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3 years ago

How do you reduce friction

1 answer:

3 0

To reduce friction between two surfaces, you have to
make the surfaces smoother.

The best way to do that is to introduce a fluid between them,
like grease, oil, or air.

You might be interested in

The mass of the sun is 1.99x10^30 kg . Jupiter is 7.79x10^8 km away from the sun and the mass of 1.90x10^27 kg

maks197457 [2]

Question is missing:

"What is the gravitational force between the Sun and Jupiter?"

Answer:

$4.16\cdot 10^{23} N$

Explanation:

The gravitational force between two objects is given by

$F=G\frac{m_1m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

In this problem, we have

$m_1 = 1.99\cdot 10^{30} kg$ is the mass of the sun

$m_2 = 1.90\cdot 10^{27} kg$ is the mass of Jupiter

$r=7.79\cdot 10^8 km = 7.79\cdot 10^{11} m$ is their separation

Solving the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(1.99\cdot 10^{30})(1.90\cdot 10^{27})}{(7.79\cdot 10^{11})^2}=4.16\cdot 10^{23} N$

6 0

3 years ago

you throw a rock with a horizontal velocity at 36.6 m/s out a window that is 29.5 m above the ground. What was it’s total time i

ANEK [815]

29.5-4.9x^2= your answer

4 0

3 years ago

When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not n

Harlamova29_29 [7]

Answer:

because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.

6 0

3 years ago

Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time–when they\'re not sleep

Anit [1.1K]

Answer:

mice total momentum (-0.000250, 0.00639) Kg m

Explanation:

To calculate the moment of the mice we must multiply their mass by their velocities, remember that the moment is a vector quantity, so we use the components of velocity

mouse 1

m1 = 0.0225 Kg

V1 = (0.869, -0.283) m / s

Px = m Vx

Px1 = 0.0225 0.869

Px1 = 0.01955 Kg m

Py = m Vy

Py1 = 0.0225 (-0.283)

Py1 = -0.006368 Kg m

P1 = (0.0196, -0.00637) Kg m

Mouse 2

m2 = 0.0223 Kg

Px2 = 0.0223 (-0.883) = -0.0196 Kg m

Py2 = 0.0223 (-0.253) = -0.00564 Kg m

P2 = (-0.0196, -0.00564) Kg m

Mouse 3

m3 = 0.0197

Px3 = 0.0197 0.345 = 0.00680 Kg m

Py3 = 0.0197 0.803 = 0.0158 Kg m

P3 = (0.00680, 0.0158) Kg m

Mouse4

m4 = 0.0127 Kg

Px4 = 0.0127 (-0.555) = -0.00705 Kg m

Py4 = 0.0127 0.205 = 0.00260 Kg m

P4 = (-0.00705, 0.00260) Kg m

To find the total momentum we must add each component of the individual moments

Px = Px1 + Px2 + Px3 + Px4

Py = py1 + Py2 + Py3 + Py4

Px = 0.0196 -0.0196 +0.00680 -0.00705

Px = -0,000250 Kg m

Py = -0.00637 -0.00564 +0.0158 +0.00260

Py = 0.00639 Kg m

P = (-0.000250, 0.00639) Kg m

7 0

3 years ago

A satellite omass1000 kg moves in a circular orbit of radius 8000 km round the earth,assumed to be a sphere of radius 6400 km. C

lubasha [3.4K]

Answer:

ΔE = 37.8 x 10^9 J

Explanation:

The energy required will increased the potential energy and increase the kinetic energy.

As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative

Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²

G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²

ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J

The centripetal force at orbit must be equal to the gravity force

mv²/R = mg'

v²/8.0e6 = 6.272

v² = (6.272(8.0e6)) = 50.2e6 m²/s²

The maximum velocity when resting on earth at the equator is about 460 m/s.

The change in kinetic energy is

ΔKE = ½m(vf² - vi²)(1000)

ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J

Total energy increase is

25e9 + 12.857e9 = 37.8e9 J

3 0

3 years ago