In order to calculate frictional force look below..
The formula given by
[tex]\\ \sf\longmapsto F_f=\mu N[/tex]
Or
[tex]\\ \sf\longmapsto F_f=\mu mg[/tex]
u is coefficient of friction
N is normal reaction.
A dry cell gives static electricity true or false?
Answer:
False
Explanation:
chứng minh mặt trời là nguồn gốc của tất cả nguồn năng lượng
An Excerpt from “Optimism”
by Helen Keller
1 Could we choose our environment, and were desire in human undertakings synonymous with
endowment, all men would, I suppose, be optimists. Certainly most of us regard happiness as
the proper end of all earthly enterprise. The will to be happy animates alike the philosopher, the
prince and the chimney-sweep. No matter how dull, or how mean, or how wise a man is, he feels
that happiness is his indisputable right.
2 It is curious to observe what different ideals of happiness people cherish, and in what singular
places they look for this well-spring of their life. Many look for it in the hoarding of riches, some
in the pride of power, and others in the achievements of art and literature; a few seek it in the
exploration of their own minds, or in the search for knowledge.
3 Most people measure their happiness in terms of physical pleasure and material possession.
Could they win some visible goal which they have set on the horizon, how happy they would be!
Lacking this gift or that circumstance, they would be miserable. If happiness is to be so
measured, I who cannot hear or see have every reason to sit in a corner with folded hands and
weep. If I am happy in spite of my deprivations, if my happiness is so deep that it is a faith, so
thoughtful that it becomes a philosophy of life,—if, in short, I am an optimist, my testimony to
the creed of optimism is worth hearing....
4 Once I knew the depth where no hope was, and darkness lay on the face of all things. Then
love came and set my soul free. Once I knew only darkness and stillness. Now I know hope and
joy. Once I fretted and beat myself against the wall that shut me in. Now I rejoice in the
consciousness that I can think, act and attain heaven. My life was without past or future; death,
the pessimist would say, “a consummation devoutly to be wished.” But a little word from the
fingers of another fell into my hand that clutched at emptiness, and my heart leaped to the
rapture of living. Night fled before the day of thought, and love and joy and hope came up in a
passion of obedience to knowledge. Can anyone who has escaped such captivity, who has felt
the thrill and glory of freedom, be a pessimist?
5 My early experience was thus a leap from bad to good. If I tried, I could not check the
momentum of my first leap out of the dark; to move breast forward is a habit learned suddenly
at that first moment of release and rush into the light. With the first word I used intelligently, I
learned to live, to think, to hope. Darkness cannot shut me in again. I have had a glimpse of the
shore, and can now live by the hope of reaching it.
6 So my optimism is no mild and unreasoning satisfaction. A poet once said I must be happy
because I did not see the bare, cold present, but lived in a beautiful dream. I do live in a
beautiful dream; but that dream is the actual, the present,—not cold, but warm; not bare, but
furnished with a thousand blessings. The very evil which the poet supposed would be a cruel
6) Read the last sentence from the text.
Only by contact with evil could I have learned to feel by contrast the beauty of truth and love and goodness.
Explain how Helen Keller develops this idea in the text. Use specific details to
support your answer.
giúp em toàn bộ nhé ............
Explanation:
nnnnkgcbchg of bbjbjk oh jjzjjzkedkkdkfjjsjdsjfkjfdkkdowkronqfojofj see j FC
Jonathan wants to separate stones, insects and other unwanted materials in his mixture of grains and corn. What technique of separating mixture is appropriate
A. Winnowing
B. physical manipulation
C. Filtering
D. Magnetism
Answer:
What is B physical manipulation
Explanation:
Physical manipulation means fertilizers that are manufactured, blended, or mixed, or animal manures or compost that have been changed from their initial physical state by manipulations such as drying, cooking, chopping, grinding, shredding, ashing, or pelleting.
Please allow me to know if my answer helped you with a thank you!
Miss Hawaii
A 500 kg car is at rest at the top of a 72 m high hill. The car rolls to the bottom of the hill. At the bottom of the hill, the car has a speed of 25.6 m/s. Calculate the mechanical energy of the car at the top and bottom of the hill. (Assume the bottom of the hill has a height of 0 m, g=9.80 ms2/).
Explanation: Solution
1.
Gravitational potential energy
U=mgh=500*9.8*50
U=245000 J
2.
Kinetic energy is present at bottom of the hill
K=(1/2)mV2=(1/2)*500*27.82
K=193210 J
3.
Work done by friction
W=193210-245000=-51790 J
The mechanical energy at the top and bottom of the hill is equal to 352800 J and 163840 J respectively.
What is the kinetic energy and potential energy?Kinetic energy (KE) can be described as the energy possessed by a moving object due to its motion. Work by a body will be done to change the kinetic energy. The kinetic energy is represented as K.E = ½mv².
Potential energy (P.E) can be described as the energy that is stored by an object due to its position and is represented in the equation as P.E = mgh, where ‘m’ is the mass, ‘g’ is the acceleration due to gravity and ‘h’ is the height.
The mechanical energy = Kinetic energy + potential energy
Given, the mass of the car, m = 500 Kg
The height of the hill, h = 72 m
The velocity of the car, v = 25.6 m/s
At the top of the hill, the mechanical energy = potential energy
The potential energy at the top of the hill, = mgh
P. E. = 500 × 9.8 ×72
P.E. = 352800 J
At the bottom of the hill, the mechanical energy = kinetic energy
The kinetic energy of the car at the bottom of the hill,
K.E. = ½ × 500 (25.6)²
K.E. = 163840 J
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25. Which of the following cannot be broken down into smaller parts through ordinary chemical means?
a. Nitrogen
b. Protein
C. Salt
d. Sugar
Answer:
the correct answer is sugar
Which theory of emotion explains the startle response
Answer:
In the present study, the startle blink reflex is used as a measure of emotion regulation to effective picture stimuli. Based on the aphasic theory of emotion, it is hypothesized that the startle response will be largest in magnitude in the presence of negative emotional stimuli (Varanasi, Spence, & Lang, 1988).
A rock falls to the ground with a force of 300N and air resistance pushes back up on the rock with a
force of 45N. Which is true of the rock?
A There are balanced forces acting on the rock and the net force is 255N down
B. There are unbalanced forces acting on the rock and the net force is 255N down
C. There are balanced forces acting on the rock and the net force is 345N down
. There are unbalanced forces acting on the rock and the net force is 345N down
Answer:
c
Explanation:
() The two forces F_{1} and F_{2} shown in Fig. 4- 39a and b (looking down) act on a 27.0-kg object on a frictionless tabletop . If F_{1} = 10.2N and F_{2} = 16.0N find the net force on the object and its acceleration for each situation , (a) and (b) .
The image shows mountains in Alaska.
Which describes the main feature of the circled area of these mountains?
A syncline is visible.
An anticline is visible.
These mountains show no evidence of folding.
These mountains likely formed from normal faults
Answer:
A syncline is visible.
Explanation:
A syncline is visible.
Answer:
These mountains show no evidence of folding.
Two horses are side by side on a carousel. Which has a greater tangential speed the one closer to the center or the one farther from the center? Explain your answer.
Answer:
The horse father from the center has a greater tangential speed. Although both horses complete one circle in the same time period, the one farther from the center covers a greater distance during that same period.
Explanation:
describe an experiment to demonstrate surface tension in a liquid
Answer:
mę břöőda I dunno this answer
Explanation:
hope you find it
please help me
7. If microsecound = 0.5, how much force must be applied to a spring (spring constant of 0.8 N/m) which is attached to a block of wood (mass = 4.0 kg) in order to just begin to move the block?
Answer:
Explanation:
Your question is quite confusing, particularly the information about microsecond = 0.5.
I'm going to ASSUME that you mean coefficient of static friction μs = 0.5
unfortunately typing a subscript "s" is very difficult and probably leads to such confusion.
I will also ASSUME that the block, and spring, and force vector are all horizontal.
If the force is slowly increased until the block slips, the spring will compress until the force on each end equals the maximum static friction force. As we are not concerned with the compression distance, only the force, we can ignore the spring constant information and simply find the maximum available static friction force.
F = μN
F = μmg
F = 0.5(4.0)(9.8)
F = 19.6 N
Not that it matters, but the spring will have extended or compressed 19.6/0.8 = 24.5 m, which is a very long and very light spring
What happened to an enzyme’s structure as it exceeds the typical human body temperature
Enzymes. ... This is because heat energy causes more collisions, with more energy, between the enzyme molecules and other molecules. However, if the temperature gets too high, the enzyme is denatured and stops working. A common error in exams is to write that enzymes are killed at high temperatures.
Which of the following is the current best hypothesis for the formation of the solar system?
A. Formed by an exploding super nova star which then collapsed and coalesced into a spinning
disk forming Sun and planets
B. Our solar system has always been here and has never changed
C. Formed from the Sun’s explosion releasing particles into space forming planets and other
objects
D. Our solar system was formed by a great collision of other stars with one another
Answer:
A
Explanation:
all galaxies exploded in order to create the sun/stars
If Brad drives 100 meters forwards and then 100 meters backwards, what is her
distance? Her displacement?
Answer:
0 because she went to exact same place as before
THE BEST ANSWER WILL BE MARKED THE BRAINLIEST!
While driving your car at a fast constant speed, you have to press down hard on your gas pedal than you do while driving at a slower constant speed.
In both cases, the net force on your car is zero. Explain why you have to apply a greater pressure to your gas pedal.
Answer:
You have to apply more pressure to the pedal so more gas runs through the car allowing it to reach higher speeds
To move, a car or any other object must be accelerated from rest to the desired speed; this necessitates that the engine force be greater than the friction force. The net force must be zero once the car is moving at constant velocity; otherwise, the car will accelerate (gain speed).
What forces act on a car at constant speed?All forces acting on a car moving at a constant speed (uniform motion) are balanced. In this case, the two opposing backward forces (air resistance and friction) perfectly balance the applied force of the wheels on the road.When we say constant speed, we mean that a body or object moves at a constant change in distance over time. For example, if we are walking in a straight line at 1 step per second, we can say we are moving at a constant speed. Our acceleration is equal to zero in this case.To learn more about : Speed
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A baseball is traveling with a velocity of 12 m/s at an angle of 45 above
horizontal. What is the velocity of the ball after two seconds?
Focus on the ball. Do you see you? Keep following it. Put all your energy into watching it and then unlock your third eye. With that you'll be able to determine your question.
1. What does the term monecious mean?
Answer:
Monecious is a plant or invertebrate animal having both the male and female reproductive organs in the same individual; hermaphrodite.
Which statement describes friction?
Answer:
include the statements pls so i can choose wich one it is and tell you
Explanation:
A stimulus that increses the occurance of a response is called
Answer:
reinforcing stimulus
Explanation:
Think its correct
Answer:
reinforcing stimulus
Explanation:
because it increases the occurence of a response
find charge and charge density on the surface of a conducting sphere of radius 15.2cm where potential at 215 v
this is the correct answer
An object of mass 1.0 kg is at rest on a smooth inclined plane with height h, length 8 m,
and which makes an angle of 30° with the horizontal. The object is allowed to move, it
slides down and onto a rough horizontal surface. After traveling a distance of 4 m it hits a spring and travels 1.3 m more, all along the rough horizontal surface. The spring
constant is 26.5 N/m. What is the coefficient of kinetic friction for the horizontal surface?
The coefficient of kinetic friction for the rough horizontal surface is 0.66.
The given parameters;
mass of the object, m = 1 kglength of the inclined plane, L = 8 mangle of inclination of the plane, θ = 30⁰ distance traveled before hitting the spring, d₁ = 4 mdistance traveled after hitting the spring, d₂ = 1.3 mthe spring constant, k = 26.5 N/mApply work-energy theorem; the work done the force of friction is equal to the energy stored in the spring.
[tex]F_kd_1 = \frac{1}{2} kd_2^2\\\\\mu_kmg cos(\theta)d_1 = \frac{1}{2} kd_2^2\\\\\mu_k(1 \times 9.8 \times cos(30)\times 4) = \frac{1}{2} \times 26.5 \times (1.3)^2\\\\33.95\mu_k = 22.39\\\\\mu_k = \frac{22.39}{33.95} \\\\\mu_k = 0.66[/tex]
Thus, the coefficient of kinetic friction for the rough horizontal surface is 0.66.
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In hockey activities, a warm hockey puck and a frozen hockey puck has a different coefficient of restitution: 0.5 for a warm hockey puck, and 0.35 for a frozen one. NHL requires the frozen pucks to be used in games. To make sure the puck can be used in the game, the referee drops the puck on its side from a height of 2.5 m. How high should the puck bounce if it is a frozen puck
If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.
Given the data in the question;
Since the hockey puck was initially in the referee's hands
Initial velocity; [tex]u = 0m/s[/tex]Distance or height from which it was dropped; [tex]h = 2.5m[/tex]Acceleration due to gravity; [tex]g = 9.8 m/s^2[/tex]Coefficient of restitution a frozen puck; [tex]0.35[/tex]First we will find the velocity of the Puck when it hits the ground
From the Third Equation of Motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.
Since the pluck is under gravity, we will have:
[tex]v^2 = u^2 + 2gh[/tex]
We substitute in our value and find "v"
[tex]v^2 = 0 + (2 \ *\ 9.8m/s^2\ *\ 2.5m )\\\\v^2 = 47.04m^2/s^2\\\\v= \sqrt{47.04m^2/s^2}\\\\v = 6.85857m/s[/tex]
Now, Velocity of the hock puck after it hits the ground and bounce back;
We know that; Coefficient of restitution [tex]= \frac{Relative\ velocity\ after\ collision}{Relative\ velocity\ before\ collision}[/tex]
Hence, Relative Velocity after collision = Coefficient of restitution × Relative Velocity before collision
we substitute in our values;
Relative Velocity after collision [tex]= 0.35 \ *\ 6.85857m/s[/tex]
Relative Velocity after collision [tex]= 2.4 m/s[/tex]
Now, to determine how high should the puck bounced back
We use the Third Equation of Motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.
Since the pluck is under gravity, we will have:
[tex]v^2 = u^2 + 2gh[/tex]
Now, since the hockey puck bounces back, it is experiencing a negative acceleration
Hence, the equation becomes
[tex]v^2 = u^2 - 2gh[/tex]
We substitute our values into the equation and find "h"
[tex](0m/s)^2 = (2.4m/s)^2 - ( 2*9.8m/s^2*h)\\\\0 = 5.76m^2/s^2 - (19.6m/s^2*h)\\\\(19.6m/s*h) = 5.76m^2/s^2 \\\\h= \frac{ 5.76m^2/s^2 }{19.6m/s^2}\\\\h = 0.3m[/tex]
Therefore, If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.
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HELPPPP
The maximum force of sliding friction between a 10 kg rubber box and the concrete
floor is 64 N. How much force should a worker push on the box with if he wants it to
move at a constant velocity?
1) A little less than 64 N
2)A little more than 64 N
3)Exactly 64 N.
4)Exactly 640 N
The force that will move the box at constant velocity must be a little more than 64 N.
The coefficient of sliding friction is obtained from the formula;
μ= F/R
Where;
F = frictional force
μ = coefficient of sliding friction
R = Normal reaction
It is necessary to note that the force that will move the body must be greater than the frictional force acting between the body and the surface in order to move the body. Hence, the force that will move the box at constant velocity must be a little more than 64 N.
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12. Suppose the polar bears were running on land instead of swimming. if the polar bear runs
at a speed of about 5.5m/s how far will it travel in 55 seconds?
the polar bear will travel 302.5m
How can wind, coiled wire, and magnets be used together to generate electricity?
Answer:
b
Explanation:
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A bowling ball, basketball, and tennis ball are all raised to the same height above the ground. Give the order of objects from the least potential energy to the most
Answer:
bowling ball, basket ball, tennis ball
Explanation:
How does the mass of the bob affect the number of swings of a pendulum?
Most gasoline engines in today's automobiles are belt driven. This means that the crankshaft, a rod which rotates and drives the
pistons, is timed to the camshaft, the mechanism which actuates the valves, by means of a belt. Starting from rest, assume it
takes t = 0.0320 s for a crankshaft with a radius of r = 3.75 cm to reach 1250 rpm. If the belt does not stretch or slip, calculate
the angular acceleration ay of the larger camshaft, which has a radius of r2 = 7.50 cm, during this time period.
The angular acceleration of the larger camshaft is 995.72 rad/s².
The given parameters;
initial angular velocity, [tex]\omega _i[/tex] = 0time of motion, t = 0.032 sradius of the crankshaft, r = 3.75 cm final angular speed, [tex]\omega _f[/tex] = 1250 rpmThe angular acceleration of the 3.75 cm camshaft is calculated as follows;
[tex]\omega _f = \omega _i + \alpha t\\\\\omega _f =0 + \alpha t\\\\\omega _f = \alpha t\\\\(1250 \ \frac{rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1\min}{60 \ s} ) = 0.032 \alpha \\\\130.92 = 0.032\alpha \\\\\alpha = \frac{130.92}{0.032} = 4091.25 \ rad/s^2[/tex]
The angular momentum of the camshaft is calculated as follows;
[tex]I_1 \alpha _1 = I_2 \alpha_2 \\\\\frac{1}{2} mr_1^2 \alpha _1 = \frac{1}{2}m R^2 \alpha_2\\\\r_1^2 \alpha _1 = R^2 \alpha_2\\\\\alpha_2 = \frac{r_1^2 \alpha _1 }{R^2} \\\\\alpha_2 =\frac{(0.037)^2 \times (4091.25)}{(0.075)^2} \\\\\alpha _2 = 995.72 \ rad/s^2[/tex]
Thus, the angular acceleration of the larger camshaft is 995.72 rad/s².
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