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Ierofanga [76]
3 years ago
15

An object moves in uniform circular motion what is true regarding the force on the object

Physics
2 answers:
MariettaO [177]3 years ago
5 0

Answer:

<u>Circular Motion:</u>

When an object is moving with a uniform circular motion, it has a net acceleration,α  which is directed towards the center of the circle.

Explanation:

<u>An object in a circular motion:</u>

When an object is moving in a circular motion, there speed is uniform across all the points on the circular path. While the change in acceleration is just because of the change in direction during the motion of the object. The force acting on the body is the centripetal force, Fc. As the force,Fc is directed towards the center of the circle and the different points on which the objects are present during the whole duration of motion is tangent to the center of the circle.

PtichkaEL [24]3 years ago
4 0
The force on the object has a constant strength, but its direction
keeps changing.  The force is always directed from the object to
the center of the circle.  It's called "centripetal force".
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3 years ago
herical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water
ki77a [65]

Answer: Below is the complete question

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)

answer:

mass transfer coefficient = 9.56 * 10^-5 m/s

Explanation:

Candy density = 1950 kg/m^3

Candy diameter = 1 cm

Velocity of water = 1 m/s

water density = 1000 kg/m^3

Viscosity of water = 1 * 10^-3 kg/m/s

diffusion coefficient of candy in water = 2 * 10^-9 m^2/s

solubility of candy = 2 kg/m^3

<u>Determine the mass transfer coefficient ( m/s ) </u>

( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3

where : Re = vdp / μ ,   Sh = KLd / Deff

attached below is the remaining solution .

mass transfer coefficient =  9.56 * 10^-5 m/s

6 0
2 years ago
A 310-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,190 A. If the conductor is copper
Zolol [24]

Answer:

t = 141.55 years

Explanation:

As we know that the radius of the wire is

r = 2.00 cm

so crossectional area of the wire is given as

A = \pi r^2

A = \pi(0.02)^2

A = 1.26 \times 10^{-3} m^2

now we know the free charge density of wire as

n = 8.50 \times 10^{28}

so drift speed of the charge in wire is given as

v_d = \frac{i}{neA}

v_d = \frac{1190}{(8.50 \times 10^{28})(1.6 \times 10^{-19})(1.26\times 10^{-3})}

v_d = 6.96 \times 10^{-5} m/s

now the time taken to cover whole length of wire is given as

t = \frac{L}{v_d}

t = \frac{310 \times 10^3}{6.96 \times 10^{-5}}

t = 4.46 \times 10^9 s

t = 141.55 years

6 0
3 years ago
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