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tresset_1 [31]
3 years ago
14

Bert is making a strawberry milkshake in his blender. A tiny, 5.0 g strawberry is rapidly spun around inside of the container wi

th a speed of 14.0 m/s, held by a centripetal force of 10.0 N. What is the radius of the container at this location?
Physics
2 answers:
alexandr402 [8]3 years ago
6 0

Answer:

Explanation:

Given that,

Mass of strawberry

m = 5g = 0.005kg

Speed at which it is spun in a circular motion

v = 14m/s

Centripetal force that held it in motion

Fc = 10N

Radius of circular motion r?

Centripetal force is given as

Fc = m•ac

Where m is mass

And ac is centripetal acceleration.

Then, centripetal acceleration is give as

ac = v²/r.

So,

Fc = mv²/r

Making r subject of formulas

r = mv² / Fc

Where,

m is mass m = 0.005kg

v is velocity v = 14m/s

Fc is centripetal force Fc = 10N

Then,

r = mv² / Fc.

r = 0.005 × 14² / 10

r = 0.098m

r = 9.8cm

The radius of the path of circular motion is 9.8cm

gulaghasi [49]3 years ago
4 0

Answer:

r = 0.098m

the radius of the container at this location is 0.098m

Explanation:

Centripetal force F = (mv^2)/r

Radius r = (mv^2)/F

Where;

r = radius

F = centripetal force = 10N

m = mass = 5g = 0.005kg

v = velocity = 14m/s

Substituting the values;

r = (0.005 × 14^2)/10

r = 0.098m

the radius of the container at this location is 0.098m

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