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3 years ago

Which of the following are equal to -40°F?

Physics

2 answers:

ValentinkaMS [17]3 years ago

8 0

Answer: -40 F and 233 K

Explanation:

Apex

Marianna [84]3 years ago

3 0

I think It may be -40degreeF is equal to -80degreeC.

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Batteries supply DC, direct current, which does not change in direction over time, though eventually they get exhausted.

SashulF [63]

My Anonymuos friend, I could not agree more with your statement.
Truer words have rarely if ever been expressed.

6 0

3 years ago

Two pieces of clay are moving directly toward each other. When they collide, they stick together and move as one piece. One piec

Wittaler [7]

Answer:

The fraction of the total initial kinetic energy is lost during the collision is $\dfrac{11}{17}\ J$

Explanation:

Given that,

Mass of one piece = 300 g

Speed of one piece = 1 m/s

Mass of other piece = 600 g

Speed of other piece = 0.75 m/s

We need to calculate the final velocity

Using conservation of energy

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v$

Put the value intro the formula

$300\times10^{-3}\times1+600\times10^{-3}\times(0.75)=(300\times10^{-3}+600\times10^{-3})v$

$v=\dfrac{00\times10^{-3}\times1+600\times10^{-3}\times(-0.75)}{(300\times10^{-3}+600\times10^{-3})}$

$v=-0.5\ m/s$

We need to calculate the total initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}\times300\times10^{-3}\times1^2+\dfrac{1}{2}\times600\times10^{-3}\times(0.75)^2$

$K.E_{i}=0.31875\ J$

We need to calculate the total final kinetic energy

Using formula of kinetic energy

$K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Put the value into the formula

$K.E_{f}=\dfrac{1}{2}\times(300\times10^{-3}+600\times10^{-3})\times(-0.5)^2$

$K.E_{f}=0.1125\ J$

We need to calculate the energy lost during the collision

Using formula of energy lost

$energy\ lost=\dfrac{0.31875-0.1125}{0.31875}$

$energy\ lost=\dfrac{11}{17}\ J$

Hence, The fraction of the total initial kinetic energy is lost during the collision is $\dfrac{11}{17}\ J$

3 0

3 years ago

A gas sample is heated from -20.0°C to 57.0°C and the volume is increased from 2.00 L to 4.50 L. If the initial pressure is 0.14

Svetach [21]

Answer:

The answer is e.

Explanation:

We take:

$T_{1}=253K$

$V_{1}=2.00 l$

$P_{1}=0.14atm$

$V_{2}=4.50 l$

$T_{2}=330K$

Taking the gas as an ideal gas, we can use the ideal gas law:

$\frac{PV}{nRT}$

⇒ $n=\frac{P_{1}V_{1}}{RT_{1}}$ ⇒ $n=0.013mol$

Then:

$P_{2}=\frac{nRT_{2}}{V_{2}}$ ⇒ $P_{2}=0.0811 atm$

Taking R=0.08205atmL/molK.

5 0

3 years ago

If the output work of a simple machine is ____ than the input work, the machine is said to have less than 100% efficiency.

bearhunter [10]

Less than. A machine should have equal or more output than input to have a high efficiency. :)

6 0

3 years ago

A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass

Talja [164]

Answer:

a)1.93 kg-m^2

b) 1.45 kg-m^2

c) = 0

d) 1.15 kg-m^2

Explanation:

mass of the bar M = 4 kg

length of the bar = 2 m

mass of balls m1= m2= 0.3 kg

moment of inertia of bar $I= \frac{ML^2}{12}$

about an axis perpendicular to the bar through its center.

a) MOI of bar + 2×m×(L/2)^2

$I= \frac{ML^2}{12}$ + $2m\frac{L}{2}^2$

now putting the values of m, M and L as above and solving we get

I= 1.93 kg-m^2

b) perpendicular to the bar through one of the balls

$I=M\frac{L^2}{3} +mL^2$

$I=4\frac{2^2}{3} +0.3\times4^2$ = 1.45 kg-m^2

c) parallel to the bar through both balls

zero as the no mass distribution along the parallel to the bar through both balls.

d) parallel to the bar and 0.500 m from it.

I= $(M+2m_1)\frac{1}{2}^2$

putting values and solving we get

1.15 kg-m^2

4 0

4 years ago