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juin [17]
3 years ago
8

A 2.0-ohm resistor is connected in a series with a 20.0 -V battery and a three-branch parallel network with branches whose resis

tance are 8.0 ohms each. Ignoring the battery’s internal resistance, what is the current in the batter? Show your work.
Physics
1 answer:
Llana [10]3 years ago
4 0

Answer:

Explanation:

Resolving the parallel resistor branches

\frac{1}{Rtotal} =\frac{1}{8.0ohm} +\frac{1}{8.0ohm} +\frac{1}{8.0ohm} =2.67Ohm

The equivalent resistor is now in series with the 2.0 Ohm resistor

so, by using Ohm's Law

I=\frac{V}{Rtotal+R} \\I=\frac{20.0V}{2.67Ohm+2.0Ohm} \\\\I= 4.28A

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B. If we drop a ball from the roof, its falls<br>downward​
frosja888 [35]

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2 years ago
Read 2 more answers
A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0
3 years ago
Q 1 . How many significant figures are in the following measurement? 0.0009(1 point)
Crazy boy [7]

Here we have some questions about experimental errors.

Q1) We want to see how many significant figures have the measure:

0.0009

The number of significant figures is the number of known digits that are not the leading zeros.

Here we can see four leading zeros, and a single-digit different than zero, which is a 9.

Then we have only one significant figure, the 9.

Q2) Here we will use the measure that is the less exact, as the error of that measure may be larger than the smaller significant figures of the other measures.

Then:

31.2 lb + 38.02lb + 45 lb

The worst measure is 45lb, so the smallest significant figure that we should use is the first one at the left of the decimal point, then we need to round the other two measures to the next whole number, we will get:

31 lb + 38 lb + 45 lb = 114lbs

Q3) We know that the measure is 11.5 seconds and the uncertainty of 1.7%, then the uncertainty will be the 1.7% of the above measure:

(1.7%/100%)*11.5 s = 0.1955 s

Notice that our measure has one significant figure after the decimal point, so we need to round the error to the same significant figure.

0.1955 s ≈ 0.2s

Then the measure is:

11.5 s ± 0.20 s

Q4) We have the measure:

312.0 mph ± 3.9 mph.

The percent uncertainty will be the quotient between the error and the measure times 100%, or:

(3.9 mph/312.0 mph)*100%  = 1.25%

This is a percent error, we do not need to round this.

If you want to learn more, you can read:

brainly.com/question/17339020

5 0
2 years ago
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