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Karolina [17]
2 years ago
13

If the weight of the ruler is one Newton ,Gc cannot have a value more than 25cm

Physics
1 answer:
nevsk [136]2 years ago
4 0

If the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm.

The given parameters:

  • Weight of the ruler = 1 N

<h3>What is center of gravity (CG)?</h3>
  • Center of gravity is the point at which the weight of an object is concentrated.

Let the length of the ruler = L

The center of the gravity of the ruler is calculated as follows;

X_{CG} = \frac{W(L_0) + W(L -X_{CG})}{W} \\\\X_{CG} = \frac{1(0) + 1(L -X_{CG})}{1}\\\\X_{CG} = L - X_{CG}\\\\X_{CG } + X_{CG} = L\\\\2X_{CG} = L\\\\X_{CG} = \frac{L}{2} \\\\when , \ L = 50 \ cm\\\\X_{CG} = \frac{50}{2} \\\\X_{CG} = 25 \ cm

Thus, if the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm. This may change if the length of the ruler changes because the center of gravity of uniform ruler depends on the length of the ruler.

Learn more about center of gravity here: brainly.com/question/6765179

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Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0
3 years ago
Plz help me plz its in the picture
natka813 [3]
Il write down the answers of the blanks which are as follows:-
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5 0
3 years ago
an object weighting 100g is thrown upwards from the ground at a speed of 100 m/s.where will the potential energy of the object b
Kay [80]

Answer:

333.3 m

Explanation:

Given

m =100g\ =\  0.1kg\\v = 100 m/s\\g = 10 m/s ^2

Potential energy =\frac{2}{3}\  of\  Kinetic\  energy......Equation(1)

We know that

Potential energy=mgh

Kinetic energy =\frac{1}{2} mv^{2}

Now From the Equation(1)

mgh=\frac{2}{3}*\frac{1}{2} mv^{2}\\  gh=\frac{v^{2} }{3} \\10 * h=\ \frac{10000}{3}\\ h=\ \frac{1000}{3} \\h=333.3\  m

3 0
3 years ago
At what rate is soda being sucked out of a cylindrical glass that is 6 in tall and has radius of 2 in? The depth of the soda dec
evablogger [386]

Answer:

The soda is being sucket out at a rate of 3.14 cubic inches/second.

Explanation:

R= 2in

S= π*R²= 12.56 inch²

rate= 0.25 in/sec

rate of soda sucked out= rate* S

rate of soda sucked out=  3.14 inch³/sec

4 0
3 years ago
10 N pushes a 10 kg crate to the right. Determine the acceleration of the crate.
oksano4ka [1.4K]
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8 0
3 years ago
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