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olchik [2.2K]
3 years ago
7

Describe what causes friction between surfaces

Physics
2 answers:
Nezavi [6.7K]3 years ago
8 0
Some things that causes friction between surfaces is a bike on a rocky road. The<span> action of one surface or object rubbing against another causing friction. Im not really sure if I helped But yeah..</span>
leva [86]3 years ago
8 0
If you look two surfaces at microscopic level, you'll see that they have tiny interlocks. When you bring two surfaces in contact, they are held together strongly because of this interlocks which makes two objects difficult to perform motion.
You might be interested in
A weight suspended from a spring is seen to bob up and down over a total distance of 20 centimeters twice each second.
Zielflug [23.3K]

<u>Answer</u>

8. 2 Hz

9. 0.5 seconds

10. 20 cm


<u>Explanation</u>

<u>Q 8</u>

Frequency is the number of oscillation in a unit time. It is the rate at which something repeats itself in a second.

In this case, the spring bob up and down 2 times per second.

∴ Frequency = 2 Hz

<u>Q 9</u>

Period is the time taken to complete one oscillation.

2 oscillations takes 1 second

1 oscillation = 1/2 seconds.

∴ Period = 0.5 seconds


<u>Q 10</u>

Amplitude is the the maximum displacement of the spring.

In this case the spring bob up 20 cm. This is it's displacement.

∴ Amplitude = 20 cm

5 0
3 years ago
A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun
Lera25 [3.4K]

Answer:

The distance is 0.53 m.

Explanation:

Given that,

Target distance = 100.0 m

Speed of bullet = 300 m/s

We need to calculate the total time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t=\dfrac{100.0}{300}

t=0.33\ sec

Now, consider vertical motion of bullet.

Initial velocity of bullet in vertical direction = 0 m/s

We need to calculate the vertically distance

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Put the value in the equation

s=0+\dfrac{1}{2}\times9.8\times(0.33)^2

s=0.53\ m

Hence, The distance is 0.53 m.

5 0
3 years ago
A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w
Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

8 0
4 years ago
A hole is punched at A in a plastic sheet by applying a 660-N force P to end D of lever CD, which is rigidly attached to the sol
olasank [31]

Answer:

hello your question lacks some data and required diagram

G = 77 GPa,  т all = 80 MPa

answer : required diameter = 252.65 * 10-^3 m

Explanation:

Given data :

force ( P )  = 660 -N force

displacement = 15 mm

G = 77 GPa

т all = 80 MPa

i) Determine the required diameter of shaft BC

considering the vertical displacement ( looking at handle DC from free body diagram )

D' = 0.3 sin∅  ,   where D = 0.015

hence ∅ = 2.8659°

calculate the torque acting at angle ∅  of  CD on the shaft BC

Torque = 660 * 0.3 cos∅

             = 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N

hello attached is the remaining part of the solution

4 0
3 years ago
A mango is dropped and fall freely from rest. What are its position and velocity after 1.0secs,2.0secs,and 3.0secs
irakobra [83]

Answer:

sxsdfsd

Explanation:

4 0
3 years ago
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