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3 years ago

The standard metric unit of volume is the _____.

Physics

1 answer:

Nataliya [291]3 years ago

4 0

Answer:

. The SI unit of volume is the cubic meter (m3), which is a derived unit.

Liter (L) is a special name for the cubic decimeter (dm3).

You might be interested in

The pressure of a box pushes down on the floor is 50 Pa if the box weighs 400 N what is the area of the base of the box

Wewaii [24]

Answer:8m^2

Explanation:

Area=force÷pressure

Area=400÷50

A=8m^2

5 0

3 years ago

You have gas in a container with a movable piston. The walls of the container are thin enough so that its temperature stays the

bearhunter [10]

Answer:

New pressure of the gas increases by 26.5% with respect to initial pressure, new volume decreases 27% with respect to initial volume and new temperature decreases 8% with respect to initial volume.

Explanation:

If we assume the gas is a perfect gas we can use the perfect gas equation:

$PV=nRT$

For Isothermal process:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$ (1)

Where subscripts 1 shows before the isothermal process and 2 after it, because isothermal means constant temperature T1=T2, and pressure increases by 10% means P2=1,1*P1, using these facts on (1) we have:

$V_{2}=\frac{V_{1}}{1.1}$ (2)

For Isobaric process:

$\frac{P_{2}V_{2}}{T_{2}}=\frac{P_{3}V_{3}}{T_{3}}$ (3)

Where subscripts 2 shows before the isobaric process and 3 after it, because isobaric means constant pressure P2=P3, and volume decreases by 20% means V3=0.8*V2, using these facts on (3) we have:

$T_{3}=0.8T_{2}$ (4)

For Isochoric process:

$\frac{P_{3}V_{3}}{T_{3}}=\frac{P_{4}V_{4}}{T_{4}}$ (5)

Where subscripts 3 shows before the isochoric process and 4 after it, because isochoric means constant volume V3=V4, and temperature increases by 15% means T4=1.15*T3, using these facts on (5) we have:

$P_{4}=1.15P_{3}$ (6)

So now because P4=1.15*P3, P2=P3 and P2=1.1*P1:

$P_{4}=1.15*1.1P_{1}=1.265P1$

This is, the new pressure of the gas increases by 26.5% with respect to initial pressure.

Similarly, we have V3=V4, V3=0.8*V2 and V1=1,1*V2:

$V_{4}=\frac{0.8}{1.1}V_{1}=0.72V1$

so the final volume decreases 27% with respect to initial volume.

T4=1,15*T3, T3=0.8*T2 and T1=T2:

$T_{4}=1.15*0.8T_{1}=0.92T1$

The new temperature decreases 8% with respect to initial volume.

3 0

3 years ago

what is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 3 m/sec

s344n2d4d5 [400]

Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)

Formula (Kinetic energy)
$E_{k} = \frac{m*s^2}{2}$

Solving:
$E_{k} = \frac{m*s^2}{2}$
$E_{k} = \frac{1*3^2}{2}$
$E_{k} = \frac{1*9}{2}$
$E_{k} = \frac{9}{2}$
$\boxed{\boxed{E_{k} = 4.5\:J}}\end{array}}\qquad\quad\checkmark$

3 0

3 years ago

Energy conservation

balandron [24]

Answer:

7.328m/s

Explanation:

Given parameters:

height of table = 0.68m

final velocity of the ball = 6m/s

Unknown:

Initial velocity of ball = ?

Solution:

To solve this problem, we are going to employ the appropriate motion equation.

We must understand that this fall occurs in the presence of gravity;

V = U + 2gH

Where;

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

H is the height of the pool table

Since U is the unknown, let us make it the subject of the expression;

U = V - 2gH

U = 6 - (2 x 9.8 x 0.68) = 7.328m/s(deceleration)

7 0

3 years ago

food product with 10 kg mass is being transported to thesurface of the moon,where the acceleration due to gravity is1.624 m/s2;

larisa [96]

Answer:

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

Explanation:

g = Acceleration due to gravity

m = Mass = 10 kg

Weight on Earth

$W=mg\\\Rightarrow W=10\times 9.81\\\Rightarrow W=98.1\ N$

Converting to lbf

$98.1\times 0.22481=22.053861\ lbf$

On Moon

$W=10\times 1.624\\\Rightarrow W=16.24\ N$

Converting to lbf

$16.24\times 0.22481=3.6509144\ lbf$

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

5 0

3 years ago