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ra1l [238]
3 years ago
9

A commonly used unit of mass in the English system is the pound-mass, abbreviated lbm, where 1 lbm = 0.454 kg. What is the densi

ty of glycerine in pound-mass per cubic foot? (Assume the density of glycerine is 1.26 ✕ 103 kg/m3.)
Physics
1 answer:
Gemiola [76]3 years ago
3 0

Answer: The density of glycerine will be 0.028lbm/foot^3

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given:

Density of glycerine= 1.26\times 10^3kg/m^3

1 lbm = 0.454 kg

1 kg =\frac{1}{0.454}=2.2lbm

Thus 0.454kg=\frac{2.2}{1}\times 0.454=0.99

Also 1m^3=35.3147foot^3

Putting in the values we get:

Density=\frac{0.99lbm}{35.3147}=0.028lbm/foot^3

Thus density of glycerine will be 0.028lbm/foot^3

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Yanka [14]
<span>Plants there are a lot. Australian plants are extremely hardy an have adapted to grow in conditions others have not. So when taken outside of Australia they do very, very well. Animals, less so. But there are a few very interesting cases.

</span>
8 0
4 years ago
32. Only a small percentage of the energy
nordsb [41]
Only a small Percentage of the energy emitted by the sun strikes earth because, since the earth is going in circles round and round, the sun only hits part of the earth and not fully. So for example if the earth is going around and the sun hits Africa then in a couple minutes it will go to the next country and that country will have sunlight. Let me know if you need anything else or if this isn't partially correct. GLAD TO HELP! :)
5 0
3 years ago
jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh
Greeley [361]

Answer:

C. 590 mph

\vert v_{cj}\vert=589.49\ mph

Explanation:

Given:

  • velocity of jet, v_j=500\ mph
  • direction of velocity of jet, east relative to the ground
  • velocity of Cessna, v_c=150\ mph
  • direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

\vec v_j=500\ \hat i\ mph

&

\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)

\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph

Now the vector of relative velocity of Cessna with respect to jet:

\vec v_{cj}=\vec v_j-\vec v_c

\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )

\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph

Now the magnitude of this velocity:

\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}

\vert v_{cj}\vert=589.49\ mph is the relative velocity of Cessna with respect to the jet.

8 0
3 years ago
A resistor of 5 Ω is placed in a circuit. The voltage drop across the resistor is 12 V. What is the current through the resistor
Kaylis [27]
<span>Data:

</span>R = 5 Ω
U = 12 V
i = ?
<span>
Formula:

</span>U = R*i → i =  \frac{U}{R}<span>

Solving:

</span>i = \frac{U}{R}
i = \frac{12}{5}
\boxed{i = 2.4A}

Answer:

<span>The current through the resistor is 2.4 Amperes</span><span>

</span>
5 0
3 years ago
How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example
MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the (velocity)^2 and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

\text { Kinetic Energy }=\frac{1}{2} m v^{2}

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}

\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

4 0
3 years ago
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