PHYSICS COLLEGE

What is the density, in Mg/m3, of a substance with a density of 0.14 lb/in3? (3 pts) What is the velocity, in m/sec, of a vehicle traveling 70 mi/hr?

Answers

Answer 1

Answer:

Answer:

$276.74* 10^8Mg/m^3$

31.29 m/sec

Explanation:

We have given density of substance $0.14lb/in^3$

We have convert this into $Mg/m^3$

We know that 1 lb = 0.4535 kg. so 0.14 lb = 0.14×0.4535 = 0.06349 kg

We know that 1 kg = 1000 g ( 1000 gram )

So 0.06349 kg = 63.49 gram

And we know that 1 gram = 1000 milligram

So 63.49 gram $=63.49* 10^3\ Mg$

We know that $1 in^3=1.6387* 10^(-5)m^3$

So $0.14in^3=0.14* 1.6387* 10^(-5)=0.2294* 10^(-5)m^3$

So $0.14lb/in^3$ =\frac{63.49\times 10^3}{0.2249\times 10^{-5}}=276.74\times 10^8lb/m^3[/tex]

In second part we have to convert 70 mi/hr to m/sec

We know that 1 mi = 1609.34 meter

So 70 mi = 70×1609.34 = 112653.8 meter

1 hour = 3600 sec

So 70 mi/hr $=(70* 1609.34meter)/(3600sec)=31.29m/sec$

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Those who support the alien theory believe that the ancient Egyptians where very primitive. What evidence about the ancient Egyptians suggests that claim is inaccurate?

Answers

Answer:

They had a well-developed agricultural system , they raised domesticated animals , they developed a writing system, and they used early forms of tools.

Final answer:

The claim that the ancient Egyptians were primitive and relied on aliens to build their monuments is inaccurate. Evidence from archaeology and history shows that the ancient Egyptians had advanced knowledge and skills in various fields. Their construction techniques and use of mathematics in building the pyramids are well-documented.

Explanation:

The claim that the ancient Egyptians were primitive and that their accomplishments, such as building the pyramids, were assisted by aliens is inaccurate. There is evidence from archaeology and history that ancient monuments were built by ancient people using their own ingenuity and capabilities. The ancient Egyptians had advanced knowledge in various fields including architecture, engineering, astronomy, and mathematics. For example, their construction techniques and use of mathematics in building the pyramids and other structures are well-documented.

Learn more about Ancient Egyptians

It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.2m from the edge.-How far are the goggles from the edge of the pool?

Answers

Answer:

the googles are 5.3 m from the edge

Explanation:

Given that

depth of pool , d = 3.2 m

Now, let i be the angle of incidence

a laser pointer 0.90 m above the edge of the pool and laser beam enters the water 2.2 m from the edge

⇒tan i = 2.2/0.9

$i=arctan(2.2/.90)$

solving we get

i = 67.8°

Using snell's law ,

n1 ×sin(i) = n2 ×sin(r)

n1= refractive index of 1st medium= 1

n2= refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

$1* sin(67.8) = 1.33* sin(r)$

r = 44.1°

Now,

distance of googles = 2.2 + d×tan(r)

distance of googles = 2.2 + 3.2×tan(44.1)

distance of googles = 5.3 m

the googles are 5.3 m from the edge

An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and a quality of 0.25. An electric heater inside the tank is turned on to heat this H2O until the pressure increases to 200 kPa. Please determine the change in total entropy of water during this process. Hint: See if you can find the electrical work consumed during this process.

Answers

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

$v_1 = vf + x(vg - vf)$

$= 0.001053 + 0.25(1.1594 - 0.001053)$

$v_1 = 0.20964 m^3/kg$

$s_1 = Sf + x(Sfg)$

$= 1.4337 + 0.25 * 5.7894 = 2.88 kJ/kg K$

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume so $v_1 - v_2 = 0.29064 m^3/kg$

$v_2 = v_1 = vf + x(vg - vf)$

$=0.29064 = x_2(0.88578 - 0.001061)$

$x_2 = 0.327$

$s_2 = 1.5302 + 0.32 * 5.5968 = 3.36035 kJ/kg K$

Change in entropy $\Delta s = m(s_2 - s_1)$

=3( 3.36035 - 2.88) = 1.44 kJ/kg

A research Van de Graaff generator has a 2.00-m diameter metal sphere with a charge of 5.00 mC on it. (a) What is the potential near its surface?
(b) At what distance from its center is the potential 1.00 MV?
(c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in MeV when the atom is at the distance found in part b?

Answers

Answer:

a) $V=49.5MV$

b) $r=49.5m$

c) $\Delta U=3*(49.5-1)=145.5 MeV$

Explanation:

a) The potential equation is given by:

$V=k(Q)/(r)$

k is the electrostatic constant ( $k=9.9*10^(9)Nm^(2)/C^(2)$ )

Q is the charge Q = 5mC

r is the radius of the sphere r = 1 m

$V=9.9*10^(9)(5*10^(-3))/(1)=49.5MV$

b) We solve it using the same equation.

Here we need to find r:

$r=k(Q)/(V)$

$r=9.9*10^(9)(5*10^(-3))/(1*10^(6))$

$r=49.5m$

c) The relation between difference potential and electrical energy is:

$\Delta U=\Delta Vq$

here q is 3e becuase oxygen atom has three missing electrons

Therefore:

$\Delta U=3*(49.5-1)=145.5 MeV$

I hope it heps you!

As the moon orbits the Earth which of the following changes (1) a. Speed b. Velocity c. Acceleration d. A, B, and C e. None

Answers

Answer:

B- Velocity

Explanation:

This means gravity makes the Moon accelerate all the time, even though its speed remains constant.

Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s. How much time does the driver of the car measure for his trip between the poles?