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2 years ago

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How did the lab activities help you answer the lesson question "How do the processes of conduction, convection, and radiation he

lp distribute energy on Earth?" What did you learn from conducting this lab?

1 answer:

Vikentia [17]2 years ago

7 0

Answer: In the lab, I conducted three experiments. And each experiment, energy in the form of he was transferred in a different way. Energy can be transferred through a material like aluminum foil through conduction. And water, energy is transferred through convection. Energy can travel from the lightbulb to the paper through radiation. I learned that conduction could move energy between earths materials, convection can transfer heat through earths Waters, and radiation can transfer of heat from the sun to earth.

Explanation:

This is the sample response on edgenuity

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In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6

stealth61 [152]

Answer:

W = 9533.09 Watt

Explanation:

given,

diameter of pipe inlet, d₁ = 10 cm

r₁ = 5 cm

diameter of pipe outlet, d₂ = 15 cm

r₂= 7.5 cm

head upto water level is to rise = 60 + 5

= 65 m

flow rate = 0.015 m³/s

we know

A₁ v₁ = A₂ v₂ = Q

π r₁² v₁ = π r₂² v₂ = 0.015

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{7.5^2}{5^2} v_2$

$v_1= 2.25 v_2$

$v_2 = \dfrac{0.015}{\pi r_2^2}$

$v_2 = \dfrac{0.015}{\pi 0.075^2}$

v₂ = 0.848 m/s

v₁ = 1.908 m/s

Applying Bernoulli's equation

$P_p = \dfrac{1}{2}\rho (v_2^2-v_1^2)+ \rho g h$

$P_p= \dfrac{1}{2}\times 1000\times (0.848^2-1.908^2)+ 1000\times 9.8\times 65$

$P_p= 635539.32 Pa$

P_p is the pump pressure

Power of the pump

W = P_p x Q

W = 635539.32 x 0.015

W = 9533.09 Watt

6 0

2 years ago

Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

$\mu_o=4\pi \times 10^{-7}\ T-m/A$

Resistivity, $\rho=1.72\times 10^{-8}\ \Omega-m$

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

$R=\rho \dfrac{l}{A}$

$R=\rho \dfrac{l}{\pi r^2}$

$r=\sqrt{\dfrac{\rho l}{R\pi}}$

$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

r = 0.00199 m

or

$r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m$

The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0

2 years ago

A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the c

g100num [7]

Explanation:

The given data is as follows.

Mass of oxygen present = 100 mg = $100 \times 10^{-3}$ g

So, moles of oxygen present are calculated as follows.

n = $\frac{100 \times 10^{-3}}{32}$

= $3.125 \times 10^{-3}$ moles

Diameter of cylinder = 6 cm = $6 \times 10^{-2}$ m

= 0.06 m

Now, we will calculate the cross sectional area (A) as follows.

A = $\pi \times \frac{(0.06)^{2}}{4}$

= $2.82 \times 10^{-3} m^{2}$

Length of tube = 11 cm = 0.11 m

Hence, volume (V) = $2.82 \times 10^{-3} \times 0.11$

= $3.11 \times 10^{-4} m^{3}$

Now, we assume that the inside pressure is P .

And, $P_{atm}$ = 100 kPa = 100000 Pa,

Pressure difference = 100000 - P

Hence, force required to open is as follows.

Force = Pressure difference × A

= $(100000 - P) \times 2.82 \times 10^{-3}$

We are given that force is 173 N.

Thus,

$(100000 - P) \times 2.82 \times 10^{-3}$ = 173

Solving we get,

P = $3.8650 \times 10^{4} Pa$

= 38.65 kPa

According to the ideal gas equation, PV = nRT

So, we will put the values into the above formula as follows.

PV = nRT

$38.65 \times 3.11 \times 10^{-4} = 3.125 \times 10^{-3} \times 8.314 \times T$

T = 462.66 K

Thus, we can conclude that temperature of the gas is 462.66 K.

6 0

2 years ago

__________ curves help lessen the effect of the force of the forward motion on your vehicle as it enters the curve.

Rainbow [258]

Answer:

Banked

Explanation:

Banked curves are formed when the inner edge is below the outer edge.

It is done in order to ensure the reliability of the frictional force as it varies when the road is wet wet or oily. Thus in order to avoid these problems the curved roads are banked.

Banking of the curve provides the necessary centripetal force, i.e., the horizontal component of the normal reaction force to keep the vehicle i motion and thus helps in reducing the effect of the forward motion force on the vehicle.

5 0

2 years ago

Hippos spend much of their lives in water, but amazingly, they don’t swim. manatees, They have, like little very body fat. The d

kenny6666 [7]

Answer:

428.59 N

Explanation:

Buoyant force, $B=Vg\rho$ where V is volume, g is gravitational constant and \rho is density

$B+F_{upward}=mg$ where $F_{upward}$ is upward force

$Vg\rho_{w}+F_{upward}=mg$

$F_{upward}=mg- Vg\rho_{w}$

$F_{upward}=g(mg- V\rho_{w})=g(m-m\frac {\rho_{w}{\rho_{hippo}}$ where $\rho_{hippo}$ is the density of hippo

$F_{upward}=mg(1-\frac {\rho_{w}}{\rho_{hippo}})$

Using g as 9.81

$F_{upward}=1500*9.81*(1-1000/1030)= 428.5922 N$

Therefore, the upward force=428.59 N

3 0

2 years ago