Given parameters:
Initial velocity = 0m/s
Acceleration = 2.857m/s²
Time = 15.5s
Unknown:
Final speed of the car = ?
Solution:
We use one of the motion equations to solve this problem;
V = U + at
Where V is the final velocity
U is the initial velocity
a is the acceleration
t is the time taken
V = 0 + 2.857 x 15.5 = 44.28m/s
3. An object with a mass of 3.2 kg has a force of 6.2 N applied to it. What is the resulting acceleration
of the object?
Answer:
The answer is 1.94 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
f is the force
m is the acceleration
From the question we have
[tex]a = \frac{6.2}{3.2} \\ = 1.9375[/tex]
We have the final answer as
1.94 m/s²Hope this helps you
If a pair of shoes weighs 0.3 N on Pluto what is the strength of gravity on Pluto
Answer:
0.6 m/s 2
Explanation:
A rectangular loop with an area of 2 m2 is placed perpendicular to a uniform magnetic field of 1 Tesla. The field’s magnitude is increased to 6 Tesla in 4 seconds. The magnitude of the induced emf is equal to:
Answer:
Induced emf = 0
Explanation:
An emf can be induced due to the change in magnetic field. It can be given by :
[tex]\epsilon=\dfrac{d\phi}{dt}\\\\\because \phi=BA\cos\theta\\\\\epsilon=\dfrac{d(BA\cos\theta)}{dt}\\\\\epsilon=A\cos\theta\dfrac{dB}{dt}[/tex]
As the loop is placed perpendicular to a uniform magnetic field of 1 Tesla. It means that [tex]\theta=90^{\circ}[/tex] and cos(90) = 0. Hence, the induced emf is equal to 0.
A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of vi = 2.60 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e).
The right end of a horizontal spring labeled k is attached to a wall. Five images show five configurations as a block labeled m approaches, compresses, and then moves away from the spring.
In figure a, the block is to the left of the spring, and an arrow above the block points to the right.
In figure b, the block is just touching the uncompressed spring, and an arrow labeled vector vi above the block points to the right.
In figure c, the block has compressed the spring by a distance d, and a label indicates vector vf = 0.
In figure d, the block is just touching the uncompressed spring, and an arrow labeled vector v above the block points to the left.
In figure e, the block is a distance D away from the spring, and a label indicates vector v = 0.
(a)
Find the distance of compression d (in m).
m
(b)
Find the speed v (in m/s) at the unstretched position when the object is moving to the left (Figure d).
m/s
(c)
Find the distance D (in m) where the object comes to rest.
m
(d)
What If? If the object becomes attached securely to the end of the spring when it makes contact, what is the new value of the distance D (in m) at which the object will come to rest after moving to the left?
m
Answer:
(a) Approximately [tex]0.335\; \rm m[/tex].
(b) Approximately [tex]1.86\; \rm m\cdot s^{-1}[/tex].
(c) Approximately [tex]0.707\; \rm m[/tex].
(d) Approximately [tex]0.228\; \rm m[/tex].
Explanation:
[tex]v_i[/tex] denotes the velocity of the object in the first diagram right before it came into contact with the spring. Let [tex]m[/tex] denote the mass of the block. Let [tex]\mu[/tex] denote the constant of kinetic friction between the object and the surface. Let [tex]g[/tex] denote the constant of gravitational acceleration.Let [tex]k[/tex] denote the spring constant of this spring.(a)Consider the conversion of energy in this object-spring system.
First diagram: Right before the object came into contact with the spring, the object carries kinetic energy [tex]\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2[/tex].
Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.
Third diagram: After the velocity of the object becomes zero, it has moved a distance of [tex]D[/tex] and compressed the spring by the same distance.
Energy lost to friction: [tex]\underbrace{(\mu \cdot m \cdot g)}_{\text{friction}} \cdot D[/tex]. Elastic potential energy that the spring has gained: [tex]\displaystyle \frac{1}{2}\,k\, D^2[/tex].The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:
[tex]\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2[/tex].
Assume that [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex]. In the equation above, all symbols other than [tex]D[/tex] have known values:
[tex]m =1.10\; \rm kg[/tex].[tex]v_i = 2.60\; \rm m \cdot s^{-1}[/tex].[tex]\mu = 0.250[/tex].[tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].[tex]k = 50.0\; \rm N \cdot m^{-1}[/tex].Substitute in the known values to obtain an equation for [tex]D[/tex] (where the unit of [tex]D\![/tex] is [tex]m[/tex].)
[tex]3.178 = 2.69775\, D + 25\, D^2[/tex].
[tex]2.69775\, D + 25\, D^2 + 3.178 = 0[/tex].
Simplify and solve for [tex]D[/tex]. Note that [tex]D > 0[/tex] because the energy lost to friction should be greater than zero.
[tex]D \approx 0.335\; \rm m[/tex].
(b)The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:
[tex]\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J[/tex].
As the object moves to the left, part of that energy will be lost to friction:
[tex](\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J[/tex].
The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:
[tex]2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J[/tex].
Calculate the velocity corresponding to that kinetic energy:
[tex]\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}[/tex].
(c)As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy ([tex]1.91\; \rm J[/tex]) would be lost to friction.
How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is [tex]\mu \cdot m \cdot g[/tex].
[tex]\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m[/tex].
(d)Similar to (a), solving (d) involves another quadratic equation about [tex]D[/tex].
Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) [tex]1.91\; \rm J[/tex].
Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.
[tex]\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2[/tex].
[tex]25\, D^2 + 2.69775\, D - 1.90811\approx 0[/tex].
Again, [tex]D > 0[/tex] because the energy lost to friction is greater than zero.
[tex]D \approx 0.228\; \rm m[/tex].
The energy transferred between the object and the spring as a closed system, therefore, conserved are;
(a) The distance of compression, d ≈ 0.3354 meters
(b) The speed in the un-stretched position wen the object is sliding to the left, v ≈ 1.8623 m/s
(c) The distance where the object comes to rest, D ≈ 0.7071 m
(d) The distance the object will come to rest attached to the spring, D ≈ 0.2278 m
The reason the above values are correct are as follows;
The known parameters are;
Mass of the object, m₁ = 1.10 kg
Coefficient of friction, μ = 0.250
The initial speed of the object, [tex]v_i[/tex] = 2.60 m/s
Force constant of the spring, K = 50.0 N/m
Distance the spring is compressed by the object = d
(a) Conservation of energy principle
[tex]Kinetic \ energy = \dfrac{1}{2} \cdot m\cdot v^2[/tex]
Work done = Force × Distance
Friction force, [tex]F_f[/tex] = W × μ
Weight, W = m·g
Weight = Mass × Acceleration
Energy transferred by object = Work done by spring + Work done by friction
[tex]Energy \ transferred \ by \ object = Kinetic \ energy = \dfrac{1}{2} \times 1.10\times 2.60^2 = 3.718[/tex]
Energy transferred by object = 3.718 J
[tex]Work \ done \ by \ spring = \dfrac{1}{2} \cdot k\cdot x^2[/tex]
[tex]Work \ by \ spring \ to \ bring \ object \ to \ rest, \ W_{spring} = \dfrac{1}{2} \times 50\times d^2[/tex]
[tex]W_{spring}[/tex] = 25·d²
Work done by friction, [tex]W_{friction}[/tex] = 1.10×9.81×0.250×d = 2.69775·d
Therefore;
3.718 = 25·d² + 2.69775·d
25·d² + 2.69775·d - 3.718 = 0
Solving gives
The distance of the compression d ≈ 0.3354 m
(b) The energy given by the spring = 25·d²
The work done by friction, [tex]W_{friction}[/tex] = 2.69775·d
Kinetic energy given to object = 0.55·v²
0.55·v² = 25·d² - 2.69775·d
0.55·v² = 25×0.3354² - 2.69775×0.3354
∴ v = √(3.4682) = 1.8623
The velocity of the object at the un stretched position, v ≈ 1.8623 m/s
(c) The kinetic energy, K.E. of the object on the way left is given as follows;
K.E. = 0.5 × 1.10 kg × 3.4682 m²/s² = 1.90751 J
The work done by friction before object comes to rest = 2.69775·D
[tex]D = \dfrac{1.90751 \, J}{2.69775 \, N} \approx 0.7071 \, m[/tex]
The distance where the object comes to rest, D ≈ 0.7071 m
(d) The work done on spring, [tex]W_{spring}[/tex] = 25·D'²
Work done on friction, [tex]W_{friction}[/tex] = 2.69775·D'
Kinetic energy of object, K.E. ≈ 1.90751 J
K.E. = [tex]W_{spring}[/tex] + [tex]W_{friction}[/tex]
1.90751 ≈ 25·D'² + 2.6775·D'
25·D'² + 2.6775·D' - 1.90751 = 0
Solving with a graphing calculator gives;
D' ≈ 0.2278 m
The new value of the distance D = 0.2278 m
Learn more about the energy conservation principle here:
https://brainly.com/question/928985
A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen when the spectacles hit the ground? Neglect the air resistance.
Answer:
[tex]h_p = 30.46\ m[/tex]
Explanation:
Free Fall Motion
A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.
The speed vf of the object when a time t has passed is given by:
[tex]v_f=g\cdot t[/tex]
Where [tex]g = 9.8 m/s^2[/tex]
Similarly, the distance y the object has traveled is calculated as follows:
[tex]\displaystyle y=\frac{g\cdot t^2}{2}[/tex]
If we know the height h from which the object was dropped, we can solve the above equation for t:
[tex]\displaystyle t=\sqrt{\frac{2\cdot y}{g}}[/tex]
The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:
[tex]\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec[/tex]
The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:
[tex]t_2=2.56 - 2 = 0.56\ sec[/tex]
Therefore, it has traveled down a distance:
[tex]\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m[/tex]
Thus, the height of the pen is:
[tex]h_p = 32 - 1.54\Rightarrow h_p=30.46\ m[/tex]
The pen is 30.52 m above the ground.
Given that the height of the stadium is h = 32m
The initial velocity of the glasses will be 0.
[tex]h=\frac{1}{2}gt^{2} \\t=\sqrt{\frac{2h}{g} } \\t=\sqrt{\frac{2*32}{9.8} }\\t=2.55s[/tex]is the time taken for the glasses to hit the ground.
Now the pen is released 2 seconds later. So by the time the glasses hit the ground the pen has spent:
[tex]t^{'}=2.55-2\\t^{'}=0.55s[/tex]in the air
distance traveled by the pen:
[tex]d=\frac{1}{2}gt^{2}\\\\d=\frac{1}{2}*9.8*0.55*0.55\\\\d=1.48m[/tex]
So the pen is [tex]h-d=32-1.48=30.52m[/tex] above the ground.
Learn More:
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A planned high-speed train between Houston and Dallas will travel a distance of 386 kilometers in 5.40 × 103 seconds. What is the average speed of this train?
The number of
• in the atom of an element determines its chemical properties.
Answer:
Yes, the number of electrons determines the chemical properties of the atom.
Explanation:
Two cars are traveling on a desert road between three consecutive poles, as shown in the
figure. After 5.1 s, they are side by side at the
next telephone pole. The distance between
the poles is 72.7 m.
1-Find the displacement of Car A after 5.1 s.(answer in units of m.)
2-Find the average velocity of Car A during 5.1
s. (answer in units of m/s.)
Answer:
SUGGGA
Explanation:
A boy and a girl are pulling a heavy crate at the same time with 7 units of firce each. What is the net force acts on the ibject? Is the object balanced or unbalanced?
Answer:
Net force= 14 units
The object is unbalanced
Explanation:
The net force refers to the sum of all forces applied to an object. However, the direction of force applied determine the net force. In this question, a boy and girl is pulling a heavy crate at the same time.
This means that the force is in the same direction, hence, the net force will be:
F(N) = 7 + 7 = 14 unit
However, since the pull is occuring at the same direction. This means that the object has a net force, therefore, will move in a particular direction. This means that the OBJECT IS UNBALANCED
Type of tissue that helps with movement.Immersive Reader
a. Epithelial
b.Muscle
c.Connective
d.Nervous
I will give you branilest!
Answer:
easy thats
13
Explanation:
Help please!!!!!!!!!!!!!
Answer:
The second choice
Explanation:
I think the answer is the second choice because if the surface is smooth, there is less friction. With a boat, it is easier to pull it on water than on the sand, because water has less friction, and thus, the answer is the second choice because rough surfaces have more friction.
Hopefully the explanation and answer helps!
A cows mass is 401 kg and a trucks mass is 832 kg. What is the difference between their weights . Answer please!!!
Answer:
403 kg is the diffrence
What causes tides to occur in the ocean?
Waves
Wind
Gravitational pull
Coriolis effect
Answer:
Gravitational pull
Explanation:
The moon pulls the tides
Answer: gravitational pull
Tides are caused by a gravitational pull from the Moon. Ocean/bay tides rise because of this pull for the gravity under the water. This can happen every day up to 6 times.
A helpful association method like remembering the Allies during World War II as BAR
(Britain, America, and Russia) is called
O an acronym
O the DAP flashcard method
O a visual image
O a mind map
Answer:
an acronym because it is shorted to remember like mvemjsun it's the planet
Part B
You should find that your Interpolated and extrapolated values are not even close to the actual recorded values for these
displacement and velocity readings. Describe the basic assumption behind Interpolation and extrapolation, then for at least
one of these values explain why the calculated value was significantly larger or smaller than the recorded value.
BI
x
Font Sizes
A- A-EIE 3
I
Characters used: 0 / 15000
Answer:
When ever we use interpolation and extrapolation in our case we use linear approximation but the displacement verses time graph as well as velocity verses time grph are not linear so that whenever we use interpolatio and extrapolation we did not get close readings to the actual recorded values.
Explanation:
When there are equal forces from opposite directions acting on an object, the forces are said to be which of the following? (4 points)
A.Positive
B.Negative
C.Balanced
D.Gravitational
Answer: C. Balanced
Explanation:
I took the test and got it right.
How are gas giants similar to one another?
Answer:
they are all made of gass and they are all giants.
Explanation:
Answer:
How are the gas giants similar to one another? dont have solid surfaces and are much larger than earth. Why do all of the gas giants have thick atmospheres? Because they are so massive, the gas giants exert a much stronger gravitational force than the terrestrial planets
Explanation:
One statement of the first law of thermodynamics is that:___________.
a. the amount of work done on a system is dependent of pathway.
b. the total work done on a system must equal the heat absorbed by the system.
c. the heat flow in or out of a system is independent of pathway.
d. the total energy flow in or out of a system is equal to the sum of the heat transferred to or from the system and the work done by or on the system.
e. in any chemical process the heat flow must equal the change in enthalpy.
Answer:
a. the amount of work done on a system is dependent of pathway
Explanation:
The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system.
ΔU = Q - W
Where;
Q, the net heat transfer into the system depends on the pathway
W, the net work done by the system also depends on the pathway
But, ΔU, the change in internal energy is independent of pathway
Therefore, the correct option is "A"
a. the amount of work done on a system is dependent of pathway
The two main types of weathering are (4 points)
A. mechanical and physical
B. physical and kinetic
C. chemical and physical
D. chemical and acidic
Answer:
b
Explanation:
Answer:
its acually c
Explanation:
electromagnetic waves and sound waves can have the same frequency. what is the wavelength of a 1.00kHz electromagnetic wave
Helpppp guys plsss help !
Answer:
with what? I can help but with what
Answer:
2
Explanation:
1) Which of the following is considered an effective treatment for someone with hearing loss based on nerve damage?
TAD
Cochlear implant
Hearing aid
OBI
No treatment available
2) Sylvester is dealing with hearing loss. The doctor informs him that his basilar membrane is damaged. What type of hearing loss is Sylvester experiencing?
Nerve deafness
Conduction hearing loss
Cochlear hearing loss
Conduction deafness
Sensory hearing loss
Answer:
For number 1 no treatment available , number 2 cochlear hearing loss
Explanation:
nerve damage is permanent
A boy pushes a box with a force of 150 N at an angle of 40 with a flat floor. What component of his force is directed downward , or into the floor . PLEASE ANSWER!!!!!
Answer:
[tex]F_y=96.4N[/tex]
Explanation:
Hello.
In this case, considering the force diagram shown on the attached picture, we can see that the component of his force is directed downwards is:
[tex]F_y=F\times sin (\theta)[/tex]
Because the other component is the horizontal one:
[tex]F_x=F\times cos(\theta)[/tex]
In this case, the y-component force turns out:
[tex]F_y=150N\times sin (40\°)\\\\F_y=96.4N[/tex]
Moreover, the x-component force is also computed if required:
[tex]F_x=150N\times cos(40\°)\\\\F_x=114.9N[/tex]
Best regards.
Consider the following debate between two students about their answer to the above question.
Student 1: I thought that whenever one object exerts a force on a second object, the second object also exerts a force that is equal in strength, but in the other direction. So even though Earth is bigger and more massive than the Moon, they still pull on each other with a gravitational force of the same strength, just in different directions.
Student 2: I disagree. I said that Earth exerts the stronger force because it is way bigger than the Moon. Because its mass is bigger, the gravitational force Earth exerts has to be bigger too. I think you are confusing Newton's third law with the law of gravity. Do you agree or disagree with either or both of the students?
Answer:
I agree with student 1
Explanation:
This is because, the magnitude of the gravitational force on both Earth and Moon depends on the product of their masses. Also, both Earth and Moon exert the same force but in opposite directions.
Student 2 is wrong because the gravitational force is the only force acting between the Earth and Moon, and from Newton's third law, it follows an action-reaction pair. But, student 2 got it wrong in the sense that the magnitude of the action and reaction forces are the same and are equal to the gravitational force.
So, the gravitational force acting on each object is the same and doesn't vary for each mass.
We should agree with the student 1.
Gravitational force:
This is to be done due to the magnitude of the force since earth and moon should be based on the masses' product. Moreover, the earth and moon exert a similar force but that should be in inverse directions. Here student 2 should be wrong because the gravitational force is that force that acted between the earth and moon. Based on this, we can say that it should be acted on each object when it is the same also it should not be changed for every mass.
Learn more about newton here: https://brainly.com/question/14379797
why a dam is thicker at the bottom than it's top
Answer: Due to water pressure.
Explanation:
As depth increases so does the pressure.
Which of the following does not discribe a mineral
Answer:
give us some further context to answer your question as well
Explanation:
A race car driver travelling at 15m*s^-1 accelerates at a constant value for 10.0s. She is now driving at a speed of 35 m*s^-1 What was her acceleration? Give your answer in m*s^-2 to one significant figure . Do not include units with your answer.
The driver speeds up with acceleration a so that
35 m/s = 15 m/s + a (10.0 s)
Solve for a :
20 m/s = a (10.0 s)
a = (20 m/s) / (10.0 s)
a = 2 m/s²
The Newton unit obtained from
A)Kg.m/s2
B) Kg. m
C) kg/m
D) kg ml
Answer:
Explanation:
The Newtons unit is kg. m/s2
Option A is the correct answer
is newton's first law true on earth?
Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
THIS LAW IS TRUE AS IT ALSO HAVE A REAL LIFE EXAMPLE.
Examples of Newton's 1st Law : If you slide a hockey puck on ice, eventually it will stop, because of friction on the ice. It will also stop if it hits something, like a player's stick or a goalpost.