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Harrizon [31]
2 years ago
11

650 N boy and a 419 N girl sit on a 150 N porch swing that is 2.0 m long. If the swing is supported by a chain at each end, what

is the tension in each chain when the boys sits .8 m from one end and the girls sits .5 m from the other end
Physics
1 answer:
Sergio039 [100]2 years ago
7 0

The tension in the two chains T1 and T2 is 676.65 N and 542.53 N respectively.

<h3>Principle of moments</h3>

The Principle of Moments states that when a body is in equip, the sum of clockwise moment about a point is equal to the sum of anticlockwise moment about the same point.

The formula for calculating moment is given below:

  • Moment = Force × perpendicular distance from the pivot

<h3>Calculating the tension in the chains</h3>

From the principle of moments:

Let tension in chain 1 be T1 and tension in chain 2 be T2.

T1 + T2 = 150 + 650 + 419

T1 + T2 =1219

Taking all distances from chain 1,

Sum of Moments = 0

419 × 0.5 + 150 × 0.85 + 650 × 0.9 = T2 × 1.7

T2 = 922/17

T2 = 542.35 N

Then, T1 = 1219 - 542.35

T1 = 676.65 N

Therefore, the tension in the two chains T1 and T2 is 676.65 N and 542.53 N respectively.

Learn more about tension and moments at: brainly.com/question/187404

brainly.com/question/14303536

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An object weighing 49 N is dropped from a height of 30 m. It is found to be moving with a velocity of 24m/s just before it hits
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The  frictional force will be 0.22N.

<h3>What is Frictional force?</h3>

Frictional force is the force generated between two surfaces that are in contact and slide against each other.

Given,

Weight=4N

mass =4.9/9.8=0.5kg

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Acceration , v²-u²=2as

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By Using conservation  of energy ,

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The force will be 0.2N

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scZoUnD [109]

If Fg=mg=ma and, Fg(planetX)=1/5Fg(earth)

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In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
muminat

Answer:

2.1\cdot 10^{21} electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

E=\frac{kQ}{r^2}

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a  charged sphere, so we have:

E_b=3.00\cdot 10^6 N/C is the maximum electric field strength tolerated by the air before breakdown occurs

r=1.00 km = 1000 m is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C

Assuming that the cloud is negatively charged, then

Q=-333.3 C

And since the charge of one electron is

e=-1.6\cdot 10^{-19}C

The number of excess electrons on the cloud is

N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}

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