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Maru [420]
3 years ago
10

How does air pressure change with altitude?

Physics
1 answer:
charle [14.2K]3 years ago
3 0
Air pressure changes with altitude because of issues related to gravity. Molecules have more weight the closer they are to the Earth and more of them move to lower elevations as a result; this causes increased pressure because there are more molecules in number and proximity. Conversely, air at higher elevations has less weight, but also forces pressure on those layers below it, resulting in the molecules closer to the Earth supporting more weight, increasing the pressure
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How does the angle of incident and reflection affect your solar oven?
JulijaS [17]
It determines whether the light beams will be concentrated or scattered.
7 0
3 years ago
DONT ANSWER IF YOU DONT KNOW. For a person who is farsighted, what would they need to have a correct focal point?
fiasKO [112]
A) Convex Lens.... hope it helps :)
5 0
3 years ago
a length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be
Natali [406]
The equivalent resistance of n resistors connected in parallel is given by
\frac{1}{R_{eq}} =  \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n} (1)

In our problem, the resulting resistance of the 5 pieces connected in parallel is R_{Eq}=2.00 \Omega, and since the 5 pieces are identical, their resistance R is identical, so we can rewrite (1) as
\frac{1}{R_{Eq} }= \frac{1}{2 \Omega}= \frac{1}{R}+ \frac{1}{R}   + \frac{1}{R}   + \frac{1}{R}   + \frac{1}{R}   =  \frac{5}{R}
From which we find R= 5 \cdot 2 \Omega = 10 \Omega.

So, each piece of wire has a resistance of 10 \Omega. Before the wire was cut, the five pieces were connected as they were in series. The equivalent resistance of a series of n resistors is given by
R_{Eq}=R_1 + R_2 + ...+R_n
So if we apply it at our case, we have
R_{eq}=R+R+R+R+R=5 R= 5\cdot 10 \Omega= 50 \Omega

therefore, the resistance of the original wire was 50 \Omega.
5 0
3 years ago
Please help with Physics Circuits!
Zigmanuir [339]
1) Let's start by calculating the equivalent resistance of the three resistors in parallel, R_2, R_3, R_4:
\frac{1}{R_{234}}= \frac{1}{R_2}+ \frac{1}{R_3}+ \frac{1}{R_4}= \frac{1}{4.5 \Omega}+ \frac{1}{1.3 \Omega}+ \frac{1}{6.3 \Omega}=1.15 \Omega^{-1}
From which we find
R_{234}= \frac{1}{1.15 \Omega^{-1}}=0.9 \Omega

Now all the resistors are in series, so the equivalent resistance of the circuit is the sum of all the resistances:
R_{eq}=R_1 + R_{234} = 5 \Omega + 0.9 \Omega = 5.9 \Omega
So, the correct answer is D) 


2) Let's start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} =  \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}
From which we find
R_{23} = 2.5 \Omega

And these are connected in series with a resistor of 10 \Omega, so the equivalent resistance of the circuit is
R_{eq}=10 \Omega + 2.5 \Omega = 12.5 \Omega

And by using Ohm's law we find the current in the circuit:
I= \frac{V}{R_{eq}}= \frac{9 V}{12.5 \Omega}=0.72 A
So, the correct answer is C).


3) Let' start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} =  \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}
From which we find
R_{23} = 2.5 \Omega
Then these are in series with all the other resistors, so the equivalent resistance of the circuit is
R_{eq}=R_1 + R_{23}+R_4 = 5 \Omega + 2.5 \Omega + 5 \Omega =12.5 \Omega

And by using Ohm's law we find the current flowing in the circuit:
I= \frac{V}{R_{eq}}= \frac{12 V}{12.5 \Omega}=0.96 A

And so the voltage read by the voltmeter V1 is the voltage drop across the resistor 2-3:
V= I R_{23} = (0.96 A)(2.5 \Omega)=2.4 V
So, the correct answer is D).


4) Again, let's start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{13 \Omega}+ \frac{1}{18 \Omega}=0.13 \Omega^{-1}
From which we find
R_{23} = 7.55 \Omega

Now all the resistors are in series, so the equivalent resistance of the circuit is:
R_{eq}= R_1 + R_{23}+R_4=8.5 \Omega+7.55 \Omega + 3.2 \Omega = 19.25 \Omega

The current in the circuit is given by Ohm's law
I= \frac{V}{R_{Eq}}= \frac{15 V}{19.25 \Omega}=0.78 A

Now we can compare the voltage drops across the resistors. Resistor 1:
V_1 = I R_1 = (0.78 A)(8.5 \Omega)=6.63 V
Resistor 2 and resistor 3 are in parallel, so they have the same voltage drop:
V_2 = V_3 = V_{23} = I R_{23} = (0.78 A)(7.55 \Omega)=5.89 V
Resistor 4:
V_4 = I R_4 = (0.78 A)(3.2 \Omega)=2.50 V

So, the greatest voltage drop is on resistor 1, so the correct answer is D).


5) the figure shows a circuit with a resistor R and a capacitor C, so it is an example of RC circuit. Therefore, the correct answer is D).

6) The circuit is the same as part 4), so the calculations are exactly the same. Therefore, the power dissipated on resistor 3 is
P_3 = I_3^2 R_3 =  \frac{V_3^2}{R_3}= \frac{(5.89 V)^2}{18 \Omega}=2.0 W
So, correct answer is B).

7) The circuit is the same as part 4), so we can use exactly the same calculation, and we immediately see that the resistor with lowest voltage drop was R4 (2.50 V), so the correct answer is B) R4.
5 0
3 years ago
Read 2 more answers
I drop a quarter from a 2 story house. It takes 1.85 to hit the ground. what was the velocity
Valentin [98]

At the instant you dropped it from your hand, its speed was zero.
The coin gained 9.8 m/s every second it fell.  At the end of 1.85 sec,
just before it went 'tink' onto the ground, its speed was

           (9.8 m/s) x (1.85 sec) = 18.13 m/s .

Its velocity at that final instant was 18.13 m/s downward.

7 0
3 years ago
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