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2 years ago

10

How much centripetal force is needed to make a body of mass 0.5kg to move in a circle of radius 50cm with speed 3ms^-1 ?

2 answers:

scoundrel [369]2 years ago

7 0

Centripetal force is given by F= mv²/r.

Given: m = 0.5 kg, v = 3 m/s, r = 0.5 m

Putting values,

F= mv²/r = 0.5× 3²/0.5 = 9 N

Step2247 [10]2 years ago

7 0

Answer:

this is the one that i accident my favorite character turn to that

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Which of the following changes would make a heat engine less efficient

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3 years ago

After the pendulum is dropped, determine the height at which the kinetic energy is equal to the potential energy.

Fofino [41]

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3 years ago

A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is appl

Taya2010 [7]

Answer:

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

Given that,

Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

$F-f=ma$

Put the value into the formula

$340-f=60\times2.0$

$f=340-60\times2.0$

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Using formula of friction force

$f= \mu mg$

$\mu=\dfrac{f}{mg}$

$\mu=\dfrac{220}{60\times9.8}$

$\mu =0.37$

We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

$F = ma$

$a=\dfrac{F}{m}$

$a=\dfrac{220}{100}$

$a=2.2\ m/s^2$

Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

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3 years ago

an object is thrown into the air with an initial velocity of 12 meters per seconds from a platform that is raised 4 meters above

butalik [34]

Answer:

1.52 seconds

Explanation:

Step 1: identity the given parameters

Initial velocity (u) = 12m/s

Height above ground (h1) = 4m

Final velocity (V) = 0

Step 2: calculate the height travelled by the object from 4m height (h2).

V^2 = U^2 -2gh

0= 12^2-2(9.8*h)

2(9.8*h) = 12^2

19.6*h = 144

h = 144/19.6

h = 7.347 m

Total height above ground (ht) = 4m +7.347m = 11.347m

Step 3: calculate the time reach ground

T = √(2h/g)

T = √(2*11.347/9.8)

T= √(22.694/9.8)

T= √2.316

T= 1.52 seconds

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