1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
levacccp [35]
3 years ago
14

In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallw

ay. If the 3.5 kg book is pushed from rest through a distance of 0.91 m by the horizontal 25 N force from the broom and then has a speed of 1.53 m/s, what is the coefficient of kinetic friction between the book and floor?
Physics
1 answer:
RSB [31]3 years ago
5 0

Answer:

The coefficient of kinetic is

u_{k}=0.59

Explanation:

The forces in the axis 'x' and 'y' using law of Newton to find coefficient of kinetic friction

ΣF=m*a

ΣFy=W-N=0

ΣFy=Fn-Fu=m*a

F_{u} =u_{k} *N\\F_{N}=25N\\N=W\\N=3.5kg*9.8\frac{m}{s^{2} }=34.3N

F_{N}-F_{u}=m*a\\F_{N}-u_{k}*N=m*a\\u_{k}*N=F_{N}-m*a\\u_{k}=\frac{F_{N}-m*a}{N}

Now to find the coefficient can find the acceleration using equation of uniform motion accelerated

v_{f} ^{2}=v_{o}^{2}+2*a(x_{f}-x_{o})\\x_{o}=0\\v_{o}=0\\v_{f} ^{2}=2*a*x_{f}\\a=\frac{v_{f} ^{2}}{2*a*x_{f}}\\ a=\frac{(1.53\frac{m}{s} )^{2}}{2*0.91m}\\a= 1.28 \frac{m}{s^{2} }

So replacing the acceleration can fin the coefficient:

u_{k}=\frac{F_{N}-m*a }{N}\\u_{k}=\frac{25N-(3.5kg*1.28\frac{m}{s^{2}} }{34.3N} \\u_{k}=0.59

You might be interested in
At what temperature do the fahrenheit and celsius scales give the same reading?
anygoal [31]
At -40.

-40 gives the same reading for Fahrenheit and Celsius scale.
7 0
3 years ago
Two cars, one of mass 1400 kg, and the
Vikki [24]

Suppose that, in the x-y plane, the first car is moving to the right so that its velocity is given by the vector

v₁ = (14 m/s) i

and the second car is moving upward so that its velocity vector is

v₂ = (20 m/s) j

Then the total momentum of two cars before their collision is

m₁v₁ + m₂v₂ = (1400 kg) (14 m/s) i + (2300 kg) (20 m/s) j

= (19,600 i + 46,000 j) kg•m/s

Their momentum after the collision is

(1400 kg + 2300 kg) v = (3700 kg) v

where v is the velocity vector of the wreckage.

By conservation of momentum,

(19,600 i + 46,000 j) kg•m/s = (3700 kg) v

Let a and b be the horizontal and vertical components of v, respectively. Then

19,600 kg•m/s = (3700 kg) a   ⇒   a ≈ 5.2973 m/s ≈ 5.3 m/s

46,000 kg•m/s = (3700 kg) b   ⇒   b ≈ 12.4324 m/s ≈ 12 m/s

so that the final speed of the wreckage is

||v|| = √(a² + b²) ≈ 13.5139 m/s ≈ 14 m/s

3 0
2 years ago
If the initial velocity of a ball is sent straight upward at 10.5m/s from the ground what will its final velocity be when it hit
Shalnov [3]

Answer: -10.08 m/s

Explanation:

Here we only need to analyze the vertical problem.

When the ball is in the air, the only force acting on it will be the gravitational force, this means that the acceleration of the ball, is equal to the gravitational acceleration, then:

a(t) = -9.8m/s^2

Where the negative sign is because gravity pulls the ball down.

To get the velocity equation we need to integrate over time, we get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity, here it is v0 = 10.5 m/s

Then the velocity equation is:

v(t) =  (-9.8m/s^2)*t + 10.5 m/s

To get the position equation, we need to integrate again over time, we get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t + p0

Where p0 is the initial position, we know that the ball is sent upward from the ground, so p0 = 0m

Then the position equation is:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

Now we need to find the value of t such that the position is equal to zero (this means that the ball hits the ground again).

Then we need to solve:

p(t) = 0 =  (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

If we divide both sides by t, we get:

0 =   (1/2)*(-9.8m/s^2)*t + (10.5 m/s)

Now we can solve it:

(1/2)*(9.8m/s^2)*t = 10.5 m/s

t = (10.5 m/s)/((1/2)*(9.8m/s^2)) = 2.14 s

This means that after 2.14 seconds, the ball will hit the ground again.

The velocity of the ball when it hits the ground is equal to:

v(2.14s) = (-9.8m/s^2)*2.14s + 10.5 m/s = -10.08 m/s

3 0
3 years ago
Two slits in an opaque barrier each have a width of 0.020 mm and are separated by 0.050 mm. When coherent monochromatic light pa
Semenov [28]

Answer:

The answer is 5

Explanation:

The maximum interference is:

m * λ = d * sinθi

Where m = 0,1,2,3,...

The first minimum diffraction is:

λ = a * sinθd

|sinθi| < sinθd

Where

(|m| * λ)/d < λ/a

|m| < d/a = 2.5

|m|max = 2

It can be concluded that coherent monochromatic light passes through the slits, therefore the maximum number of interference is 5.

5 0
3 years ago
Blank occurs when salt is evenly mixed with water
podryga [215]
I think solution because it dissolves
6 0
3 years ago
Other questions:
  • A turnip is dropped from a height of 70 meters. How long does it take to hit the ground?
    14·1 answer
  • Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.
    13·1 answer
  • Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
    9·1 answer
  • Physics Help Please:
    12·2 answers
  • (HELP QUICK!!) Air pressure is a result of _____
    6·2 answers
  • Which scenario would result in a positive population growth? Use this formula: population growth = (birth rate + immigration) –
    5·2 answers
  • Please help find the total resistance of these circuits!! I will mark brainliest if right!!
    12·1 answer
  • The low point of a transverse wave is called a
    14·1 answer
  • When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these
    5·1 answer
  • A bucket tied to a rope is moving at a constant speed of 5.0 m/s in a circle of radius
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!