Answer:
m = 170 g
Explanation:
Meter stick is suspended at 40 cm mark
So here the torque due to additional mass and torque due to weight of the spring must be counter balanced
given that
1) 220 g is suspended at x = 5 cm
2) 120 g is suspended at x = 90 cm
3) mass of the scale is acting at its mid point i.e. x = 50 cm
now with respect to the suspension point the torque must be balanced
so we have
by solving above equation we have
m = 170 g
1. $85047
2. $113,080
3. $ 75,820
4. $41595
Answer:
11250 N
Explanation:
From the question given above, the following data were obtained:
Normal force (R) = 15000 N
Coefficient of static friction (μ) = 0.75
Frictional force (F) =?
Friction and normal force are related by the following equation:
F = μR
Where:
F is the frictional force.
μ is the coefficient of static friction.
R is the normal force.
With the above formula, we can calculate the frictional force acting on the car as follow:
Normal force (R) = 15000 N
Coefficient of static friction (μ) = 0.75
Frictional force (F) =?
F = μR
F = 0.75 × 15000
F = 11250 N
Therefore, the frictional force acting on the car is 11250 N
Her <span>acceleration is 8.75 hope is helps</span>
When a satellite is moving around the Earth's orbit, two equal forces are acting on it. The centripetal and the centrifugal force. The centripetal force is the force that attracts the object toward the center of the axis of rotation. The opposite force is the centrifugal force. It draws the object away from the center. When these forces are equal, the satellite uniformly rotates along the orbit.
Centripetal force = Centrifugal force
Mass of satellite * centripetal acceleration = Mass of satellite * centrifugal acceleration
Centripetal acceleration = Centrifugal acceleration
ω^2r
where
= mass of earth
G = gravitational constant = 6.6742 x 10-11<span> m</span>3<span> s</span>-2<span> kg</span><span>-1
</span>
= radius of earth
ω = angular velocity
<span>r = radius of orbit
To convert to angular velocity:
</span>Tangential velocity = rω
ω = 5000/r
Then,
r = 2557110.465 m
Therefore, the distance of the centers of the earth and the satellite is
2.6 x 10^6 m.
<span>
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