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4 years ago

How is coulomb's law similar to newton's law of gravity?

1 answer:

4 0

Newton's law of gravitation is attractive, whereas Coulomb's law is attractive or repulsive. Both are proportional to the inverse square of distance.

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1. A record with a radius of 0.3m spins in a clockwise circle with a centripetal

Hitman42 [59]

Solve for the linear/tangential speed:

a = v²/r

where a = centripetal acceleration, v = speed, and r = radius.

4.7 m/s² = v²/(0.3 m)

v² = (0.3 m) (4.7 m/s²)

v ≈ 3.96 m/s

For every time the record completes one revolution, a fixed point on the edge of the record travels a distance equal to its circumference, which is 2π (0.3 m) ≈ 1.88 m. So if 1 rev ≈ 1.88 m, then the angular speed of the record is

(3.96 m/s) (1/1.88 rev/m) ≈ 7.46 rev/s

Take the reciprocal of this to get the period:

1 / (7.46 rev/s) ≈ 0.134 s/rev

So it takes the record about 0.134 seconds to complete one revolution.

6 0

3 years ago

gregori [183]

Answer:

1. Revolve around a point

2. Formed from dust and gas particles

3. Exoplanets and associated star orbit a common center of mass

4. Composed of gases found in Jupiter and Saturn

3 0

3 years ago

Two bicycle tires are set rolling with the same initial speed of 4.00 m/s along a long, straight road, and the distance each tra

77julia77 [94]

Answer:

The coefficient of rolling friction for the tire under low pressure is 0.0342.

Explanation:

Two bicycle tires are set rolling with the same initial speed of 4.00 m/s

Final speed of both the bicycle, speed is reduced by half is measured, v = 2 m/s.

Here,

$u_kmg = ma\\\\a=\mu g$

Using third equation of motion as :

$v^2-u^2=2as\\\\v^2-u^2=2\mu gs\\\\\mu =\dfrac{v^2-u^2}{2gs}\\\\\mu =\dfrac{4^2-2^2}{2\times 9.8\times 17.9}\\\\\mu=0.0342$

So, the coefficient of rolling friction for the tire under low pressure is 0.0342.

5 0

3 years ago

Two hockey players are about to collide on the ice, One player has a mass of

Furkat [3]

Answer:

19 kg x m/s East

Just did it!

3 0

3 years ago

Steam enters a long, horizontal pipe with an inlet diameter of D1 = 16 cm at 2 MPa and 300°C with a velocity of 2.5 m/s. Farther

polet [3.4K]

Answer:

m = 0.4005 kg/s

Q_out = 45.1 KJ/s

Explanation:

Given

Pipe inlet diameter D1 = 16 cm

Steam inlet pressure P1 = 2 Mpa

Steam inlet temperature T1 = 300 °C

Pipe outlet diameter D2 = 14 cm

Steam inlet velocity V1 = 2.5 m/s

Steam outlet pressure P2 = 1.8 MPa

Steam outlet temperature T2 = 250 °C

Required

Determine

(a) The mass flow rate of steam.

(b) The rate of heat transfer.

Assumptions

Kinetic and potential energy changes are negligible.

This is a steady flow process.

There is no work interaction.

Solution

Part a From steam table (A-6) at P1 = 2 Mpa , T1 = 300 °C

vl = 0.12551 m^3/Kg

h1 = 3024.2 KJ/Kg

The mass flow rate of steam could be defined as the following

m = 1/v1*A1*V1

m = 0.4005 kg/s

Part b We take the pipe as our system.The energy balance could be defined as the following

E_in -E_out =ΔE_sys = 0

E_in = E_out

mh_1 = Q_out + mh_2

Q_out = m(h_1-h_2)

From steam table (A-6) at P2= 1.8 Mpa T2 = 250 °C h2= 2911.7 KJ/Kg The heat transfer could be defined as the following

Q_out = m(h_1-h_2)

Q_out = 0.4005*(3024.2 -2911.7) =45.1 KJ/s

6 0

3 years ago