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
slamgirl [31]
3 years ago
12

A man attempts to push a 19.8 kg crate across a warehouse floor. He slowly increases the force until the crate starts to move at

a force of 34.8 N. He then maintains this 34.8 N force while the box accelerates at 0.357 m/s^2. What is the coefficient of static friction between the crate and the floor?
Physics
2 answers:
VARVARA [1.3K]3 years ago
7 0

Answer:

 μ = 0.18

Explanation:

Let's use Newton's second Law, the coordinate system is horizontal and vertical

Before starting to move the box

Y axis

     N-W = 0

     N = W = mg

X axis

     F -fr = 0

     F = fr

The friction force has the formula

     fr = μ N

     fr =  μ m g

At the limit point just before starting the movement

     F = μ m g

     μ = F / m g

calculate

      μ = 34.8 / (19.8 9.8)

    μ = 0.18

notka56 [123]3 years ago
3 0

Answer:

The coefficient of friction between the crate and the floor = 0.143

Explanation:

Frictional Force: This is the force that act between two surface in contact and tend to oppose their motion. it is measured in Newton (N)

F - F₁ = ma.................... Equation 1

Where F = Force of the crate, F₁ = frictional force, m = mass of the crate, a = acceleration of the crate

making F₁ the subject the equation 1

F₁ = F - ma .................... Equation 2

<em>Given: F = 34. 8 N,  m = 19.8 kg. a = 0.357 m/s².</em>

Substituting these values into equation 2

F₁ = 34.8 - (19.8×0.357)

F₁ = 34.8 - 7.07

F₁ = 27.73 N.

F₁ = μR ............... equation 3

making μ the subject of formula in equation 3

μ = F₁/R.............. Equation 4

Where F₁ = Frictional Force, μ = coefficient of static friction, R = Normal reaction.

R = mg,  

where g = 9.8 m/s², m = 19.8 kg

R = 19.8( 9.8) = 194.04 N,

R = 194.04 N, F₁ = 27.73 N

Substituting these values into equation 4

 μ  = 27.73/194.04

μ  = 0.143.

therefore the coefficient of friction between the crate and the floor = 0.143

You might be interested in
Properties of most medals include
Aleonysh [2.5K]

Answer:

Metals are lustrous, malleable, ductile, good conductors of heat and electricity. Other properties include: State: Metals are solids at room temperature with the exception of mercury, which is liquid at room temperature (Gallium is liquid on hot days).

3 0
3 years ago
Two hockey players have a total momentum of +200 kg x m/s before a collision (+ is to the right). After the collision, they move
Juliette [100K]

Total momentum after the collision: +200 kg m/s to the right

Explanation:

We can answer this question by using the law of conservation of momentum, which states that for an isolated system (=no external forces acting on the system), the total momentum is conserved.

Mathematically,

p_i=p_f

where

p_i is the total momentum before the collision

p_f is the total momentum after the collision

In this problem, the system consists of two hockey players. Before the collision, their total momentum is

p_i = +200 kg m/s (to the right)

Therefore, according to the law of conservation of momentum, their total momentum after the collision must be the same:

p_f = +200 kg m/s

And given that the sign is +, the direction is still the same, therefore to the right.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

8 0
4 years ago
Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa
zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, \rho _o = 926 kg/m³

density of the wood, \rho _{wood} = 974 kg/m³

density of water, \rho _w = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood}  -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood}  -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm

Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.

6 0
3 years ago
A basketball is tossed upwards with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m
Hatshy [7]

Answer:

The last one

Explanation:

4 0
3 years ago
Calculate the power to move a 9j box in 20 seconds​
dalvyx [7]

Answer:

please find attached pdf

Explanation:

Download pdf
3 0
3 years ago
Other questions:
  • An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Doe
    10·2 answers
  • While playing baseball with your friends your hands begin to sting after you ctach several fast balls.
    11·1 answer
  • Someone please help??
    6·1 answer
  • Do you wont to be friends
    9·1 answer
  • The alkali metals all react vigorously with water. Which alkali metal would react to the greatest extent if you have equal amoun
    10·1 answer
  • How many inches are in 6 meters?
    15·1 answer
  • What are the two <br>factors in which weight of object depends?​
    9·2 answers
  • BRAINLIST A wave travels at a constant speed. How does the wavelength change if the
    8·1 answer
  • HELPpPpo ASAP
    5·1 answer
  • imagine you are outside enjoying the warm sunshine with friends. as you briefly glance up at the sun, the part of the sun that y
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!