All categories

3 years ago

Oil can be used to reduce friction. true or fales take 100 mark for solving gg

Physics

2 answers:

GalinKa [24]3 years ago

8 0

Answer:

This is true.

Oil has a good consistency for reduction in tight areas of something. If you've ever played a recorder for band class or just for fun, you may have realized that every time you take it apart and put it together again it gets harder, and normally the recorder comes with a small container of oil that you rub between the two connecting pieces.

Naddika [18.5K]3 years ago

4 0

yes because it has to reduce frition by its own method called fellding

You might be interested in

Which is the only mechanism that can introduce new alleles in a gene pool? genetic drift random mating mutation

Vikentia [17]

Mutation because they change the genes

8 0

3 years ago

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

gayaneshka [121]

Force applied on the car due to engine is given as

$F_1 = 300 N$ towards right

Also there is a force on the car towards left due to air drag

$F_2 = 150 N$ towards left

now the net force on the car will be given as

$\vec F_{net} = \vec F_1 + \vec F_2$

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

$F_{net} = F_1 - F_2$

$F_{net} = 300 - 150$

$F_[net} = 150 N$

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0

3 years ago

A car starting from rest has a constant acceleration of 4 meters per second per second. How fast will it be going in 5 seconds?

irinina [24]

Answer:

18 m

Explanation:

Given : vo = 0 m/s ; t = 3 s; a = 4 m/s^2 ; d = ? m ; average velocity = ? m/s ; fonal velocity = ? m/s

solving for the final velocity, v

v = a * t

v = 4 m/s^2 * 3 s

v = 12 m / s

Solving for the average velocity. avg v

avg v = (vo + v) / 2

avg v = (0 m / s + 12 m/s) / 2

avg v = 6 m / s

Solving for the distance traveled after 3 s

d = avg v * t

d = 6 m / s * 3 s

d = 18 meters

In the first 3s the car travels 18 meters.

4 0

2 years ago

A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha

Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

7 0

3 years ago

Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radiu

Gnesinka [82]

The orbiting velocity of the satellite is 4.2km/s.

To find the answer, we need to know about the orbital velocity of a satellite.

<h3>What's the expression of orbital velocity of a satellite?</h3>

Mathematically, orbital velocity= √(GM/r)
r = radius of the orbital, M = mass of earth

<h3>What's the orbital velocity of a satellite orbiting earth with a radius 3.57 times the earth radius?</h3>

M= 5.98×10²⁴ kg, r= 3.57× 6.37×10³ km = 22.7×10⁶m
Orbital velocity= √(6.67×10^(-11)×5.98×10²⁴/22.7×10⁶)

=4.2km/s

Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.

Learn more about the orbital velocity here:

brainly.com/question/22247460

#SPJ1

8 0

2 years ago