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
Stolb23 [73]
4 years ago
10

A book of mass 7 kg rests on a plank. You tilt one end of the plank and slowly increase the angle of the tilt. The coefficient o

f static friction between the book and the plank is 0.32. What is the maximum angle of tilt for which the book will remain stationary and not slide down the plank
Physics
2 answers:
VARVARA [1.3K]4 years ago
5 0

Answer:

17.74°

Explanation:

For an inclined surface,

F = μmgcosΦ................ Equation 1

F' = mgsinΦ................. Equation 2

Where F = friction force, F' = down slope force, m = mass of the book, g = acceleration due to gravity, μ = coefficient of static friction, Φ = maximum angle of tilt.

For the book to remain stationary and not slide.

F = F'

μmgcosΦ = mgsinΦ

μ = cosΦ/sinΦ

μ = tanФ

make Ф the subject of the equation

Ф = tan⁻¹(μ).................... Equation 3

Given: μ = 0.32

Substitute into equation 3

Ф = tan⁻¹(0.32)

Ф = 17.74°

MAXImum [283]4 years ago
5 0

To solve this problem we will apply the concepts related to the Friction force and the force induced by gravity. Since the displacement is in angular mode, the component of the horizontal force of friction will be equivalent to the component of the vertical force of gravity. For balance to exist, both must be equal

F_{fx} = F_{gy}

Where,

F_{fx} = \mu mg cos\theta

F_{gy} = mg sin \theta

m = Mass

g = Gravitational acceleration

\mu = Constant of friction

Then,

m g sin \theta = \mu m g cos \theta

sin \theta = mu cos \theta

tan \theta = mu

tan\theta = 0.32

\theta = 17.74\°

Therefore the maximum angle  of tilt for which the book will remain stationary and not slide down the plank is 17.74°

You might be interested in
Are we real?Should we believe in hell and heaven?Should we take life seriously?Why do we feel like we are being pulled down from
nlexa [21]
Uhhhhh just sleep? Are you ok?
7 0
3 years ago
Read 2 more answers
A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.15 T over a region of radius 7.4 m. The charge on
klio [65]

Explanation:

It is given that,

Magnetic field, B = 0.15 T

Charge on a proton, q=1.60218\times 10^{-19}\ C

Mass of a proton, m=1.67262 \times 10^{-27}\ kg

The cyclotron frequency is given by :

f=\dfrac{qB}{2\pi m}

f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}

f = 2286785.40 Hz

or

\omega=14368296.44\ rad/s

\omega=1.43\times 10^7 rad/s

Hence, this is the required solution.

8 0
3 years ago
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
How much force is needed to accelerate a 1-kilogram toy car at a rate of 2 meters per second per second?
kozerog [31]
F = ma,    where m = mass in kg, a = acceleration in m/s², F = Force in Newton

F = 1 * 2

F = 2 N

Force needed is 2 Newtons.
5 0
3 years ago
Read 2 more answers
A ball is thrown horizontally with a velocity of 12 m/s. How
miskamm [114]

Answer:

is there a picture of it

7 0
3 years ago
Other questions:
  • The diameter of Mars is 6794km, and its minimum distance from the earth is 5.58×10^7 km. When Mars is at this distance, find the
    10·1 answer
  • Point charge A is located at point A and point charge B is at point B. Points A and B are separated by a distance r. To determin
    5·1 answer
  • Based on their composition materials can be divided into pure and
    12·1 answer
  • In which of the following cases is work being done on an object? Question 2 options: Pushing against a locked door Carrying a bo
    12·1 answer
  • If a species can adapt to a changing environment, or ____, its descendants will survive.
    5·1 answer
  • Q. Two balls are released simultaneously from a certain
    10·1 answer
  • Please help PHYSICS
    7·2 answers
  • A ball is thrown up at 30m/s from the ground. What is it s maximum height?<br> How long did it take?
    5·2 answers
  • the diameter of the wheels on your car ( including the tires) is 25 inches. you are going to drive 250 miles today. each of your
    12·1 answer
  • Tearing of paper is said to be a change that cannot be reversed. What about paper recycling? Explain.
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!