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3 years ago

Matter is defined as anything that has weight and takes up space.

Physics

1 answer:

serg [7]3 years ago

5 0

Answer:

true matter takes up space and is anything that is around us

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Air, considered an ideal gas, is contained in an insulated piston-cylinder assembly outfitted with a paddle wheel. It is initial

Maru [420]

Our data are,

State 1:

$P_1= 10psi=68.95kPa\\V_1 = 1ft^3=0.02831m^3\\T_1 = 100\°F = 310.93K$

State 2:

$P_2 =5psi=34.474kPa\\V_2 = 3ft^3=0.0899m^3$

We know as well that $3BTU=3.16kJ/K$

To find the mass we apply the ideal gas formula, which is given by

$P_1V_1=mRT_1$

Re-arrange for m,

$m= \frac{P_1V_1}{RT_1}\\m= \frac{68.95*0.02831}{(0.287)310.9}\\m=0.021893kg=0.04806lbm\\$

Because of the pressure, temperature and volume ratio of state 1 and 2, we have to

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Replacing,

$T_2 = \frac{P_2V_2}{P_1V_1}T_1\\T_2 =\frac{34.474*0.0844}{68.95*0.02831}*310.93\\T_2 = 464.217K=375.5\°F$

For conservative energy we have, (Cv = 0.718)

$W = m C_v = 0.718 \Delta T +dw\\dw = W - mv\Delta T\\dw = 3.16-(0.0218*0.718)(454.127-310.93)\\dw = 0.765kJ=0.72BTU$

3 0

3 years ago

Can science answer every question

Alla [95]

No because your opinion and beliefs answers many questions

5 0

2 years ago

Read 2 more answers

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

2 years ago

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

aliina [53]

Answer:

9.43 m/s

Explanation:

First of all, we calculate the final kinetic energy of the car.

According to the work-energy theorem, the work done on the car is equal to its change in kinetic energy:

$W=K_f - K_i$

where

W = -36.733 J is the work done on the car (negative because the car is slowing down, so the work is done in the direction opposite to the motion of the car)

$K_f$ is the final kinetic energy

$K_i = 66,120 J$ is the initial kinetic energy

Solving,

$K_f = K_i + W = 66,120 + (-36,733)=29,387 J$

Now we can find the final speed of the car by using the formula for kinetic energy

$K_f = \frac{1}{2}mv^2$

where

m = 661 kg is the mass of the car

v is its final speed

Solving for v, we find

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.43 m/s$

3 0

3 years ago

A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,

alexgriva [62]

The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
$S= \frac{1}{2} a t^2$
where a is the acceleration. Using the data of the problem, we can find a:
$a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2$
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J$

and so the power output is
$P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W$

b) power output at t=4.0s
The velocity at t=4.0 s is
$v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J$

and so the power output is
$P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W$

c) Power output at t=6.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J$

and so the power output is
$P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W$

8 0

2 years ago