This question is incomplete
Complete Question
m1 is 10kg, m2 is 4.0kg. The coefficient of static friction between m1 and the horizontal surface is 0.50. and the Coefficient of kinetic friction is 0.30.
a) if the system is released from rest what will be its acceleration
Answer:
0.7 m/s²
Explanation:
The coefficient of static friction between m1 and the horizontal surface is 0.50. and the coefficient of kinetic friction is 0.30.
(a) if the system is released from rest what will be its acceleration
g = acceleration due to gravity = 9.81 m/s²
Coefficient of Kinetic Friction = μk = 0.30
m1 = 10kg
m2 = 4.0kg
The formula to solve question a is given as:
a = acceleration at rest
m2g- μk m1g = (m1+ m2) a
Making a the subject of the formula:
a = (m2g- μk×m1g )/(m1+ m2)
a = [(4.0 kg × 9.81m/s²) – (0.30 ×9.81 × 10) ]/(10+4)
a = 0.7 m/s²
What do mammoths and tigers need energy for
The boys are finally old enough to compete in the box car derby race at the local fair. They have been working on their cars since the conclusion of the race last year. One boy's car raced down the track and placed 2nd in his race. However, the other boy's car started well but half-way through the race a wheel came off and his car came to a complete stop. The boy was very disappointed and the other boy felt horrible for his friend. Which of the following graphs best represents the motion of boy's car that stopped?
It took 50 joules to push a chair 5 meters across the floor. With what force was the chair pushed?
Answer:
The chair was pushed with 10 N.
Explanation:
The chair was pushed with 50 Joules.
Work = Force * Distance
50 J = F * 5m
F = 50 / 5 = 10N
The chair was pushed with 10 N.
The chair was pushed with 10 N force.
What is Work done?Work is defined as the measure of energy transfer that occurs when an object is moved over a distance by an external force, at least part of which is applied in the direction of displacement.
If the force is constant then work can be calculated by multiplying the length of the path by the component of the force acting along the path, which is expressed mathematically as work W equal to the force f over a distance d, or W = fd.
So, for above given information,
Work done= 50 joules
Distance covered by the chair = 5m
Then, Force= W/d
=50/5= 10N
Thus, the chair was pushed with 10 N force.
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A .05 kg rubber ball is dropped and hits the floor with an initial velocity of 10 m/s. It rebounds away from the floor with a final speed of 7 m/s after being in contact with the floor for .01 seconds. Find the magnitude of the force exerted by the floor on the rubber ball.
Answer:the answer is 3
Explanation:
The Earth's magnetic field is modeled as that of a bar magnet with the geographic poles being Magnetic poles of the bar magnet, Based on our definitions of Magnetic Poles, if you were to go to the Earth's Geographic North Pole, you would be at a Magnetic _______________ of the bar magnet.
Answer:
South pole
Explanation:
In a bar magnet, field lines go from the North Pole to the South Pole (outside the magnet).
As the earth magnetic field lines go from South Pole (geographic) to the North one, this means that the North pole (geographic) really behaves as a South Pole (magnetic).
If 0.5 kg of this material is used in a transformer core, how long would it have to operate at a frequency of 60 cps to heat up 1
Answer:
Hello your question is incomplete attached below is the complete question
answer : 49 seconds
Explanation:
considering only Hysteresis loss
we have to calculate the Area affected/under the Hysteresis loss
= volume * area
= 4 * ( 1.5 * 20 ) = 120 tesla. A/m
next we calculate the volume of the material
= mass of material / density
= 500 grams / 7.9 g/cm^3 = 6.33 * 10^-5 m^3
next we calculate the heat lost per cycle
= 6.33 * 10^-5 m^3 * 120 = 0.007596 joules
The total heat required to raise temperature by 1°c
= Cp * ΔT * n
= 3R * n * ΔT = 3(8.314) * 8.95 * 1 = 223.23 Joules
where n = number of moles = 500grams / 55.85 = 8.95moles
ΔT = 1
Therefore the time required to have to operate at a frequency of 60 cps
= Total heat required / heat lost per cycle
=( 223.23 / 0.007596 ) 60 cps
= 489.796 seconds ≈ 49 seconds
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop. (a) What is the kinetic energy of the ball just bef
Answer:
(a) The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the bowling ball is 113.272 joules.
Explanation:
The statement is incomplete. The complete question is:
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.
(a) What is the kinetic energy of the ball just before it hits the mattress?
(b) How much work does the gravitational force of the earth do on the ball as it falls, for the first part of the fall (from the moment you drop it to just before it hits the mattress)?
(c) How much work does the gravitational force do on the ball while it is compressing the mattress?
(d) How much work does the mattress do on the ball? (You’ll need to use the results of parts (a) and (c)）
(a) Based on the Principle of Energy Conservation, we know that ball-earth system is conservative, so that kinetic energy is increased at the expense of gravitational potential energy as ball falls:
[tex]K_{1}+U_{g,1} = K_{2}+U_{g,2}[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Kinetic energies at top and bottom, measured in joules.
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Gravitational potential energies at top and bottom, measured in joules.
Now we expand the expression by definition of gravitational potential energy:
[tex]U_{g,1}-U_{g,2} = K_{2}-K_{1}[/tex]
[tex]K_{2}= m\cdot g \cdot (z_{1}-z_{2})+K_{1}[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the bowling ball, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the bowling ball, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex], [tex]z_{2} = 0\,m[/tex] and [tex]K_{1} = 0\,J[/tex], the kinetic energy of the ball just before it hits the matress:
[tex]K_{2} = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1.5\,m-0\,m)+0\,m[/tex]
[tex]K_{2} = 102.974\,J[/tex]
The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The gravitational work done by the gravitational force of Earth ([tex]\Delta W[/tex]), measured in joules, is obtained by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,1}-U_{g,2}[/tex]
[tex]\Delta W = m\cdot g\cdot (z_{1}-z_{2})[/tex] (Eq. 2)
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex] and [tex]z_{2} = 0\,m[/tex], then the gravitational work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.5\,m-0\,m)[/tex]
[tex]\Delta W = 102.974\,J[/tex]
The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) The work done by the gravitational force of Earth while the bowling when mattress is compressed is determined by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,2}-U_{g,3}[/tex]
Where [tex]U_{g,3}[/tex] is the gravitational potential energy of the bowling ball when mattress in compressed, measured in joules.
[tex]\Delta W = m\cdot g \cdot (z_{2}-z_{3})[/tex]
Where [tex]z_{3}[/tex] is the height of the ball when mattress is compressed, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2}= 0\,m[/tex] and [tex]z_{3} = -0.15\,m[/tex], the work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0\,m-(-0.15\,m)][/tex]
[tex]\Delta W = 10.298\,J[/tex]
Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the ball equals the sum of kinetic energy just before mattress compression and the work done by the gravitational force when mattress is compressed:
[tex]\Delta W' = K_{2}+\Delta W[/tex]
([tex]K_{2} = 102.974\,J[/tex], [tex]\Delta W = 10.298\,W[/tex])
[tex]\Delta W' = 113.272\,J[/tex]
The work done by the mattress on the bowling ball is 113.272 joules.
Which exerts more force, the Earth pulling on the moon or the moon pulling on the Earth? Explain.
Answer: the earth
Explanation: Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth. Over a very long time, the moon's rotations created fiction with the Earth's tugging back, until the moon's orbit and rotational locked with Earth.
and that's why the earth pulls the moon
A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it going after that acceleration?
Answer:68.15m/s
Explanation:
Given:
v₁=15m/s
a=6.5m/s²
v₁=?
x=340m
Formula:
v₁²=v₁²+2a (x)
Set up:
=[tex]\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)[/tex]
Solution:68.15m/s
If Mary runs 5 miles in 50 minutes, what is her speed with the correct
label?
Silly Goose falls 1.0 m to the floor. How long does the fall take
Answer:You need to give more explanation sorry
Explanation:
Answer:
4.20 seconds
Explanation:
Supposing that silly goose weighs 69 pounds, we need to start on the math.
Simple maths, truly and really. 69/1=69, of course.
Therefore it will take 4.20 seconds for silly goose to hit the ground. if he is going to be a silly goose though, he can just go in the pond, instead of wasting his time.
Examine the diagram below.
Which of the above objects is moving the fastest?
A.
All 3 objects are moving at the same speed.
B. B
C. C
D. A
Answer:
Correct choice: D. Object A is the fastest
Explanation:
In a distance vs time graph, the distance is plotted vertically, and the time is plotted horizontally.
The diagram shows three graphs of objects A, B, and C.
The graph of A shows the object traveled 12 meters in 3 seconds, for a speed of 12/3= 4 m/s.
The graph of B shows the object traveled 8 meters in 4 seconds for a speed of 8/4=2 m/s.
Finally, the object C travels 4 meters in 4 seconds, for a speed of 4/4= 1 m/s
Thus, the fastest object is A.
Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birmingham—1963. Explain how they are similar or different in a few sentences.
Answer:
they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta
Explanation:
use in your own words teachers know when your not trust me.
Help me out on this?
PLEASE HELP EASY MULTIPLE CHOICE!!!!!!!!!!!
Answer:
options C is correct
Explanation:
asking questions is super in this education life
Answer:
option c should be the answer
Your TV has a resistance of 10 ohms and a wall voltage of 120 V. How much current and power does it use
Answer:
Current used is 12 ampere.
Power is 1440 watts.
Explanation:
To find the current used by the TV.
Current (I) = voltage/resistance
Current= 120/10
Current is 12Ampere.
To get power used by the TV,
Power = voltage × current.
= 120× 12
Power = 1440 watts.
Two charged objects are separated by distance, d. The first charge has a larger magnitude (size) than the second charge. Which one exerts the most force?
Answer:
The two charged objects will exert equal and opposite forces on each other.
Explanation:
Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of charges on the objects and inversely proportional to the square of the distance between the two objects.
This force of attraction or repulsion between the two charged objects is always equal and opposite.
Therefore, the two charged objects will exert equal and opposite forces on each other.
A
6. All other changeable factors that must
be kept the same to ensure a fair test
(what you keep the same).
Answer:
a constant variable?
Explanation:
A constant variable is any aspect of an experiment that a researcher intentionally keeps unchanged throughout an experiment.
Experiments are always testing for measurable change, which is the dependent variable. You can also think of a dependent variable as the result obtained from an experiment. It is dependent on the change that occurs
plzzz helppp
You are pushing a box North in the hallway, at 20n, and a friend gets in front of the box and goes in the opposite direction, at 30n. What direction is the box going at? How much force does the box have going in that direction?
Answer:
the box is going south at 10n
Explanation:
A vector of components (−23, −22) is multiplied by the scalar value of −6. What is the magnitude and direction of the resultant vector?
Answer:
(1,)
Explanation:
Answer:
magnitude: 21.6; direction: 33.7°
Explanation:
Grass and plants get energy from
А
the sun.
B
eating food.
с
windmills.
D
electrons.
Answer:
From the Sun
Explanation:
Plants can't eat any food. They don't ue or need windmills to get energy. They are plants so they don't have any electrons. The only way that they can recive energy from is the sun. Sometimes plants die when they don't get enough sun because they don't have any energy to live.
If a rock is skipped into a lake at 24 m/s2, with that what force was the rock thrown if it was 1.75kg?
Answer: f= M×A
1.75kg×24= 42N
Explanation:
Because to find force you do Mass times acceleration so I did 1.75 kg times 24 would equal 42 Newtons!
A car stops in 130 m. If it has an acceleration of -5 m/s2 what was the cars starting velocity?
after
Variables:
Equation and Solve:
Answer:
We are given:
displacement (s) = 130 m
acceleration (a) = -5 m/s²
final velocity (v) = 0 m/s [the cars 'stops' in 130 m]
initial velocity (u) = u m/s
Solving for initial velocity:
From the third equation of motion:
v² - u² = 2as
replacing the variables
(0)² - (u)² = 2(-5)(130)
-u² = -1300
u² = 1300
u = √1300
u = 36 m/s
A person has a mass of 1000g and an acceleration of 20 m/s/s. What is the force on the person
Answer:
20000
Explanation:
Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000
A student uses a microwave oven to heat a meal. The wavelength of the radiation is 8.97 cm. What is the energy of one photon of this microwave radiation? Multiply the answer you get by 1025 to be able to input a number more easily into canvas. Enter to 2 decimal places.
Answer:
The energy of one photon is 2.21x10⁻²⁴ J. Multiplied by 10²⁵ is 22.10 J.
Explanation:
The energy (E) of a photon is:
[tex] E = h\frac{c}{\lambda} [/tex]
Where:
h: is the Planck's constant = 6.62x10⁻³⁴ J.s
λ: is the wavelength of the radiation = 8.97 cm
c: is the speed of light = 3.00x10⁸ m/s
[tex] E = h\frac{c}{\lambda} = 6.62 \cdot 10^{-34} J.s\frac{3.00\cdot 10^{8} m/s}{8.97 \cdot 10^{-2} m} = 2.21 \cdot 10^{-24} J [/tex]
Hence, the energy of one photon is 2.21x10⁻²⁴ J.
Now, if we multiply the answer by 10²⁵ we have:
[tex] E = 2.21 \cdot 10^{-24} J \cdot 10^{25} = 22.10 J [/tex]
I hope it helps you!
The energy of one photon is 2.21x10⁻²⁴ J. Multiplied by 10²⁵ is 22.10 J.
Calculation of energy:We know that
[tex]E = h\frac{c}{\lambda}[/tex]
Here
h be the Planck's constant = 6.62x10⁻³⁴ J.s
λ be the wavelength of the radiation = 8.97 cm
c be the speed of light = 3.00x10⁸ m/s
Now
Here we need to multiply the answer 10^25 so that the correct answer could come.
[tex]E = 6.62.10^{-34} \frac{3.00.10^{8}}{8.97.10^{-2}}[/tex]
= 2.21x10⁻²⁴ J.
= 22.10 J.
Hence, the energy of one photon is 2.21x10⁻²⁴ J. Multiplied by 10²⁵ is 22.10 J.
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An object is rolled at 12 m/s down a table. It stops
after 15s. What was its acceleration?
Variables:
Equation and Solve:
Answer:
We are given:
initial velocity (u) = 12 m/s
final velocity (v) = 0 m/s
time taken (t) = 15 seconds
acceleration (a) = a m/s²
Solving for acceleration:
from the first equation of motion
v = u + at
replacing the variables
0 = 12 + (a)(15)
0 = 15a + 12
a = -12 / 15
a = -4 / 5 m/s²
What is the key for a successful relationship? and Why?
Answer:
communication, if you don't talk you'll never know what's going on.
Explanation:
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp. How much of its original total energy (in J) survives as KE when it reaches the ground
This question is incomplete, the complete question is;
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.
How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°
Answer:
the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J
Explanation:
Given that;
m = 9.9 kg
h = 4.9 m
d = 5 m
μ = 0.3
θ = 36.87°
Now from conservation of energy, the energy is;
Et = mgh
we substitute
Et = 9.9 × 9.8 × 4.9
= 475.398 J
Also the loss of energy i
E_loss = (umg cosθ) d
we substitute
E_loss = 0.3 × 9.9 × 9.8 × cos36.87° × 5
= 116.423 J
so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be
E = Et - E_loss
E = 475.398 J - 116.423 J
E = 358.975 J
the radius of the earth social
Why do you feel that you are being thrown upward out of your seat when going over an arced hump on a roller coaster
Answer: The options are not given.
Here are the options.
a) There is an additional force lifting up on you.
(b) At the top you continue going straight and the seat moves out from under you.
(c) You press on the seat less than when the coaster is at rest.Thus the seat presses less on you. (
d) Both b and c are correct.
(e) a, b, and c are correct.
The correct option Is D.
B.At the top you continue going straight and the seat moves out from under you. C.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less.
Explanation:
At the top you continue going straight and the seat moves out from under you.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less because it is as a result of a phenomenon called Weightlessness. This occur when there is no force or little force is acting on your body. At the top you continue going straight and the seat moves out from under you because there is no force acting on your body and when the body is in free fall i.e acceleration due to gravity , the person is not supported by any thing at.
That is the scenarion that occur...