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denis23 [38]
3 years ago
5

Assume that a resistor is connected between the 150 V terminal and the common terminal. The voltmeter is then connected to an un

known voltage using the common terminal and the 300 V terminal. The voltmeter reads 288 V.What is the unknown voltage
Physics
1 answer:
Lena [83]3 years ago
7 0

Answer: 316.8V

Explanation:

given data:

metre moving current = 0.96mA

meters voltage = 288V

or  0.96*300V = 288V

<u><em>Solution:</em></u>

<u><em /></u>v1 = (0.96mA*150)<u><em /></u>

<u><em /></u>= 144V<u><em /></u>

<u><em /></u>

i1 = \frac{144v}{750}

= 0.192mA

i2 = imovement + i1

i2= 0.96mA+0.192mA

= 1.152mA

Vmeasured = 144V+(150)(1.152mA)

=316.8V

the unknown voltage is 316.8V

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3 years ago
Given the information below, estimate the total distance travelled during these 6 seconds using a left endpoint approximation. t
Nutka1998 [239]

Answer:

184 feets

Explanation:

Given the data:

time (sec) __ velocity (ft/sec)

0 __________30

1 __________ 54

2 __________56

3 __________34

4 __________ 8

5 __________ 2

6 __________22

Using left end approximation:

(0,1) ___ f(0) = 30

(1,2) ___ f(1) = 54

(2,3) ___f(2) = 56

(3,4) ___f(3) = 34

(4,5) ___f(4) = 8

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Hence, the Total distance traveled during the 6 second interval is:

Change ; dT = 1

1 * (30 + 54 + 56 + 34 + 8 + 2) = 184

4 0
2 years ago
A car drives off a cliff next to a river at a speed of 30 m/s and lands on the bank on theother side. The road above the cliff i
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Answer:1.301 s

Explanation:

Given

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Time taken to cover 8.3 m

h=ut+\frac{at^2}{2}

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8.3=\frac{gt^2}{2}

t^2=1.69

t=1.301 s

Horizontal distance

R=u\times t

R=30\times 1.301=39.04 m

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0.36+0.07

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