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Svetradugi [14.3K]

2 years ago

7

A body travels a distance of 20m in the 7th second and 24m in the 9th second. How much distance shall it travel in the 15th seco

nd?

2 answers:

DaniilM [7]2 years ago

4 0

Answer:

<u>36 m</u>

Explanation:

We can consider this to be an AP.

Then,

a₇ = 20
a₉ = 24

<u>Subtract a₇ from a₉.</u>

a + 8d - a + 6d = 24 - 20
2d = 4
d = 2

a + 6(2) = 20
a = 8

<u>Finding a₁₅</u>

a₁₅ = a + 14d
a₁₅ = 8 + 14(2)
a₁₅ = 8 + 28
a₁₅ = <u>36 m</u>

KengaRu [80]2 years ago

4 0

f(7)=20
f(9)=24

Find rate of change

f(9)-f(7)/9-7
24-20/9-7
4/2
2m/s

So

Distance travel in 6 s

Speed×Time
6(2)
12m

Distance in 15s:-

12+24
36m

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Two ropes support a load of 478 kg. The two ropes are perpendicular to each other, and the tension in the first rope is 2.2 time

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Answer:

T₂ = 1937.68 N

Explanation:

First, we will calculate the weight of the object:

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Now, we will calculate the resultant tension in the ropes. Since the ropes are perpendicular. Therefore,

$T = \sqrt{T_1^2+T_2^2}\\$

where,

T = Resultant Tension

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According to the given condition tension in the first rope is 2.2 times the tension in the second rope:

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$T = W\\2.42T_2 = 4689.18\ N\\\\$

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Using the superposition method, calculate the current through R5 in Figure 8-71

Natasha2012 [34]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

$r_1 = \frac{2.2 * 1}{2.2 + 1}$

$r_1 = 0.6875 k ohm$

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

$r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}$

$r_2 = 0.95 k ohm$

now the current flowing through the battery is

$i = \frac{2}{1 + 0.95} = 1.02 mA$

now this will divide into R3 and R2 so current flowing in R3 will be

$i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA$

now this will again divide in R4 and R5

so current in R5 will be

$i_5 = \frac{R_4}{R_4 + R_5}* i_1$

$i_5 = 0.18 mA$

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

$r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm$

also after its series with R3

$r_2 = 1 + 0.6875 = 1.6875 k ohm$

now it is in parallel with R5 on other side

$r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm$

now current through the battery will be given as

$i = \frac{3}{1 + 0.95} = 1.53 mA$

now it is divide in r2 and R5

so current in R5 is given as

$i_5 = \frac{r_2}{r_2 + R_5}*i$

$i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53$

$i_5 = 0.67 mA$

now the total current in R5 will be given by super position which is

$i = 0.67 + 0.18 = 0.85 mA$

so there is 0.85 mA current through R5 resistance

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3 years ago