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Aleks04 [339]
3 years ago
8

Why does friction increase as speed increases??

Physics
1 answer:
kozerog [31]3 years ago
7 0
Friction is directly related to air particles.

When we say that friction is high, it means that you're colliding with lots of air particles, and hence you can't speed up as easily.
Thus, the more air particles you encounter, the higher the friction.

The faster you go, the more particles you will encounter in a given time; hence at higher speeds, the friction is higher.
You might be interested in
You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0
3 years ago
If the weight of the ruler is one Newton ,Gc cannot have a value more than 25cm
nevsk [136]

If the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm.

The given parameters:

  • Weight of the ruler = 1 N

<h3>What is center of gravity (CG)?</h3>
  • Center of gravity is the point at which the weight of an object is concentrated.

Let the length of the ruler = L

The center of the gravity of the ruler is calculated as follows;

X_{CG} = \frac{W(L_0) + W(L -X_{CG})}{W} \\\\X_{CG} = \frac{1(0) + 1(L -X_{CG})}{1}\\\\X_{CG} = L - X_{CG}\\\\X_{CG } + X_{CG} = L\\\\2X_{CG} = L\\\\X_{CG} = \frac{L}{2} \\\\when , \ L = 50 \ cm\\\\X_{CG} = \frac{50}{2} \\\\X_{CG} = 25 \ cm

Thus, if the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm. This may change if the length of the ruler changes because the center of gravity of uniform ruler depends on the length of the ruler.

Learn more about center of gravity here: brainly.com/question/6765179

4 0
2 years ago
You take the same 3 kg metal block and slide it along the floor, where the coefficient of friction is only 0.4. You release the
spin [16.1K]

Answer:

1.52905 seconds

4.58715 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

\mu = Coefficient of friction = 0.4

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration = \mu g

Equation of motion

v=u+at\\\Rightarrow t=\frac{v-u}{\mu g}\\\Rightarrow t=\frac{0-6}{-0.4\times 9.81}\\\Rightarrow t=1.52905\ s

It will take 1.52905 seconds for the block to slow down

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2\mu g}\\\Rightarrow s=\frac{0^2-6^2}{2\times 0.4\times -9.81}\\\Rightarrow s=4.58715\ m

The block will travel 4.58715 m before it stops

7 0
3 years ago
Calculate the weight in Newtons of a person who has a mass of 69 kg.
BaLLatris [955]
I believe the answer would be 676.65885 rounded to 676.7
5 0
3 years ago
Read 2 more answers
What are the units for impulse?<br>A. kg•m/s<br>B. kg/s <br>C. kg•m^2/s<br>D. kg m/s^2​
ICE Princess25 [194]

Answer:

A. kg•m/s

Explanation:

6 0
3 years ago
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