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4 years ago

What must be the centripetal force that keeps the lady bug moving in a circle?

1 answer:

3 0

The centripetal force that keeps the lady bug moving in a circle is FRICTION. The Centripetal Force is a net force acting on an circling object that sustains constant circular motion. It opposes the motion by acting in opposite direction. A ladybug sits at rest on a turntable that is rotating, during this motion, the ladybug's acceleration vector points toward the center of the turntable.

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A stone whirled on a string experiences a centripetal acceleration of 10 m/s². If the string were shortened to half its length (

LUCKY_DIMON [66]

Answer:

a = 80 m/s²

Explanation:

10 = v²/r

a = (2v)²/(r/2)

a = 4v²•(2/r)

a = 8(v²/r)

a = 8(10)

a = 80 m/s²

4 0

3 years ago

A woman drops a vibrating tuning fork, which is vibrating at 513 Hertz, from a tall building. Through what distance (in m) has t

umka21 [38]

Answer:

h = 15.34 m

Explanation:

given,

tuning fork vibration = 513 Hz

speed of sound = 343 m/s

frequency after deflection = 489 Hz

the source (the fork) moves away from the observer, its speed increases and hence the apparent frequency decreases

$f_{apparent} = \dfrac{v}{v+u}f_0$

$489 = \dfrac{343}{343+u}\times 513$

$0.953 = \dfrac{343}{343+u}$

$343+u = \dfrac{343}{0.953}$

$343+u = 359.92$

u = 16.92 m/s

height of the building

v² = u² + 2 g s

16.92² = 2 x 9.8 x h

h = 14.61 m

time taken by sound to reach observer

$t = \dfrac{14.61}{343}$

$t =0.0426\ s$

in this time tuning fork has fallen one more now,

$h' = u t + \dfrac{1}{2}gt^2$

$h' = 16.92\times 0.0426 + \dfrac{1}{2}\times 9.8 \times 0.0426^2$

h' = 0.7296 m = 0.73 m

total distance

h = 14.61 + 0.73

h = 15.34 m

6 0

4 years ago

Why is it that when you throw a Frisbee, it flies better when it is normally thrown and not upside down?

Katena32 [7]

Because the air gets stuck in the curves of the frissbee

6 0

3 years ago

How much power does it take to lift a 5.00 kg mass at a constant speed by 3.00 m in 2.50 s? Unit=W

nydimaria [60]

The weight/force of the object is F = mg = (5.00 kg)(9.81 m/s^2) = 49.05 N
The work done in moving the object is W = Fd = (49.05 N)(3.00 m) = 147.15 J
The power required to accomplish this work in 2.50 s is P = W/t = 147.15 J / 2.50 s = 58.86 Watts

Power requirement increases with increased weight and increased distance, but decreases with increased time.

4 0

4 years ago

Now explore friction force. Set the piece of plastic or wood on the table and push it steadily across the tabletop using your fi

photoshop1234 [79]

Answer: Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are several types of friction: Dry friction is a force that opposes the relative lateral motion of two solid surfaces in contact

5 0

3 years ago