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3 years ago

Could someone do this for me? 50 points for the whole page

Physics

1 answer:

labwork [276]3 years ago

6 0

1. How does the distance from an eye chart affect the number of letters that are recognized on a line?

Constant/Controlled variables: Eye chart, position of viewer and the light in the room.

2. How does the amount of light affect the growth of a plant?

Constant/Controlled variables: Type of plant, amount of water, type/amount of soil

3. How does the amount of oxygen in the water affect the oyster population?

Constant/Controlled Variables: Temperature of water, location, time of year.

4. How will the amount of fertilizer used on a field affect the number of earthworms found there?

Constant/Controlled Variables: type of soil, temperature, amount of water.

5. How does the length of a string affect the number of times a pendulum will swing back and forth in ten seconds?

Constant/Controlled Variables: mass of pendulum, distance pulled back, height of pendulum

6. How does the size of a bicycle tire affect the distance it will travel when it is pedaled in a given number of times?

Constant/Controlled Variables: inflation of tire, force of pedal, terrain (grass vs. blacktop)

Hope this helped :) You can find out the Hypothesis/Prediction on your own from the other things I've included. It will make it look like your own work better that way.

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Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

For aluminium:

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

For copper:

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

Aluminium wire:

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

Copper wire:

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

3 0

4 years ago

Protons have a positive charge A.True B.False

PIT_PIT [208]

Answer:

TRUE

Explanation:

Protons have a positive charge. Electrons have a negative charge. The charge on the proton and electron are exactly the same size but opposite. Neutrons have no charge.

5 0

3 years ago

Read 2 more answers

If a horse does 4000 J of work over 20 m, how much force does the horse use?

Eduardwww [97]

If you take a look at the definition of work, you will get this formula:

$W=Fd \\ F= \frac{4000}{20} \\ \boxed {F=200~N}$

8 0

3 years ago

Modern atomic theory states that the atom is a diffuse cloud surrounding a small, dense nucleus. Identify which particle is in t

dalvyx [7]

note-Unlike the planetary model, our model view is that the electrons move around the nucleus in an electron cloud. Modern atomic theory states that the atom is a diffuse cloud surrounding a small, dense nucleus. Identify which particle is in the nucleus. note-Neutron is a particle in the nucleus.

8 0

3 years ago

Read 2 more answers

Sitting in a second-story apartment, a physicist notices a ball moving straight upward just outside her window. The ball is visi

gregori [183]

Answer:

1.013 s

Explanation:

You can solve this problem using the equations for constant acceleration motion. The velocity at the bottom of the window can be found using this expression:

$(x-x_o) = \frac{1}{2} at^2 + v_ot = -\frac{1}{2}gt^2 + v_ot$

the gravity is negative as it opposes the movement.

$(x-x_o) = -\frac{1}{2}gt^2 + v_ot\\v_o = \frac{(x-x_o)+\frac{1}{2}gt^2}{t} = \frac{1.19m+\frac{1}{2} 9.81m/s^2(0.2s)^2}{0.2s} = 6.931 m/s$

Now, the time elapsed before the ball reappears is 2 times the time that it takes for the ball to go from the bottom of the window, reach maximum height, and reach again the bottom of the window, minus 2 times the time that it takes for the ball to travel from the top to the bottom of the window. The time that takes to the ball to reach maximum height, or in other words, to time that takes for the velocity of the ball to go from vo to 0m/s:

$t_1 = \frac{0m/s - v_o}{g} = \frac{-6.931 m/s}{-9.81m/s^2} = 0.706 s$

Then:

$t = 2t_1 - 2*0.2s = 2*(0.706s) - 0.4s = 1.013 s$

7 0

3 years ago