form from either sedimentary or igneous rocks.

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2 years ago

What is kepler-10b ?

Stolb23 [73]

Answer:

is the first confirmed terrestial planet to have been discovered outside the solar system by the kepler space telescope

Explanation:

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3 years ago

Read 2 more answers

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

2 years ago

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

Tpy6a [65]

Answer:

1)m=89.01 g

2)V(max) = 97.3 cm/s

Explanation:

Given that

K= 15 N/m

The maximum amplitude ,A=7.5 cm = 0.075 m

Given that 31 oscillations in 15 seconds ,this means that frequency f

f=\dfrac{31}{15}

f=2.066 Hz

lets take mass of the ball = m kg

$f=\dfrac{1}{2\pi}\times \sqrt{\dfrac{K}{m}}$

$m=\dfrac{1}{4\times \pi^2}\times{\dfrac{K}{f^2}}$

$m=\dfrac{1}{4\times \pi^2}\times{\dfrac{15}{2.066^2}}$

m=0.08901 kg

m=89.01 g

The maximum speed

V(max)= ω x A

ω = 2π f= 2 x π x 2.066 = 12.98 rad/s

V(max) = 12.98 x 0.075 =0.973 m/s

V(max) = 97.3 cm/s

3 0

3 years ago