Answers
Explanation:
a) Here is the free-body diagram. Note that I included the components of the weight mg (shown in dotted arrows) for use in the other parts of the problem.
b) The component of the weight parallel to the plane (shown in the diagram as a dotted arrow along the x-axis) is [tex]mg\sin15[/tex] and it is equal to
[tex]mg\sin15 = (25\:\text{kg})(9.8\:\text{m/s}^2)\sin15 = 63.4\:\text{N}[/tex]
c) Applying Newton's 2nd law to the y-axis, we can write
[tex]y:\;\;\;N - mg\cos15 = 0 \Rightarrow N = mg\cos15[/tex]
[tex]N = (25\:\text{kg})(9.8\:\text{m/s}^2)\cos15 = 236.7\:\text{N}[/tex]
d) The component of the weight mg into the plane is the same as the normal force, hence it's also 236.7 N.
e) To solve for the coefficient of friction, we apply Newton's 2nd law to the x-axis:
[tex]x:\;\;\;mg\sin15 - F_f = 0[/tex]
[tex]\Rightarrow F_f = mg\sin15\;\;(2)[/tex]
where [tex]F_f[/tex] is the frictional force defined as [tex]F_f = \mu N[/tex] so we can use Eqn(1) on Eqn (2) to write
[tex]\mu (mg\cos15) = mg\sin15[/tex]
Solving for [tex]\mu,[/tex] we get
[tex]\mu = \dfrac{\sin15}{\cos15} = \tan15 = 0.27[/tex]
Related Questions
Mechanical energy conservation states that
The total amount of energy will eventually be destroyed.
Potential energy will be conserved, but kinetic energy will be destroyed.
The total amount of energy, kinetic plus potential, remains the same.
Kinetic energy will be conserved, but potential energy will be destroyed.
Pls hurry I’ll give 50 points
Answers
Answer:
The total amount of energy, kinetic plus potential, remains the same.
Explanation:
You are angry at Dr. Anderson for this exam, so you throw a 0.30-kg stone at his car with a speed of 44 m/s. How much kinetic energy does the stone have
Answers
Answer:
Explanation:
KE = ½mv²
KE = ½(0.30)44²
KE = 290 J rounded to 2 s.d.
This is an image of a satellite traveling around Earth. Explain what are the two forces that are keeping the satelite around Earth without flying off or hitting the ground.
Answers
Answer:
One force will be gravity & inertia.
Explanation:
Bioth are combine to keep Earth in orbit around the sun, and the moon in orbit around Earth
which statement about metals is true?
Answers
All metals are solid at (our) living temperatures except for mercury which is a liquid. Metals are shiny for a while at least when properly finished. Metals are good conductors of both heat and electricity. This is due to the excess electrons in their valence clouds that facilitate energy transfer.What are 5 characteristics of metal?
Metals are lustrous, malleable, ductile, good conductors of heat and electricity
Silver conducts electricity better than any other metal.
Brass is an alloy made from zinc and copper.
Pure gold is too soft for many things so most gold is combined with other metals to make it stronger. ...
World time periods are often listed by the metal used.
Identify the type of chemical reaction:
CaCO3 -->CaO + CO2
Answers
Explanation:
decomposition reaction.....
Answer:
caco₃--cao+co₂
calcium oxide + carbon die oxide gives us calcium carbonate
this is the reaction of Acidic oxide(cao) and Basic oxide (co₂) to form salt.
The primary evidence that has led astronomers to conclude that the expansion of the universe is accelerating comes from __________.
Answers
The Universe is often studied by Scientist. The primary form of evidence that has led astronomers to conclude that the expansion of the universe is accelerating is the Observations of white dwarf supernovae.
There was the discovery of the faintness of high-redshift Type I supernovae by a team of scientist which showed that the expansion of the universe is accelerating.
The key evidence for that made scientist to talk about the expansion of the universe is accelerating comes from viewing and studying of white dwarf supernovae. It has been found that most distant stars all orbit at approximate speed as stars found about 30,000 light-years from their center.
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An object is released from height of 17m.
The object will hit the ground approximately in
Answers
[tex]\text{Given that,}\\\\\text{Height, h = 17 m}\\\\\\\text{We know that,}\\\\h = v_0t + \dfrac 12 gt^2\\\\\implies h = \dfrac 12 gt^2\\\\\implies t^2 = \dfrac{2h}g\\\\\implies t =\sqrt{\dfrac{2h}g} = \sqrt{\dfrac{2(17)}{9.81}} = 1.87 ~ \text{sec}[/tex]
how far from the lens is the image of the house if the house is 16 ft from the thin convex lens with a focal length of 8 ft
Answers
This question involves the concepts of the thin lens formula, focal length, and image distance.
The image of the house is "16 ft" away from the lens.
According to the thin lens formula:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where,
f = focal length = 8 ft
p = object distance = 16 ft
q = image distance = ?
Therefore,
[tex]\frac{1}{8\ ft}=\frac{1}{16\ ft}+\frac{1}{q}\\\\\frac{1}{q}=\frac{1}{8\ ft}-\frac{1}{16\ ft}\\\\\frac{1}{q}=0.125\ ft^{-1}-0.0625\ ft^{-1}\\\\q=\frac{1}{0.0625\ ft^{-1}}\\\\[/tex]
q = 16 ft
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write 2 situations in which the energy changes mentioned occur
Answers
Answer:
The types of energy is bond breaking and bond forming in chemical energy.
Explanation:
During Chemical reaction energy is required either for breaking up bonds in case of reactants and building bonds to form products.
The chemical reaction in which energy is released is called exothermic reactions, which is released due to making up the bonds.
The chemical reaction in which energy is absorbed is called endothermic reactions, in which energy is absorbed for breaking up the bonds.
1 point
Kinetic friction is defined as a force that acts between moving surfaces. A
body moving on the surface experiences a force in the opposite direction
of its movement. A student is investigating the motion of a block sliding
down a ramp onto the floor. The diagram below shows the block at five
points during the investigation. The block is at rest at point V. The student
releases the block so that it slides down the ramp and stops at point Z.
Which of the following best explains where kinetic friction is acting on the
block?
Answers
Answer:
Explanation:
Not sure what your options are but anything that says something like
"at the block surface in contact with the ramp along the line from V to Z" is probably a good shot.
the turns ratio for a transformer with 225 turns of wire in its primary winding and 675 turns in the secondary is: n
Answers
The ratio of the primary turns to the secondary turns is 1/3
The correct answer to the question is Option A. 1/3
From the question given above, the following data were obtained:
Primary turn (Nₚ) = 225 turnsSecondary turn (Nᵣ) = 675 turns Ratio of primary to secondary =?Ratio = Nₚ/Nᵣ
Nₚ/Nᵣ = 225 / 675
Nₚ/Nᵣ = 1/3
Therefore, the ratio of the primary turns to the secondary turns is 1/3
Complete question:
See attached photo
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BECAUSE OF THEIR A PAIR OF SUNGLEASSES ON THE DASHBOARD WILL CONTINUE MOVING FORWARD WHEN THE CAR TURNS SHARPLY
a. Acceleration
b. inertia
c. velocity
d. weight
Answers
Answer:
b. inertia
Explanation:
BECAUSE OF THEIR inertia A PAIR OF SUNGLEASSES ON THE DASHBOARD WILL CONTINUE MOVING FORWARD WHEN THE CAR TURNS SHARPLY.
K
Mission CG9: Weightlessness
Consider the several locations along a roller coaster
track. In which location(s) would the riders feel less
than their normal weight? Select all that apply.
Location A
Location B
Location C
a
=-10 m/s/s, dn
--2 m/s/s, up
a--6 m/s/s, dn
Location D
Location E
x=-12 m/s/s, dn
---6 m/s/s, up
Answers
The locations where the riders feel less than their normal weight are Location A, Location C and Location D.
The given parameters;
Location A, a = 10 m/s² downLocation B, a = 2 m/s² upLocation C, a = 6 m/s² downLocation D, a = 12 m/s² downLocation E, a = 6 m/s² upThe normal weight of the riders is calculated by applying Newton's second law of motion as follows;
W = mg
W = 9.8m
The apparent weight of the riders for the upward acceleration is calculated as follows;
[tex]R = m(g + a)[/tex]
The apparent weight of the riders for the downward acceleration is calculated as follows;
[tex]R = m(g - a)[/tex]
The apparent weight of the riders at location A is calculated as follows;
[tex]R_ A = m(9.8 - 10)\\\\R_ A = -0.2 m[/tex]
The apparent weight of the riders at location B is calculated as follows;
[tex]R_B = m(9.8 + 2)\\\\R_B = 11.8 m[/tex]
The apparent weight of the riders at location C is calculated as follows;
[tex]R_C = m(9.8 - 6)\\\\R_C = 3.8 m[/tex]
The apparent weight of the riders at location D is calculated as follows;
[tex]R_D = m(9.8 - 12)\\\\R_D = -2.2 m[/tex]
The apparent weight of the riders at location E is calculated as follows;
[tex]R_E = m(9.8 + 6)\\\\R_E = 15.8 m[/tex]
Thus, the locations where the riders feel less than their normal weight are;
Location ALocation CLocation D.Learn more about Newton's second law and reading of a scale here: https://brainly.com/question/11603452
The diagram shows the velocity-time graph for a car travelling in a straight line along a road. Calculate the acceleration between t = 2.0 s and t = 5.0 s.
Answers
Answer:
a = Δv/Δt = (0 - 20) / (5 - 2) = -6⅔ m/s²
The 0.15kg baseball has a speed of v=30 m/s just before it is struck by the bat. It then travels along the trajectory shown before the outfielder catches it. Determine the magnitude of the average impulsive force imparted to the ball if it is in contact with the bat for 0.75 ms
Answers
The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N
The mass of the baseball, m = 0.15 kg
The speed at which it moves, v = 30 m/s
Time at which the baseball was in contact with the bat, t = 0.75 ms
t = 0.75/1000 s
t = 0.00075 s
The impulsive force is given by the formula:
[tex]F=\frac{mv}{t}[/tex]
Substitute m = 0.15 kg, v = 30, and t = 0.00075s into the formula above:
[tex]F=\frac{0.15 \times 30}{0.00075} \\\\F=6000N[/tex]
The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N
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in a compoumd are atoms physically or chemically combined
Answers
Answer:
They are...if I'm correct Chemically combined, sorry if I'm wrong.
Would you please help me with this? I can't figure it out, please! I need to know what the E means!
Answers
Answer:
Without the full content of your question, I will have to GUESS at the context and assume
E = Energy
released when glucose is broken down.
PLEASE HELP FOR PHYSICS!
All objects exert a gravitational force on all other objects. This force is given by, F = GMm r2 , where the value of G = 6.673 × 10–11 N–m2/kg2 , M is the mass of the heavier object, m is the mass of the lighter object, and r is the distance between the two objects.
What is the force of gravity between two balls of mass 50 kg each if the distance between them is 25 m. Assume that there is no interference from any other gravitational field.
Answers
Hi there!
Recall Newton's Law of Universal Gravitation:
[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]
Where:
Fg = Force of gravity (N)
G = Gravitational Constant
m1, m2 = masses of objects (kg)
r = distance between objects (m)
Plug in the given values stated in the problem:
[tex]F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}[/tex]
A person sitting in a chair with wheels stands up, causing the chair to roll backward across the floor. The momentum of the chair
a. was zero while stationary and increased when the person stood.
b. was greatest while the person sat in the chair.
c. remained the same.
d. was zero when the person got out of the chair and increased while the person sat.
Answers
Answer:
a. was zero while stationary and increased when the person stood.
Explanation:
momentum is mass times velocity.
initial velocity was zero
final velocity was NOT zero.
Which feature of a balanced chemical equation demonstrates the law of
conservation of mass?
O A. It has the same types of atoms on both sides of the reaction
arrow.
O B. It shows the reactants of a chemical reaction to the left of the
reaction arrow.
O C. It has coefficients to show how much of each substance a
chemical reaction uses.
Thing
D. It shows the products of a chemical reaction to the right of the
reaction arrow.
Answers
Answer: A) It has the same types of atoms on both sides of the reaction
arrow.
Explanation: A balanced equation demonstrates the conservation of mass by having the same number of each type of atom on both sides of the arrow.
A cyclist rides in a circle with speed 8.1 m/s. What is his centripetal
acceleration if the circle has a radius of 27 m?
Answers
Explanation:
We know that the tangent velocity is 8.1 m/s. We also know that the tangent velocity can be written in the following way:
Vt = ωr with ω being the angular velocity.
We now calculate ω:
ω = Vt/r = 8.1 m/s / 27m = 0.3 rad/s
Now that we have ω we can calculate the centripetal aceleration:
a = ω^2 * r = ( 0.3 )^2 * 27 = 2.43 m/s^2
3. A ball is dropped from the roof of a building 55 meters tall. What is the approximate time of fall?
(Neglect air resistance and round to 2 decimal places).
Answers
Answer:
3.35 seconds
Explanation:
Use one of the equations of accelerated motion:
Δd = v1Δt+1/2aΔt^2
and rearrange for Δt which is time
Δt = √(2Δd)/a
now we can substitute in the values
a= 9.8 (acceleration due to gravity) and Δd= 55 as that is the height of the building
Δt = √(2*55)/9.8
Δt = 3.3503s
48.36
g.
MgSO4 to motes
Answers
Answer:120.3676
Explanation: using the molecular calculator and molar mass of MgSO4. hope this helps!
What is the approximate value of k when 30 = e^5k?
Answers
Answer:
Explanation:
30 = e^5k
ln30 = lne^5k
ln30 = 5k
k = ln30/5
k = 0.68023947...
round to your heart's content.
A 10kg object is 15 meters up a hill. Find its potential energy
Answers
Answer:
Explanation:
Relative to an origin at the bottom of the hill,
PE = mgh = 10(9.8)(15) = 1470 J
Three particles are placed in the xy plane. A 30-g particle is located at (3, 4) m, and a 40-g particle is located at (-2, -2) m. Where must a 20-g particle be placed so that the center of mass of the three-particle system is at the origin?
Answers
Answer:
Explanation:
30(3) + 40(-2) + 20(x) = 0(20 + 30 + 40)
x = -0.5
30(4) + 40(-2) + 20(y) = 0(20 + 30 + 40)
y = -2
(-0.5, -2)
Light and Reflection
Diagram Skills
E
STI
500
Mirrot
Flat Mirrors
1. The point of a 20.0 cm
D
pencil is placed 25.0 cm
from a flat mirror. Its
eraser is 15.0 cm from
the mirror. Three of the
light rays from the
pencil's point hit the
mirror with incident
angles of 0°, 20°, and
50° at points A, B, and C as shown.
a. Use a protractor to draw the reflected rays from points A, B, and C.
b. Where do reflected rays or their extensions intersect?
Mirror
B
c. What is the distance between the pencil's head and its image?
d. Would a person's eye located at point D perceive one of the reflected rays
drew? Will the person be able to see the image? Explain.
e. What if the eye is located at point E?
f. Draw incident rays from the eraser of the pencil to point A and to poin
Answers
The law of reflection allows to find the results for the questions about ray reflection in a plane mirror are:
a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.
b) The extension of the reflected rays is what forms the image.
c) The image's distance is 20 cm behind the flat mirror.
d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C
e) the Rays at points A, B, C cannot perceive in the point E.
f) attachment we see the rays that come out from the pencil eraser.
g) The image is behind the mirror at 15 cm.
The geometric interaction describes the interaction of light rays with surfaces, looking for where the rays are directed, it is described by two phenomenological laws:
Refraction. Establishes a relationship between incident rays and those transmitted by material means. Reflection. It establishes that the angle of incidence and reflection of the rays is the same.[tex]\theta_i = \theta_{r}[/tex]
From these two general laws, geometric optics establishes a relationship for the formation of the image, called the constructor's equation.
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
Where f is the focal length, p and q are the distance to the object and the image, respectively.
In this exercise, the medium is a mirror, which is why it must comply with the law of reflection.
a) In the attachment we see a diagram of the incident and reflected rays for the three points.
According to the law of reflection, the incident and reflected angles are equal.
b) From the diagram we can see that the extension of the reflected rays is what forms the image, which is called virtual and is located behind the mirror.
c) In the diagram we see two rays to form the image, we see that the distance to the object is equal to the distance to the image.
From the constructor's equation a plane mirror has an infinite radius.
p = -q
Therefore the image's distance is 20 cm behind the flat mirror. Therefore the distance to the object and the image are the same, the negative sign indicates that the image is behind the mirror.
d) A person located at point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C since their angle of reflection is not equal to the incident angle.
To perceive a ray it must have an angle of incidence of 25º.
e) Point E is located very far from the pencil, so the incident angle increases as does the reflected angle.
the Rays at points A, B, C cannot perceive.
f) In the attachment we see the rays that come out from the pencil eraser, they indicate that the distance to the plane mirror is 15.0 cm,
g) The image is behind the mirror at 15 cm.
In conclusion using the law of reflection we can find the results for the questions are:
a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.
b) The extension of the reflected rays is what forms the image.
c) The image's distance is 20 cm behind the flat mirror.
d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C
e) the Rays at points A, B, C cannot perceive in the point E.
f) attachment we see the rays that come out from the pencil eraser.
g) The image is behind the mirror at 15 cm.
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Physics!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
What is the formula for calculating distance?
QA: Speed x Time -- Speed/Time -- Time/Speed
Answers
Answer:
x=v.t
The answer: Distance= Speed x Time
And also
Time = Distance/Speed
Speed= Distance/Time
Motion Velocity
Reference point Speed
1. An object is in __________ when its distance from a(n) ________ is changing.
2. Speed is given direction is called _______________
3. ____________ can be calculated if you know the distance that an object travels in one unit of time.
Answers
Answer:
1. An object is in motion when its distance from another object is changing.
2.Speed is given direction is called velocity.
Speed can be calculated.......
A 1200 kg car moves due north with a speed of 15m/s. An identical car moves due east with the same speed of 15m/s what are the direction and the magnitude of the system’s total momentum
Answers
a.
The direction of the total momentum is 45°
The momentum of the first car is given by p = mv where m = mass of car = 1200 kg and v = velocity of car = 15 m/sj (since it moves due north).
So, p = mv
= 1200 kg × (15 m/s)j
= (18000 kgm/s)j
Also, the momentum of the identical car, p' = mv' where m = mass of car = 1200 kg and v' = velocity of car = (15 m/s)i (since it moves due east).
So, p' = mv'
= 1200 kg × (15 m/s)i
= (18000 kgm/s)i
So, the total momentum of the system P = p + p'
= (18000 kgm/s)j + (18000 kgm/s)i
= (18000 kgm/s)i + (18000 kgm/s)j
The direction of the total momentum of the system P is gotten from
tanФ = p'/p
= 18000 kgm/s ÷ 18000 kgm/s
= 1
Ф = tan⁻¹(1)
= 45°
The direction of the total momentum is 45°
b.
The magnitude of the total momentum of the system is 25455.84 kgm/s
The magnitude of the total momentum of the system P = √(p'² + p²)
= √[(18000 kgm/s)² + (18000 kgm/s)²]
= (18000 kgm/s)√(1 + 1)
= (18000 kgm/s)√2
= 25455.84 kgm/s
The magnitude of the total momentum of the system is 25455.84 kgm/s
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