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Aleks [24]
1 year ago
12

The platofm on the school stage is 8 3/4 feet wide. each chair is 1 5/12 feet wide. how many chairs will fit acroos the platform

?
Physics
1 answer:
wariber [46]1 year ago
3 0

6 chairs will fit across the platform.

<h3>Calculation</h3>

We have a school stage of 8 3/4 feet wide.

It's also given that each chair is 1 5/12 feet wide.

To find out the number of chairs possible we can simply divide the total length available by width of the chair.

8 3/4 = 35/4

1 5/12 =17/12

So, 35/4 / 17/12

So, the final answer comes up to 6.

so, the total number of chairs that would fit in are 6.

To know more about random number intervals, visit:

brainly.com/question/13657605

#SPJ4

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With countercurrent flow, diffusion happened in all regions of the filter. Explain why
jeka94

Answer:

When the blood and the dialysate are flowing in the same direction, as the the dialysate and the blood move away from the region of higher concentration of the urea, to a region distant from the source, the concentration of urea in the blood stream and in the dialysis reach equilibrium and diffusion across the semipermeable membrane stops within the higher filter regions such as II, III, IV or V

However, for counter current flow, as the concentration of the urea in the blood stream becomes increasingly lesser the, it encounters increasingly unadulterated dialysate coming from the dialysate source, such that diffusion takes place in all regions of the filter

Explanation:

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4 years ago
What causes one plate to sink under another <br> 1 Mass<br> 2 Shape<br> 3 Density
solong [7]
It is either the mass or density, I believe it is 3 density tho but it could be mass good luck:)
3 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
Consider a circuit with a main wire that branches into two other wires. If the current is 10 A in the main wire and 4 A in one o
k0ka [10]

If the primary wire's power is 10 A and one branch's power is 4 A, another branch's power will be 6A.

According to Kirchhoff's current law (KCL), the total current flowing through a parallel route circuit's junction equals the total current flowing away from it.

Provided that one of the two branches through which power exits the intersection has a flow of 4A, and also that the junction's overall flow entering it is 10A, the entire current going the junction should be 10A.

Consequently, the second wire's power may be expressed as;

I = I1+ I2 [ where I= total current (10A);

                I1= current in one branch (4A) &

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⇒I2 = I - I1

⇒I2 = 10A - 4A

⇒I2 = 6A

Therefore, it can be concluded that when the primary wire bears 10A power having 4A in one of its branches, another branch carries 6A power.

Learn more about Kirchhoff's law here:

brainly.com/question/6417513

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2 years ago
What is the definition of mass.
kozerog [31]
A coherent, typically large body of matter with no definitive shape
Hope this helps good luck
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3 years ago
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