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nata0808 [166]
2 years ago
12

Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. Th

e boys pull themselves together along the rod. When they meet the 40-kg boy will have moved what distance?.
Physics
1 answer:
zlopas [31]2 years ago
4 0

When they meet the 40-kg boy would have moved a distance of 6 m.

<h3>Distance moved by the 40 kg boy</h3>

Apply the principle of center mass;

Take the 40 kg mass as the reference point;

X(40 kg) = (40kg x 0  + 60kg x 10 m)/(40 kg + 60 kg)

X(40 kg) = (600)/(100)

X(40 kg) = 6 m

Thus, when they meet the 40-kg boy would have moved a distance of 6 m.

Learn more about distance here: brainly.com/question/2854969

#SPJ1

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3 years ago
Urgently!
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Answer:

0.003333 s to 0.000125s or from 3.33ms to 0.125ms wher m is for milli

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Explanation:

T= 1/f=

if f= 300Hz then T = 1/300 =0.003333 s

if f= 8000 then T= 1/8000 = 0.000125s

now v=f×wave length

or wavelength = speed/ frequency

when f = 300 Hz

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note : i have taken speed of sound as 330 m/s you can take any value given in between 330m/s to 340m/s

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Jennifer and Katie are leaning on each other. Jennifer weighs 150 and Katie weighs 120. Which one is pushing harder on the other
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Read 2 more answers
If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli
Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N==\frac{m(V)^{2}}{R}=m*(w)^2*R

Friction force\\Ff = k*N\\Ff= k*m*w^2*R

Gravitational force \\Fg = m*g

These must be balanced (the net force on the people will be 0) so set them equal to each other.

Fg = Ff

m*g = k*m*w^2*R

g=k*w^{2}*R

w^2 =\frac{g}{k*R}

w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}

There are 2*pi radians in 1 revolution so:

RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88

So you need about 30 RPM to keep people from falling out the bottom

7 0
3 years ago
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