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
Answers
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
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Related Questions
Which best describes a difference between laser light and regular light?
Laser light is less concentrated.
Laser light has a beam that spreads out.
O Laser light is made of one specific color.
O Laser light pumps out light particles one at a time.
Answers
Answer: the correct answer should be B
Explanation:
The statement that, best describes a difference between laser light and regular light is laser light is made of one specific color.
What is laser light?A laser light is a light produced by a laser
Laser itself, is a device that emits light through a process of optical amplification based on the stimulated emission of electromagnetic radiation.
Difference between laser light and regular lightRegular light is non-directional and inconsistent, while laser light shows directional and highly consistent distribution.Regular light is a mixture of electromagnetic waves of different wavelengths while Laser light is monochrome (one colour).Thus, the statement that, best describes a difference between laser light and regular light is laser light is made of one specific color.
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The cheetah, the fastest land animal, can sprint at 30 m/s. Could the cheetah get a speeding
ticket on a highway where the speed limit is 55 mph?
Answers
Answer:
#1, cheetas can't go 30 miles per second. #2 no
Explanation:
No because animals can't get tickets.
Answer:
yes
Explanation:
it would be going 67.108 mph,
Which of the following is not a part of the appendicular skeleton
a. femur
b. scapula
c.phalanges
d.hyoid
Answers
Answer:
Hyoid
Explanation:
The hyoid is located in the neck area, not the limbs.
A yo-yo is attached to the end of a string. A professional yo yo is spinning it by the string around in a circular
motion. All of a sudden he lets go of the string. What direction is the yo yo moving in initially?
Answers
Answer:
counter-clockwise
Explanation:
Sand is made of tiny pieces of rock that have been worn
down by wind and water. Explain why the formation of
sand is a physical change.
Answers
Answer:
Sand is a form of physical change because it shows how one object is being worn out (broken down) into another object by nature.
Explanation:
A power cycle operating between hot and cold reservoirs at 500 K and 300 K, respectively, receives 1000 kJ by heat transfer from the hot reservoir. The magnitude of the energy discharged by heat transfer to the cold reservoir must satisfy
Answers
Answer:
The value is [tex]Q_l \ge 600\ k J[/tex]
Explanation:
From the question we are told that
The temperature of hot is [tex]T_h = 500 \ K[/tex]
The temperature of cold is [tex]T_c = 300 \ K[/tex]
The energy received is [tex]E = 1000 \ kJ = 1000 *10^{3 } \ J[/tex]
Generally the maximum thermal efficiency of the engine is mathematically represented as
[tex]\eta = \frac{T_h - T_c}{T_h}[/tex]
=> [tex]\eta = \frac{500 - 300}{500}[/tex]
=> [tex]\eta = 0.4[/tex]
Generally the thermal efficiency of the engine is
[tex]\eta_t = \frac{Q - Q_l}{Q}[/tex]
Here [tex]Q_l[/tex] is the heat energy rejected
Generally the thermal efficiency must be less than or equal to the maximum thermal efficiency
So
[tex]\frac{Q - Q_l}{Q} \le 0.4[/tex]
=> [tex]\frac{1000 *10^{3} - Q_l}{1000 *10^{3} } \le 0.4[/tex]
the change in inequality sign is because [tex]1000*10^{3}[/tex] which was dividing started multiplying
=> [tex]Q_l \ge 1000*10^{3} - 400*10^{-3}[/tex]
=> [tex]Q_l \ge 600*10^{3} \ J[/tex]
=> [tex]Q_l \ge 600\ k J[/tex]
Suppose a rocket-propelled motorcycle is fi red from rest horizontally across a canyon 1.00 km wide. (a) What minimum constant acceleration in the x-direction must be provided by the engines so the cycle crosses safely if the opposite side is 0.750 km lower than the starting point? (b) At what speed does the motorcycle land if it maintains this constant horizontal component of acceleration? Neglect air drag, but remember that gravity is still acting in the negative y-direction.
Answers
x
=
a
t
2
2
x=
2
at
2
y
=
−
g
t
2
2
y=−
2
gt
2
Initial conditions
g
=
9.8
m
s
2
g=9.8
s
2
m
the cycle crosses safely if the...
1000
=
a
t
2
2
1000=
2
at
2
−
750
=
−
g
t
2
2
−750=−
2
gt
2
From this we have acceleration
a
=
1000
⋅
9.8
750
=
13.1
m
s
2
a=
750
1000⋅9.8
=13.1
s
2
m
What statements accurately describe sunspots? Check all that apply.
Sunspots are storms on the Sun's surface.
Sunspots are marked by intense magnetic activity.
Sunspots produce solar flares and hot gassy ejections.
Sunspots can affect Earth's climate.
Sunspots are cool areas where the Sun is covered by clouds.
Answers
Answer:
sunspots are storms on the Suns surface
Sunspots are marked by intense magnetic activity
Sunspots produce solar flares and hot gassy ejections.
Sunspots can affect Earth’s climate.
Explanation:
I just did this lesson
The statement that describe sunspot is Sunspots are storms on the Sun's surface.
What are sunspot?Sunspot are region of the earth that are more darker than other surfaces. They have temporary spots that are darker because the temperature of that area is cooler than other surfaces. They have a reduced temperature.
Therefore, The statement that describe sunspot is Sunspots are storms on the Sun's surface.
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1)
You travel east at 50 km/hr and then north at 75 km/hr. What is your Resultant Velocity in km/h?
(Remember the angle) (5pts)
Answers
Answer:
15
Explanation:
A stone is thrown vertically upward with the velocity of 25 m/s. How long day
it take to reach the maximum height? Also calculate the height
Answers
Answer:
The stone takes 2.55 sec to reach a maximum height of 31.89 m
Explanation:
Vertical Motion
When an object is thrown vertically up with a speed vo, it loses speed because of the effect of the gravity and gains height until the speed is eventually zero. The height attained at that time is called maximum height, and can be calculated as follows:
[tex]\displaystyle h_m=\frac{v_o^2}{2g}[/tex]
Since vo=25 m/s
[tex]\displaystyle h_m=\frac{25^2}{2\cdot 9.8}=31.89\ m[/tex]
Also, the time taken to reach the maximum height is calculated by:
[tex]\displaystyle t_m=\frac{v_o}{g}[/tex]
[tex]\displaystyle t_m=\frac{25}{9.8}=2.55\ s[/tex]
The stone takes 2.55 sec to reach a maximum height of 31.89 m
PLEASE HELP 50 POINTS!!!!!!!!!!!!!!!
Andrew and Deshawn are teammates on a high school soccer team. They’ve been playing soccer together for years. They both play the center forward position. Some years, Andrew is better than Deshawn and serves as the team’s starter. Other years, Deshawn is the stronger player and gets the most playing time. They are both very competitive with each other. Although they know each other well, they have never been friends.
In practice yesterday, the team was running drills in the rain. Deshawn slipped in the mud and collided with Andrew. Andrew hurt his knee badly in the fall and will have to sit out for several weeks.
The next day, Andrew told everyone at school that Deshawn hurt him on purpose to get more playing time. Deshawn retaliated by sharing private information about Andrew on social media. Andrew’s best friend Mateo saw the collision. He knows it was a complete accident.
Pick the role of Andrew, Deshawn, or Mateo.
What are the main issues/problems from this person’s perspective?
What ethical issues are involved for this person?
Why are these ethical issues relevant for this person?
How should the person you chose handle the situation?
Answers
Answer:
1. I pick the role of Mateo.
2. The main problems from Mateo's perspective are: He knows that the collision was a complete accident and that Deshawn did not injure Andrew on purpose. Because Andrew told everyone that Deshawn injured him on purpose, Deshawn did something unethical by sharing private information about Andrew on social media.
3. The ethical issues are: Andrew took his hurt and frustration of being injured and missing play time out on Deshawn, so he slandered him by saying that Deshawn hurt him on purpose. Deshawn then retaliated by sharing personal information about Andrew on social media, which is also unethical.
4. These issues are relevant to Mateo because he is Andrew's best friend AND he saw the incident happen and knows it was an accident.
5. If I was Mateo, I would talk to Andrew and explain to him that I know he is mad that he is injured and can't play for several weeks, but telling everyone that Deshawn did it on purpose wasn't right or good sportsmanship. I would encourage Andrew to meet with Deshawn with me and have Andrew apologize to Deshawn for saying it was his fault, and then encourage Andrew to go and tell everyone the real story. I would also encourage Deshawn to apologize for posting private information about Andrew, have Andrew ask him to take it down, and then ask Deshawn to post another post saying why he did it and that he was sorry. I would also encourage them to work together, since they are both star players.
Explanation:
Hellllllpppppp what happens
Answers
Answer:
u can write ur answer and submit it
its a test
I am so close to wanting to drop out of physics
I'm in college and idk why physics has to be this tough. I took physics back in high school, and it was super easy & I had a really nice teacher.
But now that I'm in college, my professor does no help and literally 60 people dropped out of his class bc his class is hard. I am going to fail my midterm this week. idk what to do and Im sick and tired but its a requirement.
I literally wanna drop out of college bc of physics
Answers
Answer: Dont drop out i know school can be a headache but if you try you can accomplished anything.
Explanation:
Answer:
college vs high school.
Explanation:
hello! first, i see the comparison between college and highschool. remember that college professors are much more strict. they dont give participation awards or guide you. this is the time to push yourself! your so close to your goal, why give up from one class? you got this! every question you get wrong or right is a step closer to understanding it better. dont give up. i hope i could help!
Two wires A and B made of the same material and having the same lengths are connected across the same voltage source. If the power supplied to wire A is six times the power supplied to wire B, what is the ratio of their diameters
Answers
Answer:
sqrt (3) : sqrt (2)
Explanation:
Given that Two wires A and B made of the same material and having the same lengths are connected across the same voltage source.
Since they are both made of the same material, they will have the same resistivity.
Resistance is directly proportional to the length of the wire and inversely proportional to the cross sectional area.
Resistance R =1/ πr^2
If the power supplied to wire A is six times the power supplied to wire B, that is,
Pa = 6Pb
Where
Power P = IV
Since the same voltage passes through both wires,
P = V^2/R
Substitutes resistance R into the equation .Therefore,
V^2 ÷ 1 / πr^2 = 6( V^2 ÷ 1 / πr^2 )
V will cancel out , leaving
πr^2 = 6πr^2
Pi( π) will also cancel out. And since diameter d = 2r
r = d/2
Substitutes r into the equation
(d/2)^2 = 6( d/2)^2
d^2/4 = 6( d^2/4)
d^2/4 = 3d^2/2
da^2/2 = 3db^2
(Da/Db)^2 = 3/2
Da : Db = sqrt (3) : sqrt (2)
therefore, the ratio of their diameters is
sqrt (3) : sqrt (2)
The ratio of the diameter of Wire A to Wire B is; √6 : 1
What is the power in the wire?Since both wires are made of the same material, they will definitely have the same resistivity.
Formula for the resistance with area is;
R = 1/A = 1/(πr²)
We are told that the power supplied to wire A is six times the power supplied to wire B. Thus;,
Pa = 6Pb
Since the same voltage passes through both wires, then formula for power in both cases is; P = V²/R. Thus;
V² ÷ (1/πr_a²) = 6(V² ÷ (1/πr_b²))
V and π will cancel out to give;
r_a² = 6r_b²
We know that radius; r = d/2
Thus;
(d_a/2)²= 6(d_b/2)²
d_a²/4 = 6(d_b²/4)
(d_a/d_b)² = 6
d_a/d_b = √6 : 1
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The path of a projectile fired at an angle above the horizontal is best described as
Answers
1.What is the mass of an object that weighs 98 N?
Answers
Answer:
10 kg
Explanation:
10 kg newton
The Plateau of Tibet is a dry place. Do you think the wide-leafed plant fossils found on the plateau could grow there today? Explain your answer.
Answers
No, I dont think that they could grow there now in this day and age becuase the Plateu of Tbiet is very dry and also a little bit dusty. They barely even have grass down there. Wide leaf plants susually grow somewhee like the rainforest where it is warm, sunny, and rainy but in a place like Tibet, I don't think that they would be able to survive there.
Answer:
No. The leaves are similar to those of plants found in lower elevations with plenty of rainfall, like the plants seen at the lower elevations of the Himalayas. Because the Plateau of Tibet has risen over time, it is not likely that these plants could live in these colder climates.
Explanation:
Sample answer from pluto
show answer No Attempt 50% Part (b) Calculate the non-relativistic speed of these electrons v in m/s.
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is [tex]v = 9.18 *10^{6} \ m/s[/tex]
Explanation:
From question we are told that
The potential difference is [tex]\Delta V = 0.24 kV = 0.24 *10^{3} \ V[/tex]
Generally the the non-relativistic speed of these electrons is mathematically represented as
[tex]v = \sqrt{\frac{2 * e * \Delta V}{m} }[/tex]
Here m is the mass of an electron with value [tex]m = 9.11 *10^{-31 } \ kg[/tex]
e is the charge on an electron with value [tex]e = 1.60 *10^{-19} \ C[/tex]
So
[tex]v = \sqrt{\frac{2 * 1.60 *10^{-19} * 0.24 *10^{3} }{9.11*10^{-31}} }[/tex]
=> [tex]v = 9.18 *10^{6} \ m/s[/tex]
Which Statement describes Newton's law of universal gravitation?
Mass has little effect on gravity between objects
gravity pushes objects away from Earth's center
gravity does not act between Earth and Moon
every object in the universe attracts every other objects
>You will get 100 points for answering this!
Answers
Answer:
The answer is D, every object in the universe attracts every other objects
Explanation:
Newton's law of universal gravitation is described as every object in the universe attracts every other objects.
What is gravitation?Gravitation is the force in physics that draws two masses together. Unbelievably, every single particle of matter in the cosmos attracts every other particle through gravity. The phrases gravitation and gravity are sometimes used synonymously to refer to the attraction that exists between everything with mass or energy.
According to Newton's Law of Universal Gravitation, every particle in the cosmos is drawn to every other particle with a force that is directly proportional to the product of their masses and inversely proportional to their distance from one another.
So, every object in the universe attracts every other objects describes Newton's law of universal gravitation.
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4. A 65 kg woman is inside an elevator. Calculate her apparent weight (Normal Force) for the following cases:
a. The elevator moves at constant speed of 10 m/s downward (2 pts)
b. The elevator accelerates upward at a constant rate of 2.4 m/s2 (2 pts)
c. If the scale read 450 N. Find the magnitude and direction of the elevator’s acceleration. (3 pts)
Answers
Answer:
Explanation:
a ) Let her apparent weight be W₁
Net force on her = W₁ - mg
acceleration a = 0
W₁ - mg = 0
W₁ = mg = 65 x 9.8 = 637 N
b )
a = 2.4 m /s²
W₁ - mg = ma
W₁ = m(g + a )
= 65 x ( 9.8 + 2.4 )
= 793 N
c )
W₁ = 450 N
a = ?
W₁ - mg = ma
450 - 65 x 9.8 = 65 x a
65 a = - 187
a = - 2.87 m /s²
acceleration is negative therefore , it must be downwards .
The elevator was accelerating downward at a constant rate of -3.1 m/s²
Let the acceleration due to gravity (g) = 10 m/s²
a) When moving at constant speed of 10 m/s downward:
Net force = mass * g = 65 kg * 10 m/s² = 650N
b) Net force = Mass * g + mass * acceleration
Net force = (65 * 10) + (65 * 2.4) = 806 N
c) Net force = Mass * g + mass * acceleration
450 = 65 * 10 + (65*a)
65a = -200
a = -3.1 m/s²
The elevator was accelerating downward at a constant rate of -3.1 m/s²
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Your are part of a team to help design the atrium of a new building.Your boss, the manager of the project, wants to suspend a 6.2 kg sculpture high over the room by hanging it from the ceiling using thin, clear fishing line (string) so that it will be difficult to see how the sculpture is held up. The only place to fasten the fishing line is to a wooden beam which runs around the edge of the room at the ceiling.The fishing line that she wants to use will hold 10 kg so she suggests attaching two lines to the sculpture to be safe. Each line would come from the opposite side of the ceiling to attach to the hanging sculpture. Her initial design has the first line making an angle of 12.67∘with the ceiling and the second line making an angle of 16.15∘ with the ceiling. She knows you took physics, so she asks you if her design can work.
How much force would the first line exert on the beam holding it up?
Assume that g=9.8 m/s−2.
Answers
Answer:
The answer is below
Explanation:
Given that the mass of the sculpture (m) = 6.2 kg, [tex]\theta_1=12.67^o,\theta_2=16.15^o[/tex]
[tex]Sum\ of\ horizontal\ force\ is\ zero\ hence:\\\\T_1cos(\theta_1)=T_2cos(\theta_2)\\\\T_1=\frac{T_2cos(\theta_2)}{cos(\theta_1)}\\ \\Sum\ of\ vertical\ force\ is\ zero:\\\\T_1sin(\theta_1)+T_2sin(\theta_2)=mg\\\\\frac{T_2cos(\theta_2)}{cos(\theta_1)}sin(\theta_1)+T_2sin(\theta_2)=mg\\\\T_2=\frac{mg}{\frac{cos(\theta_2)}{cos(\theta_1)}sin(\theta_1)+sin\theta_2} \\\\T_2=\frac{6.2*9.81}{\frac{cos(16.15)}{cos(12.67)}sin(12.67)+sin16.15} =123.1\ N\\\\[/tex]
[tex]T_1=\frac{T_2cos(\theta_2)}{cos(\theta_1)}=121.2\ N[/tex]
What President is responsible for eliminating most vending machines from the cafeterias of public schools ?
Answers
Answer:president obama
Explanation:Why this is because he took a of mayor Bloomebergs amti obeity book and made new rules for snack foods.
Which landform represents a piece of land that is formed when a river splits before flowing into the ocean?
Delta
Glacier
Lake
Mountain
Answers
Answer:
delta
Explanation:i just got it right
PLZ HELP!!!!!!! ILL GIVE U BRAINLIEST
Answers
Answer:
Claim: Heart Rate for Sarah and Matt (this would be the homeostasis AT REST and throughout the graph time)
Evidence: The Graph and the HR/S and its evidence towards the heart rate/beats every 15 seconds, due to the homeostasis being a specific time limit and due to the specific heart rate of Matt and Sarah ONLY.
Explanation: A/B Checkpoint throughout the homeostasis (this would be heart rate/trial throughout the time of the graph)
Conclusion: Matt was further away ONLY BECAUSE Sarah has a higher Heart Rate during the time on the graph, which can represent how long they did something to increase/decrease from and where the homeostasis would be located.
Answer:
Claim: Heart Rate for Sarah and Matt (this would be the homeostasis AT REST and throughout the graph time)
Evidence: The Graph and the HR/S and its evidence towards the heart rate/beats every 15 seconds, due to the homeostasis being a specific time limit and due to the specific heart rate of Matt and Sarah ONLY.
Explanation: A/B Checkpoint throughout the homeostasis (this would be heart rate/trial throughout the time of the graph)
Conclusion: Matt was further away ONLY BECAUSE Sarah has a higher Heart Rate during the time on the graph, which can represent how long they did something to increase/decrease from and where the homeostasis would be located.
What is velocity in science
Answers
Answer:
Velocity is the rate at which the position changes. The average velocity is the displacement or position change (a vector quantity) per time ratio.
Hope this helps! ^-^
Answer:
Velocity is the speed at which something moves in one direction.
Explanation:
Hope this helps : )!!
Question 3 of 10
A 0.057 kg tennis ball and a tennis racket collide. The racket has an initial
momentum of -1.75 kg-m/s and a final momentum of -1.25 kg-m/s. The ball
has an initial momentum of 0.00684 kg-m/s. If you assume the collision is
elastic, what is the final velocity of the ball?
A. -0.50 m/s
B. -52.48 m/s
C. -2.99 m/s
d
D. -8.65 m/s
Answers
If we assume the collision to be elastic the final velocity of the ball will be -8.65 m/s, therefore the correct option is D.
What is elastic collision?It is a type of Collison for which the momentum, as well as the kinetic energy after and before the collision, is constant. there is no loss of energy in a perfectly elastic collision.
By using the law of conservation of momentum
momentum before collision = momentum after the collision
the initial momentum of rocket +initial momentum, of ball = final momentum of rocket + final momentum of the ball
By substituting the respective values
-1.75 + 0.00684 = -1.25 + m*v
-1.75 + 0.00684 = -1.25 + 0.057*v
v = -8.65 m/s
Thus, the final velocity of the ball is -8.65 m/s
Learn more about elastic collision
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Question 11 of 15
When hydrogen is attached to the more highly electronegative oxygen atom in
a water molecule,
A. the electronegative atom becomes strongly positive
B. the hydrogen atom becomes partially positive
O C. the oxygen atom becomes partially negative
If answer is right WILL GIVE BRAINLIEST
D. the hydrogen atom becomes partially negative
Answers
Answer:
I'm leaning twards A
Explanation:
explain why a wrecking ball can destroy a building, but a yo-yo cant. Use the term kinetic energy in your explanation.
Answers
momentum increases as force increases.
momentum= mass x velocity
the wrecking ball has a bigger mass and weight.
the mass of the weight of the wrecking ball is constant, the velocity of the weight as it swings must be increased if momentum is going to be increased.
The kinetic energy of a body determines it's level of impact on the object in which it comes in contact with. Hence, the much larger kinetic energy exhibited by a wrecking ball compared to a yo-yo means that is has a much larger impact on a building than a yo-yo.
Kinetic Energy = 0.5mv²The kinetic energy of a body is a factor of it's velocity and mass as they are directly proportional.
The wrecking ball has a very large mass which is thousands of times larger than that of a yo-yo. Also, the velocity at which a wrecking ball is launched is higher than the velocity of a yo-yo.
This means that the kinetic energy of a wrecking ball is much higher than that of yo-yo. Hence, having much greater impact on a building compared to a yo-yo.
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50 POINTS PLSS
Your car is initially at rest when your hit that gas and the car begins to accelerate at a rate of 2.857 m/s/s. The acceleration lasts for 15.5 s. What is the final speed of the car and how much ground does it cover during this acceleration?
Answers
Answer:I think the answer is 5.43 I don’t rlly know
Explanation:
Your car is initially at rest when your hit that gas and the car begins to accelerate at a rate of 2.857 m/s/s. The acceleration lasts for 15.5 s. What is the final speed of the car and how much ground does it cover during this acceleration?
Answers
Answer:
Vf = 44.56 [m/s]
Explanation:
In order to solve this problem we must use the following expression of kinematics.
[tex]v_{f} = v_{i}+(a*t)[/tex]
where:
Vf = final velocity [m/s]
Vi = initial velocity = 0
a = acceleration = 2.857 [m/s^2]
t = time = 15.5 [s]
Note: the initial velocity is equal to zero as the car begins its movement from rest, or with an initial velocity equal to zero.
Vf = 0 + (2.875*15.5)
Vf = 44.56 [m/s]