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3 years ago

14

A 35 kg child moves with uniform circular motion while riding a horse on a carousel. The horse is 3.2 m from the carousel's axis

of rotation and has a tangential speed of 2.6 m/s.
What is the child's centripetal acceleration?

2 answers:

musickatia [10]3 years ago

7 0

A=ω²r=0.8²*3.2=2.0 m/s²
v=ωr, then ω=v/r;
ω=2.6/3.2=0.8 rad/s

Lana71 [14]3 years ago

4 0

The answer is 74 N on edge.

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one group of atoms on the periodic tables are known as "noble gases". why have they been given this name?

maria [59]

This group is called “noble gases” because they do not react with other elements. This is because they have a full valence shell.

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3 years ago

Given that coal used by electric power plants has a heating value of 27.5 million btus metric ton (25 million btus per ton), det

fredd [130]

Answer:

• 36.4 kg of coal.

• 80 pounds of coal.

Explanation:

Using proportionality constant,

Mass of coal = 1,000,000/27,500,000 btus/metric ton

= 0.0364 metric tons of coal

Mass of coal = 1,000,000/25,000,000 btus/ton

= 0.04 tons of coal.

Converting metric tons to kilogram,

1 metric ton = 1000kg,

0.0364 metric ton;

= 36.4 kg of coal.

Converting tons to pounds,

1 ton = 2000 pounds,

0.04 metric ton;

= 80 pounds of coal.

3 0

3 years ago

Psuedopod in a sentence

mixas84 [53]

Does it have to be that exact word. cause it is just another term for psuedopodium

5 0

3 years ago

A cart travels 4.00 meters east and then4.00 meters north. determine the magnitude ofthe cart’s resultant displacement.

gogolik [260]

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Square root of (4^2 + 4^2) = 4*squareRoot(2)

7 0

3 years ago

The Coulomb force between two charges q1 and q2 at separation r in the air is 10N. If half of the separation is filled with medi

adoni [48]

Answer:

The value of new coulomb force is 1.43 N.

Explanation:

Given;

Coulomb's force in vacuum (air), $F_v$ = 10 N

dielectric constant, K = 7

The Coulomb's force between two charges separated by a distance r in a vacuum is given as;

$F_v = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2}$

The Coulomb's force between two charges separated by a distance r in a medium with dielectric constant is given as

$F_m = \frac{1}{4\pi K\epsilon_0} \frac{q_1q_2}{r^2}$

Take the ratio of the two forces;

$\frac{F_v}{F_m} = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2} \ \times \ \frac{4\pi K\epsilon_0 r^2}{q_1q_2} = K\\\\\frac{F_v}{F_m} = K\\\\\frac{10}{F_m} = 7\\\\F_m = \frac{10}{7} \\\\F_m = 1.43 \ N$

Therefore, the value of new coulomb force is 1.43 N.

5 0

3 years ago