a method of reducing friction​

Answers

Answer 1

Answer:

Lubrication

Explanation:

People oil/lubricate bicycle chains because the chain turns around the cogs and rub together so this help with friction.

Hope this helps :)

Answer 2

Answer:

The method of reducing friction are :

i) In moving parts of machine friction can be reduced by using a ball bearing between the moving surfaces

ii) The bodies of aeroplane ,ship ,boat etc are made streamlined to reduce friction.

iii) Friction can be reduced by polishing rough surfaces. For example : carrom boards are highly polished to reduce friction.

I hope this help you:)


Related Questions

A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?

Answers

Answer:

The mass of the second block=0.457 kg

Explanation:

We are given that

m1=1.5 kg

v1=1.3m/s

v2=4.3 m/s

V=2.0 m/s

We have to find the mass of the second block.

[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]

Let m2=m

Substitute the values

[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]

[tex]1.95+4.3m=3+2m[/tex]

[tex]4.3m-2m=3-1.95[/tex]

[tex]2.3m=1.05[/tex]

[tex]m=\frac{1.05}{2.3}[/tex]

[tex]m=0.457 kg[/tex]

Hence,  the mass of the second block=0.457 kg

the plane of a 5.0 cm by 8.0 cm rectangular loop wire is parallel to a 0.19 t magnetic field. if the loop carries a current of 6.2 amps, what is the magnitude of the torque on the loop

Answers

here is the state ment :0.2 loop

A 0.160 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s. It has a head-on collision with a 0.296 kg glider that is moving to the left with a speed of 2.23 m/s. Suppose the collision is elastic.

Required:
a. Find the magnitude of the final velocity of the 0.157kg glider.
b. Find the magnitude of the final velocity of the 0.306kg glider.

Answers

The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.

Momentum is conserved, so the total momentum of the system is the same before and after the collision:

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

==>

(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'

==>

-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'

where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.

Kinetic energy is also conserved, so that

1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' + 1/2 m₂ (v₂'

or

m₁ v₁² + m₂ v₂² = m₁ (v₁' + m₂ (v₂'

==>

(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' + (0.296 kg) (v₂'

==>

1.55 kg•m²/s² ≈ (0.160 kg) (v₁' + (0.296 kg) (v₂'

Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is

v₁' ≈ -3.11 m/s

v₂' ≈ -0.167 m/s

and take the absolute values to get the magnitudes.

If you want to instead use the masses from the "Required" section, you would end up with

v₁' ≈ -3.18 m/s

v₂' ≈ -0.236 m/s

Proper physical exercise makes bones _[blank 1]_.
People with stronger muscles and bones have better _[blank 2]_.

Which option shows the words that correctly fill in blank 1 and blank 2, in that order?


longer, flexibilitylonger, flexibility , ,

stronger, posturestronger, posture , ,

longer, posturelonger, posture , ,

stronger, flexibility

Answers

stronger, posturestronger, posture

hope that helped

Suppose a 60-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.325 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T

Answers

Answer:

 emf = 312 V

Explanation:

In this exercise the electromotive force is asked, for which we must use Faraday's law

           emf =  [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt

           Ф = B. A = B A cos θ

bold type indicates vectors.

They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values ​​1

It also indicates that the area is reduced from  a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear

            emf = -N B [tex]\frac{dA}{dT}[/tex]

            emf = - N B (A_f - A₀) / Dt

we calculate

           emf = - 60 1.60 (0 - 0.325) /0.100

           emf = 312 V

The direction of this voltage is exiting the page

5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?​

Answers

Answer:

gshshs

Explanation:

hshsksksksbsbbshd

A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.

Answers

Answer:

Explanation:

An impulse results in a change of momentum.

The impulse is the product of a force and a distance. This will be represented by the area under the curve

a) W = ½(4.00)(3.00) = 6.00 J

b) W = (11.0 - 4.00)(3.00) = 21.0 J

c) W = ½(17.0 - 11.0)(3.00) = 9.00 J

d) ASSUMING the speed at x = 0 is in the direction of applied force

½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00

v₄ = 2.05 m/s

½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00

v₁₇ = 4.92 m/s

If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.

Open the sash half way up, take the beaker containing the dry ice / water out of the hood, and slowly move it from right in front of the hood all the way down to the floor. At what point do the fumes stop getting sucked up by the fume hood?

Answers

Answer:

The fumes stop getting sucked up by the fume hood once the beaker is pulled out of the hood.

A capacitor consists of two parallel conducting plates, each of area 0.4 m2 and separated by a distance of 2.0 cm. Assume there is air between the plates. While connected to a battery the electric field within the plates is 500 N/C. The potential difference between the plates is: ________
a) 5.0 V
b) 10 V
c) 30 V
d) 20 V

Answers

Answer:

check photo

Explanation:

The cation that is reabsorbed from the urine in response to aldosterone

Answers

Answer:

If decreased blood pressure is detected, the adrenal gland is stimulated by these stretch receptors to release aldosterone, which increases sodium reabsorption from the urine, sweat, and the gut. This causes increased osmolarity in the extracellular fluid, which will eventually return blood pressure toward normal.

Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. Part A When she wears 2.0 D reading glasses, what is her near point

Answers

Answer:

The right answer is "0.273 m".

Explanation:

Given:

Power (P),

[tex]\frac{1}{f} = 2D[/tex]

Near point,

u = 0.6 m

As we know,

⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]

By substituting the values, we get

⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]

            [tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]

            [tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]

            [tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]    

By applying cross-multiplication, we get

          [tex]0.6=2.2 \ v[/tex]

            [tex]v = \frac{0.6}{2.2}[/tex]

      [tex]S_{near} = 0.273 \ m[/tex]

S.I unit for distance =______

(A) m (B)cm

(c) km (d) mm

Answers

Answer:

opinion a

Explanation:

the si units of distance is metre (m)

Answer:

A

Explanation:

b. Projectile on cliff (range)
An object of mass 5 kg is projected at an angle of 25° to the horizontal with a speed of 22 ms-1 from the top of the cliff.
The height of the cliff is 21 m. Take g, the acceleration due to gravity, to be 9.81 ms2
How far horizontally (to 1 decimal place) from the base of the cliff does the object land?

Answers

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

         cos 25 = v₀ₓ / v₀

         sin 25 = Iv_{oy} / v₀

         v₀ₓ = v₀ cos 25

         v_{oy} = v₀ sin 25

         v₀ₓ = 22 cos 25 = 19.94 m / s

         v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

         y = y₀ + v_{oy} t - ½ g t²

     

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

         0 = 21 + 0.0192 t - ½ 9.81 t²

         4.905 t² - 0.0192 t - 21 = 0

         t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

        t = [tex]\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}[/tex]

        t = [tex]\frac{0.003914 \ \pm 4.13828}{2}[/tex]

        t₁ = 2.07 s

        t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

        t = 2.07 s

now we can look up the distance traveled

         x = v₀ₓ t

         x = 19.94  2.07

         x = 41.28 m

I’m a photoelectric effect, which property of the incident light determines how much kinetic energy the ejected electrons have ?
A) brightness
B) frequency
C) size of the beam
D) none of the above

Answers

Answer:

b = frequency

What type of wave is a microwave?
O heat
O longitudinal
sound
transverse

Answers

Answer:

Microwave is a types of a electromagnetic radiation

Answer:

Transvers

Explanation:

Because microwave is electromagnetic  waves and all electromagnetic waves are transvers.

Why we use semiconductor instead of metal in thermopile.

Answers

Answer:

Semiconductors are not normal materials. They have special properties which conductors/metals cannot exhibit. The main reason for the behavior of semiconductors is that they have paired charge carriers-the electron-hole pair. This is not available in metals.

Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.00133 m^2. Calcular la fuerza que el gas ejerce sobre el embolo del pistón.

Answers

Answer:

F = 1.128 10⁸ Pa

Explanation:

Pressure is defined by

         P = F / A

If the gas is ideal for equal force eds on all the walls, so on the piston area we have

        F = P A

We reduce the pressure to the SI system

       P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa

we calculate

       F = 150 10³ / 0.00133

       F = 1.128 10⁸ Pa

Explain how blood circulation takes place in humans?​

Answers

Blood comes into the right atrium from the body, moves into the right ventricle and is pushed into the pulmonary arteries in the lungs. After picking up oxygen, the blood travels back to the heart through the pulmonary veins into the left atrium, to the left ventricle and out to the body's tissues through the aorta.

Hope it helps you

Mark my answer as brainlist

have a nice day

if you jog at a speed of 1.5m/s for 20 seconds how far di you travel

Answers

Answer: 30m

Explanation:

Given:

Speed: 1.5m/s

Time: 20 seconds

Distance = speed × time

Distance = 1.5 × 20

= 30m

Therefore you will travel 30m

Must click thanks and mark brainliest

In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue ball bounces off with a speed of 0.8 m/s at an angle of 20', as shown in the diagram below. Both balls have a mass of 0.6 kg.
a) what is the momentum of the system before the collision
b) what is the momentum after the collision
c) what angle dose the right ball travel after the collision
d) what is the magnitude of the eight balls velocity after the collision

Answers

Answer:

a)  p₀ = 1.2 kg m / s,  b) p_f = 1.2 kg m / s,  c)   θ = 12.36, d)  v_{2f} = 1.278 m/s

Explanation:

For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

 

a) the initial impulse is

         p₀ = m v₁₀ + 0

         p₀ = 0.6 2

         p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

        p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

 

we write the final moment for each axis

X axis

         p₀ₓ = 1.2 kg m / s

         p_{fx} = m v1f cos 20 + m v2f cos θ

         p₀ = p_f

        1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

         1.2482 = v_{2f} cos θ

Y axis  

        p_{oy} = 0

        p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

        0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

        0.2736 = v_{2f} sin θ

we write our system of equations

         0.2736 = v_{2f} sin θ

         1.2482 = v_{2f} cos θ

divide to solve

         0.219 = tan θ

          θ = tan⁻¹ 0.21919

          θ = 12.36

let's look for speed

            0.2736 = v_{2f} sin θ

             v_{2f} = 0.2736 / sin 12.36

            v_{2f} = 1.278 m / s

a point object is 10 cm away from a plane mirror while the eye of an observer(pupil diameter is 5.0 mm) is 28 cm a way assuming both eye and the point to be on the same line perpendicular to the surface find the area of the mirror used in observing the reflection of the point

Answers

Answer:

1.37 mm²

Explanation:

From the image attached below:

Let's take a look at the two rays r and r' hitting the same mirror from two different positions.

Let x be the distance between these rays.

[tex]d_o =[/tex] distance between object as well as the mirror

[tex]d_{eye}[/tex] = distance between mirror as well as the eye

Thus, the formula for determining the distance between these rays can be expressed as:

[tex]x = 2d_o tan \theta[/tex]

where; the distance between the eye of the observer and the image is:

[tex]s = d_o + d_{eye}[/tex]

Then, the tangent of the angle θ is:

[tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex]

replacing [tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex] into [tex]x = 2d_o tan \theta[/tex], we have:

[tex]x = 2d_o \Big( \dfrac{R}{d_o+d_{eye}}\Big)[/tex]

[tex]x = 2(10) \Big( \dfrac{0.25}{10+28}\Big)[/tex]

[tex]x = 20\Big( \dfrac{0.25}{38}\Big) cm[/tex]

x = (0.13157 × 10) mm

x = 1.32 mm

Finally, the area A = π r²

[tex]A = \pi(\frac{x}{2})^2[/tex]

[tex]A = \pi(\frac{1.32}{2})^2[/tex]

A = 1.37 mm²

calculate the length of wire.

Answers

Answer:

L = 169.5 m

Explanation:

Using Ohm's Law:

V = IR

where,

V = Voltage = 1.5 V

I = Current = 10 mA = 0.01 A

R = Resistance = ?

Therefore,

1.5 V = (0.01 A)R

R = 150 Ω

But the resistance of a wire is given by the following formula:

[tex]R = \frac{\rho L}{A}[/tex]

where,

ρ = resistivity = 1 x 10⁻⁶ Ω.m

L = length of wire = ?

A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²

A = 1.13 x 10⁻⁶ m²

Therefore,

[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]

L = 169.5 m

A friend lends you the eyepiece of his microscope to use on your own microscope. He claims that since his eyepiece has the same diameter as yours but twice the focal length, the resolving power of your microscope will be doubled. Is his claim valid? Explain.

Answers

Answer:

The resolving power remains same.

Explanation:

The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.

As the diameter is same but the focal length is doubled so the resolving power remains same.

why the stone moves away when the string is broken rotation​

Answers

Answer:

When a stone is going around a circular path, the instantaneous velocity of stone is acting as tangent to the circle. When the string breaks, the centripetal force stops to act. Due to inertia, the stone continues to move along the tangent to circular path. So, the stone flies off tangentially to the circular path

Answer:

when the string's rotation is broken, there will be no centripetal force to keep the stone stationary. Thus, the stone will flung away when the rotation is stopped

. A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?

Answers

Answer:

4

Explanation:

Good evening everyone Help me i n my hw ,The wall of cinema hall are covered with sound absorbing materials. Why?Answer it ASAP.Good day ​

Answers

what do you mean about it

A space ship has four thrusters positioned on the top and bottom, and left and right as shown below. The thrusters can be operated independently or together to help the ship navigate in all directions.
Initially, the Space Probe is floating towards the East, as shown below, with a velocity, v. The pilot then turns on thruster #2.

Select one:

a.
Space ship will have a velocity to the West and will be speeding up.

b.
Space ship will have a velocity to the East and will be speeding up.

c.
Space ship will have a velocity to the East and will be slowing down.

d.
Space ship will have a velocity to the West and will be slowing down.

e.
Ship experiences no change in motion.

Answers

Answer:

The correct answer is - c.  Spaceship will have a velocity to the East and will be slowing down.

Explanation:

In this case, if turned on thruster #2 then it will exert force on the west side as thruster 2 is on the east side and it can be understood by Newton's third law that says each action has the same but opposite reaction.

As the spaceship engine applies force on the east side then according to the law the exhauster gas applies on towards west direction. It will try to decrease the velocity of the spaceship however, the direction of floating still be east side initally.

Why is it advised not to hold the thermometer by its bulb while reading it?

Answers

Because your body heat might change the temperature

Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 220 m/s^2 for 20 msms, then travels at constant speed for another 30 ms.

Required:
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?

Answers

Solution :

We know,

Distance,

[tex]$S=ut+\frac{1}{2}at^2$[/tex]

[tex]$S=ut+0.5(a)(t)^2$[/tex]

For the first 20 ms,

[tex]$S=0+0.5(220)(0.020)^2$[/tex]

S = 0.044 m

In the remaining 30 ms, it has constant velocity.

[tex]$v=u+at$[/tex]

[tex]$v=0+(220)(0.020)[/tex]

v = 4.4 m/s

Therefore,

[tex]$S=ut+0.5(a)(t)^2$[/tex]

[tex]$S'=4.4 \times 0.030[/tex]

S' = 0.132 m

So, the required distance is = S + S'

                                              = 0.044 + 0.132

                                              = 0.176 m

Therefore, the tongue can reach = 0.176 m or 17.6 cm

Answer:

The total distance is 0.176 m.

Explanation:

For t = 0 s to t = 20 ms

initial velocity, u = 0

acceleration, a = 220 m/s^2

time, t = 20 ms

Let the final speed is v.

Use first equation of motion

v = u + at

v = 0 + 220 x 0.02 = 4.4 m/s

Let the distance is s.

Use second equation of motion

[tex]s = u t + 0.5 at^2\\\\s = 0 + 0.5 \times 220 \times 0.02\times 0.02\\\\s = 0.044 m[/tex]

Now the distance is

s' = v x t

s' = 4.4 x 0.03 = 0.132 m

The total distance is

S = s + s' = 0.044 + 0.132 = 0.176 m

Q)what are convex mirrors?​

Answers

Answer:

A curved mirror is a mirror with a curved reflecting surface. The surface may be either convex or concave. Most curved mirrors have surfaces that are shaped like part of a sphere, but other shapes are sometimes used in optical devices.

A convex mirror (or lens) is one constructed so that it is thicker in the middle than it is at the edge.

Other Questions
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