answer.
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
pentagon [3]
2 years ago
7

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

think the coefficient of friction between the sled and the snow is. You've been walking at a steady 1.5\;{\rm m}/{\rm s}, and the rope pulls up on the sled at a 42.0^\circ angle. You estimate that the mass of the sled, with your friend on it, is 73.0 kg and that you're pulling with a force of 87.0 N.?

Physics
1 answer:
Juli2301 [7.4K]2 years ago
5 0

Answer:

0.0984

Explanation:

From the first diagram attached below; a free flow diagram shows the interpretation of this question which will be used  to solve this question.

From the diagram, the horizontal component of the force is:

F_X = F_{cos \ \theta}

Replacing 42°  for θ and 87.0° for F

F_X =87.0 \ N \ *cos \ 42 ^\circ

F_X =64.65 \ N

On the other hand, the vertical component  is ;

F_Y = Fsin \ \theta

Replacing 42°  for θ and 87.0° for F

F_Y =87.0 \ N \ *sin \ 42 ^\circ

F_Y =58.21  \ N

However, resolving the vector, let A be the be the component of the mutually perpendicular directions.

The magnitude of the two components is shown in the second attached diagram below and is now be written as A cos θ and A sin θ

The expression for the frictional force is expressed as follows:

f = \mu \ N

Where;

\mu is said to be the coefficient of the friction

N = the  normal force

Similarly the normal reaction (N) = mg - F sin θ

Replacing F_Y \ for \ F_{sin \  \theta}. The normal reaction can now be:

N = mg \ - \ F_Y

By balancing the forces, the horizontal component of the force equals to frictional force.

The horizontal component of the force is given as follows:

F_X = \mu \ ( mg - \ F_Y)

Making \mu the subject of the formular in the above equation; we have the following:

\mu \ = \ \frac{F_X}{mg - F_Y}

Replacing the following values: i.e

F_X \ = \ 64.65 \  N

m = 73 Kh

g  = 9.8 m/s²

F_Y = \ 58.21 N

Then:

\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}

\mu = 0.0984

Thus, the coefficient of friction is = 0.0984

You might be interested in
A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved​ along, a marble was fir
bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.  

So with y(1) = v - 4.9 = 0 we have  

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now,  the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

3 0
2 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
2 years ago
Taylor places a nail on a bar magnet. The nail sticks to the magnet when lifted up off the table. She touches a paperclip to the
Ratling [72]
Prior to touching the bar magnet, the magnetic domains in the nail were pointing in random directions. When Taylor touched the nail to the bar magnet the magnetic fields of the magnetic domains aligned and it became a temporary magnet.
5 0
2 years ago
Read 2 more answers
Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.
Rufina [12.5K]

Answer: 1. decreasing the mass of both objects

2. decreasing the mass of one of the objects

3. increasing the distance between the objects

Explanation: Hope that helped! (:

4 0
2 years ago
You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w
Anit [1.1K]

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg  mass of the box

d= 5.4m  length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof  

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof  and the y-axis in the direction perpendicular to it.

W: Weight of the box  : In vertical direction

N : Normal force : perpendicular to the direction the  roof

fk : Friction force: parallel to the direction to the roof

Calculated of the weight  of the box

W= m*g  =  (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay    ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86  = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34  m/s²

Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  = 5.4 m

v₀: initial speed  

vf: final speed  = 0

a : acceleration of the box = -1.34  m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

v_{o} =\sqrt{14.472}

v₀ = 3.8 m/s

7 0
2 years ago
Other questions:
  • A cave explorer travels 3.0 m eastward, then 2.5 m northward, and finally 15.0 m westward. use the graphical method to find the
    8·2 answers
  • Which best describes a similarity between power plants that use water as an energy source and those that use wind as an energy s
    10·2 answers
  • The inner and outer surfaces of a 5m x 6m brick wall of thickness 30 cm and thermal conductivity 0.69 w/m.0 c are maintained at
    7·1 answer
  • Use scientific (exponential) notation to express the following quantities in terms of the SI base units in
    6·1 answer
  • A glider of mass 0.240 kg is on a frictionless, horizontal track, attached to a horizontal spring of force constant 6.00 N/m. In
    14·1 answer
  • A ball hangs on the end of a string that is connected to the ceiling so that it swings like a pendulum. You pull the ball up so
    5·1 answer
  • Lynn rubs a balloon with a piece of wool, which causes the balloon to pick up some of the electric charges from the wool. Lynn t
    10·2 answers
  • A particle is located on the x axis at x = 2.0 m from the origin. A force of 25 N, directed 30° above the x axis in the x-y plan
    8·1 answer
  • The wad of clay of mass m = 0.36 kg is initially moving with a horizontal velocity v1 = 6.0 m/s when it strikes and sticks to th
    7·1 answer
  • Because of your knowledge of physics, you have been hired as a consultant for a new James Bond movie, "Oldfinger". In one scene,
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!