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3 years ago

when forces on an object are acting in the same direction, how do you add them together to find the net force?

Physics

1 answer:

AysviL [449]3 years ago

4 0

If two forces act on an object in the same direction, the net force is equal to the sum of the two forces

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It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T

tekilochka [14]

Answer:

ω = 0.05 rad/s

Explanation:

We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

$Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\$

where,

ω = angular velocity of cylinder = ?

g = required acceleration = 9.8 m/s²

r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m

Therefore,

$\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\$

ω = 0.05 rad/s

7 0

3 years ago

A solid ball with a mass of 4.25 kg and a

Liono4ka [1.6K]

7.20 units long yea

8 0

2 years ago

A motorist travels at 270 km in p hours at a certain average speed. an expression for the distance that this motorist travelled

wel

Average speed = total distance / time ⇒ total distance = average speed * time

Average speed = 270 km / p hours
distance = d
hours = x

d = 270/p * x

5 0

3 years ago

A honey bee flaps its wings 200 times per second. How much time is required for one wingbeat? Give your answer in milliseconds.

kakasveta [241]

A bees wings move so rapidly that studying them, even seeing them, has proved difficult.
The honeybees have a rapid wing beat honeybee flaps its wings 230 times every second.

Therefore, a bee flaps its wings over 200 times in about 1ms

3 0

3 years ago

Read 2 more answers

A fan that is rotating at 960 rev/s is turned off. It makes 1500 revolutions before it comes to a stop. a) What was its angular

Evgesh-ka [11]

Answer:

α = 1930.2 rad/s²

Explanation:

The angular acceleration can be found by using the third equation of motion:

$2\alpha \theta=\omega_f^2-\omega_i^2$

where,

α = angular acceleration = ?

θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s

Therefore,

$2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}$

α = - 1930.2 rad/s²

negative sign shows deceleration

5 0

3 years ago