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jeyben [28]
3 years ago
8

Two children want to balance horizontally on a seesaw. The first child is sitting one meter to the left of the pivot point locat

ed at the center of mass of the seesaw. The second child has one-half the mass of the first child. Where should the second child sit to balance the seesaw?
a. 1m to the right of the pivot
b. 0.5m to the right of the pivot
c. 4m to the right of the pivot
d. 2m to the right of the pivot
Physics
1 answer:
Natali [406]3 years ago
4 0

Answer:

d. 2m to the right of the pivot

Explanation:

m1 = m

m2 = 0.5m

d1 = 1m

d2 = ?

from principle of moment,

CWM = ACWM

m × 1 = 0.5m × d2

d2 = m/0.5m

= 1/0.5

= 2m

The 2nd child will have to sit 2m to the right

The turning effect of a force is known as the moment. It is the product of the force multiplied by the perpendicular distance from the line of action of the force to the pivot or point where the object will turn.

The principle of moments states that when in

equilibrium the total sum of the anti clockwise

moment is equal to the total sum of the

clockwise moment.

When a system is stable or balance it is said to be in equilibrium as all the forces acting on the system cancel each other out.

In equilibrium

Total Anticlockwise Moment = Total

Total Anticlockwise Moment = TotalClockwise Moment

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Answer:

Explanation:

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3 years ago
food product with 10 kg mass is being transported to thesurface of the moon,where the acceleration due to gravity is1.624 m/s2;
larisa [96]

Answer:

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

Explanation:

g = Acceleration due to gravity

m = Mass = 10 kg

Weight on Earth

W=mg\\\Rightarrow W=10\times 9.81\\\Rightarrow W=98.1\ N

Converting to lbf

98.1\times 0.22481=22.053861\ lbf

On Moon

W=10\times 1.624\\\Rightarrow W=16.24\ N

Converting to lbf

16.24\times 0.22481=3.6509144\ lbf

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English units 22.053861 lbf, 3.6509144 lbf

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An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu
Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

\alpha = \frac{\omega_f - \omega_i}{t}

where we have

\alpha = -80.0 rad/s^2 is the angular acceleration (negative since the rotor is slowing down)

\omega_f is the final angular speed

\omega_i = 2000 rad/s is the initial angular speed

t = 10.0 s is the time interval

Solving for \omega_f, we find the final angular speed after 10.0 s:

\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

\alpha = \frac{\omega_f - \omega_i}{t}

If we re-arrange it for t, we get:

t = \frac{\omega_f - \omega_i}{\alpha}

where here we have

\omega_i = 2000 rad/s is the initial angular speed

\omega_f=0 is the final angular speed

\alpha = -80.0 rad/s^2 is the angular acceleration

Solving the equation,

t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s

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