Answers
Answer:
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
Explanation:
The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:
[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]
[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.
[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.
And we expand the equation above by definitions of elastic potential energy and kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the block, measured in kilograms.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.
[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.
If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:
[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]
[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
Related Questions
correct me if im wrong
Answers
If a net force of 15N is applied to a 3kg box, what is the acceleration of the box?
Group of answer choices
5 m/s2
45 m/s2
0.2 m.s2
18 m/s2
Answers
Answer:
5
Explanation:
Bextra in bf x vi d sj by
The Intensity level of a loud saw is 100 db at a distance of 5m. At what distance would the level be 80 db
Answers
Answer:
50 m
Explanation:
The relationship between the intensity of sound in dB and distance is given by the formula:
[tex]B_2=B_1+20log(\frac{R_1}{R_2} )\\\\Where \ B_2\ is \ the\ sound\ intensity\ at\ distance\ R_2\ and\\B_1\ is \ the\ sound\ intensity\ at\ distance\ R_1\ \\\\Given\ that: B_1=100\ dB, R_1=5\ m, B_2=80\ dB\\\\B_2=B_1+20log(\frac{R_1}{R_2} )\\\\80=100+20log(\frac{5}{R_2} )\\\\-20=20log(\frac{5}{R_2} )\\\\log(\frac{5}{R_2} )=-1\\\\\frac{5}{R_2}=10^{-1}\\\\\frac{5}{R_2}=0.1\\\\R_2=5/0.1=50\ m[/tex]
What is the change in internal energy (in J) of a system that absorbs 0.523 kJ of heat from its surroundings and has 0.366 kcal of work done on it
Answers
Answer:
The change in internal energy of the system is 2,054 J
Explanation:
The first law of thermodynamics relates the work and the transferred heat exchanged in a system through internal energy. This energy is neither created nor destroyed, it is only transformed.
Taking into account that the internal energy is the sum of all the energies of the particles that the system has, you have:
ΔU= Q + W
where U is the internal energy of the system (isolated), Q is the amount of heat contributed to the system and W is the work done by the system.
By convention, Q is positive if it goes from the environment to the system, or negative otherwise, and W is positive if it is carried out on the system and negative if it is carried out by the system.
In this case:
Q= 0.523 kJ (because the energy is absorbed, this is,it goes from the environment to the system)W= 0.366 kcal= 1.531 kJ (because the work is done on the system, and being 1 kcal= 4.184 kJ)Replacing:
ΔU= 0.523 kJ + 1.531 kJ
Solving:
ΔU= 2.054 kJ = 2,054 J (being 1 kJ=1,000 J)
The change in internal energy of the system is 2,054 J
Matching type. Send help please. ASAP!
Answers
Answer:
46-D
47-C
48-F
49-A
50-B
I am not very sure I am right about those answers though.
Bird A, with a mass of 2.2 kg, is stationary while Bird B, with a mass of 1.7 kg, is moving due north from Bird A at 3 m/s. What is the velocity of the center of mass for this system of two birds
Answers
Answer:
1.3 m/s
Explanation:
It is given that,
Mass of bird A, [tex]m_A=2.2\ kg[/tex]
Mass of bird B, [tex]m_B=1.7\ kg[/tex]
Initial speed of bird A is 0 as it was at rest
Initial speed of bird B is 3 m/s
We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,
[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]
So, the center of mass for this system is 1.3 m/s.
An object moving 20 m/s
experiences an acceleration of 4 m/s' for 8
seconds. How far did it move in that time?
Variables:
Equation and Solve:
Answers
Answer:
We are given:
initial velocity (u) = 20m/s
acceleration (a) = 4 m/s²
time (t) = 8 seconds
displacement (s) = s m
Solving for Displacement:
From the seconds equation of motion:
s = ut + 1/2 * at²
replacing the variables
s = 20(8) + 1/2 * (4)*(8)*(8)
s = 160 + 128
s = 288 m
What is the answer to this ?
Answers
The equation that governs the period of a pendulum’s swinging. T=2π√L/g
Where T is the period, L is the length of the pendulum and g is a constant, equal to 9.8 m/s2. The symbol g is a measure of the strength of Earth’s gravity, and has a different value on other planets and moons.
On our Moon, the strength of earth’s gravity is only 1/6th of the normal value. If a pendulum on Earth has a period of 4.9 seconds, what is the period of that same pendulum on the moon?
Answers
Answer:
The period of that same pendulum on the moon is 12.0 seconds.
Explanation:
To determine the period of that same pendulum on the moon,
First, we will determine the value of g (which is a measure of the strength of Earth's gravity) on the Moon. Let the value of g on the Moon be [tex]g_{M}[/tex].
From the question, the strength of earth’s gravity is only 1/6th of the normal value. The normal value of g is 9.8 m/s²
∴ [tex]g_{M}[/tex] = [tex]\frac{1}{6} \times 9.8 m/s^{2}[/tex]
[tex]g_{M}[/tex] = 1.63 m/s²
From the question, T=2π√L/g
[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]
We can write that,
[tex]T_{E} = 2\pi \sqrt{\frac{L}{g_{E} } }[/tex] .......... (1)
Where [tex]T_{E}[/tex] is the period of the pendulum on Earth and [tex]g_{E}[/tex] is the measure of the strength of Earth's gravity
and
[tex]T_{M} = 2\pi \sqrt{\frac{L}{g_{M} } }[/tex] .......... (2)
Where [tex]T_{M}[/tex] is the period of the pendulum on Moon and [tex]g_{M}[/tex] is the measure of the strength of Earth's gravity on the Moon.
Since we are to determine the period of the same pendulum on the moon, then, [tex]2\pi[/tex] and [tex]L[/tex] are constants.
Dividing equation (1) by (2), we get
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
From the question,
[tex]T_{E} = 4.9secs[/tex]
[tex]g_{E}[/tex] = 9.8 m/s²
[tex]g_{M}[/tex] = 1.63 m/s²
[tex]T_{M}[/tex] = ??
From,
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
[tex]\frac{4.9}{T_{M} } = \sqrt{\frac{1.63}{9.8} }[/tex]
[tex]\frac{4.9}{T_{M} } = 0.40783[/tex]
[tex]T_{M} =\frac{4.9}{0.40783 }[/tex]
[tex]T_{M} = 12.01 secs[/tex]
∴ [tex]T_{M} = 12.0secs[/tex]
Hence, the period of that same pendulum on the moon is 12.0 seconds.
Answer:
The period of that same pendulum on the moon is 12.0 s
Explanation:
Given;
period of a pendulum’s swinging, T=2π√L/g
the strength of earth’s gravity on moon, g₂ = ¹/₆(g₁)
period of pendulum on Earth, T₁ = 4.9 s
period of pendulum on moon, T₂ = ?
The length of the pendulum is constant, make it the subject of the formula;
[tex]T = 2\pi \sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\\\\(\frac{T}{2\pi} )^2 =\frac{L}{g}\\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\ L = \frac{gT^2}{4\pi^2}\\\\L_1 = L_2\\\\\frac{g_1T_1^2}{4\pi^2}= \frac{g_2T_2^2}{4\pi^2}\\\\g_1T_1^2 = g_2T_2^2\\\\T_2^2 = \frac{g_1T_1^2}{g_2} \\\\T_2 = \sqrt{\frac{g_1T_1^2}{g_2}}\\\\ T_2 = \sqrt{\frac{g_1T_1^2}{g_1/6}}\\\\ T_2 = \sqrt{\frac{6*g_1T_1^2}{g_1}}\\\\T_2 = \sqrt{6T_1^2}\\\\ T_2 = T_1\sqrt{6} \\\\T_2 = (4.9)\sqrt{6}\\\\ T_2 = 12.0 \ s[/tex]
Therefore, the period of that same pendulum on the moon is 12.0 s
SOH-CAH-TOA is used to solve for the ________
velocities in a full/angled projectile.
a. final (x and y)
b. overall
c. initial (x and y)
d. resultant
Answers
Answer:
c. initial (x and y)
Explanation:
When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.
Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"
Thus, this method resolves the initial x and y velocities.
What are the significant transitions middle adulthood?
Answers
Answer:
Making the transition from young adulthood to middle adulthood can be difficult for some people. There are many changes which affect areas in a person’s biology, their psychology, their social life and their spiritual relationship. There are multiple stages of development which may affect an individual during middle adulthood which can be defined between either 30-65 years old or 40-65 years old.
Explanation:
In a spy movie, the hero, James, stands on a scale that is positioned horizontally on the floor. It registers his weight as 810 N . Unknown to our hero, the floor is actually a trap door, and when the door suddenly disappears, James and the scale fall at the acceleration of gravity, down towards an unknown fate. As James falls, he looks at the scale to see his weight. What does he see
Answers
Answer:
His weight would be zero on the scale i.e he is weightless at that instance.
Explanation:
weight = mg
where m is the mass of the object, and g is the acceleration of gravity.
⇒ 810 = mg
During free fall, the weight of an object can be determined by:
W = mg - ma (provided that acceleration of gravity is greater than acceleration of the object)
where a is the acceleration of the object.
But since James fall at the acceleration of gravity, then:
g = a
mg = ma = 810 N
So that;
W = 810 - 810
= 0 N
Therefore though the weight of James is 810 N, but the scale reads 0 N. this condition is referred to as weightlessness.
What type of force holds atoms together in a crystal?
Answers
Answer:
Covalent Bond
Explanation:
i took the test , mark me brainliest.
Answer: Electrical
Explanation: Atoms are tied together by electrical bonding forces.
Notice that the electromagnet in the virtual simulation is made up of a battery and a wire. What item could you add to the electromagnet to make it even stronger?
Answers
Answer:
Explanation:
Have y’all seen steeleflag19 at all on here?
calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun
Answers
Answer:
Hello your question is incomplete below is the complete question
Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000
answer : V = 1.624* 10^-5 m/s
Explanation:
First we have to calculate the value of a
a = 93 * 10^6 mile/m * 1609.344 m
= 149.668 * 10^8 m
next we will express the distance between the earth and the sun
[tex]r = \frac{a(1-E^2)}{1+Ecos\beta }[/tex] --------- (1)
a = 149.668 * 10^8
E (eccentricity ) = ( 1/60 )^2
[tex]\beta[/tex] = 90°
input the given values into equation 1 above
r = 149.626 * 10^9 m
next calculate the Earths velocity of approach towards the sun using this equation
[tex]v^2 = \frac{4\pi^2 }{r_{c} }[/tex] ------ (2)
Note :
Rc = 149.626 * 10^9 m
equation 2 becomes
([tex]V^2 = (\frac{4\pi^{2} }{149.626*10^9})[/tex]
therefore : V = 1.624* 10^-5 m/s
first correct answer gets brainliest
Answers
Answer:
electrical energy transforming into sound energy in speaker
Answer:
the first one. Electrical energy transforming into sound energy in a speaker
An airplane, starting at rest, takes off on a 600. m long runway accelerating at a rate of 12 m/s/s. How many seconds does it take to reach the end of the runway?
Answers
x = 1/2 at²
where x = length of runway, a = acceleration, and t = time.
600 m = 1/2 (12 m/s²) t²
t² = (1200 m) / (12 m/s²)
t² = 100 s²
t = 10 s
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
Answers
This question is incomplete, the complete question is;
The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
Answer:
the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Explanation:
Given that;
P₁ = 1.00 atm
P₂ = ?
V₁ = 1 L
V₂ = 1.60 L
the temperature of the gas is kept constant
we know that;
P₁V₁ = P₂V₂
so we substitute
1 × 1 = P₂ × 1.60
P₂ = 1 / 1.60
P₂ = 0.625 atm
Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Newton's first law states that objects do not change their motion unless acted upon by a net force. What does the word 'net' mean in this context?
A woven net, such as a fishing net or basketball net
B To catch or ensnare
C Remaining or left over after everything has been accounted for
D To cover, such as with mosquito netting
Answers
how far will a brick starting from rest fall freely in 3.0 seconds?
Answers
Answer: It will be about 44.1m
Explanation:
A 30%-efficient car engine accelerates the 1300 kg car from rest to 10 m/s . How much energy is transferred to the engine by burning gasoline
Answers
Answer:
The Energy transferred to the engine by burning gasoline = 216.67 KJ
Explanation:
The parameters given are:
The efficiency of the car engine, E = 30% = 0.3
Mass, m = 1300 kg
Initial velocity, u = 0, since the car is from rest
The final velocity, v = 10 m/s
Since the car was moving, we calculate its kinetic energy.
kinetic energy = ((1/2) (m) (v^2)
((1/2) (1300 kg) (10 m/s^2)
= 65,000 j
The Energy, Q transferred to the engine by burning gasoline in this case
= potential energy / The efficiency of the car engine, E
Q = 65,000 j / 0.3
= 216,666.66 J
Converting Joule to kilojoule
where 1KJ = 1000j
216,666.66 J = 216.67 KJ
what is the summary for Electrons and protons
Answers
Explanation:
the link enjoy
HELP PLS7. A steel ball is dropped from a height of 100 meters. Which velocity-time graph best describes the
motion of the ball?
Answers
Answer:
Option C.
Explanation:
To know which velocity-time graph best describes the motion of the ball, let us calculate the velocity of the ball and the time taken for the ball to get the ground. This can be obtained as follow:
1. Determination of the velocity.
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 100 m
Final velocity (v) =.?
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 100)
v² = 0 + 1960
v² = 1960
Take the square root of both side.
v = √(1960)
v = 44.27 m/s
2. Determination of the time taken.
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 100 m
Time (t) =.?
h = ½gt²
100 = ½ × 9.8 × t²
100 = 4.9 × t²
Divide both side by 4.9
t² = 100 / 4.9
Take the square root of both side
t = √(100 / 4.9)
t = 4.52 s
From the above illustration,
Initial time (t1) = 0 s
Final time (t2) = 4.52 s
Initial velocity (u) = 0 m/s
Final velocity (v) = 44.27 m/s
Thus, we can see that as the time increase, the velocity also increase. Therefore, option C gives the correct answer to the question.
What kind of variation is there in the mechanical energy as the cart rolls down the ramp? Does this agree with your prediction? Explain.
Answers
Answer:
Em₀ = U = m g h , Em_{f} = K = ½ m v²
Explanation:
When a car is on a ramp it has a certain amount of mechanical energy. At the highest point of the ramp the mechanical energy is fully potential given by
Em₀ = U = m g h
As part of this energy descends down the ramp, part of this energy is transformed into kinetic energy and has one part of each, even though the sum remains the initial energy
Em = K + U = ½ m v² + mg y
y <h
when it reaches the bottom of the ramp it has no height therefore there is no potential energy, all of it has been transformed into kinetic energy
Em_{f} = K = ½ m v²
This energy transformation is in the case that the friction force is zero.
If there is a friction force, it performs work against the low car, it is reflected in an increase in the internal energy (temperature) of the car. In this case the energy in the lower part is less than the initial one by a factor
[tex]W_{nc}[/tex] = - fr L
therefore the numeraire values of the velocity are lower, due to the energy lost by friction.
A ray is incident at at 50 degrees angle on a plane mirror. What will be the deviation after reflection from the mirror?
Answers
Answer:
Explanation:
If the ray were not deviated, it would travel straight through the mirror. Due to the mirror, the incident ray is reflected at 30°. The ray travels 30° + 30° = 60°. The angle of deviation is 180° - 60° = 120°.
Super Mario and Bowser Jr. are racing around a track when Baby Bowser launches a green shell at Mario, bringing him to rest. Bowser Jr. then passes Mario at his top speed of 30 blocks/h, moving down the track in a straight line. Mario quickly accelerates and reaches his top speed of 40 blocks/h in order to catch back up and pass Bowser Jr., but by this point he has opened up a 75 block head start down the track. Mario world measure distance using the units of blocks, with 1 block = 0.47 m.
a) What are Mario and Bowser Jr.'s speeds in m/s?
Assuming both Mario and Bowser Jr. race to the finish in a straight line at their top speeds,
b) How long does it take for Mario to catch Bowser Jr.?
c) How far down the track is Mario from the point at which he reaches his top speed?
Answers
Answer:
(a). Mario's speed in m/s = 5.2 × 10^-3 m/s.
Bowser Jr.'s speeds in m/s = 3.92 × 10^-3 m/s.
(b). 27001.2 seconds(s)..
(c). 141 metre(m).
Explanation:
So, the following data or parameters or information was given in the question above. These informations are going to help us in solving this question or problem;
=>" Bowser Jr. then passes Mario at his top speed = 30 blocks/h.
=> " Mario quickly accelerates and reaches his top speed of 40 blocks/h in order to catch back up and pass Bowser Jr., but by this point he has opened up a 75 block head start down the track."
=> "Mario world measure distance using the units of blocks, with 1 block = 0.47 m"
Therefore, the solution is given below;
(1). For the first part, we are to determine or calculate Mario and Bowser Jr.'s speeds in m/s.
Therefore, Mario's speed in m/s = 40 × 0.47) ÷ 3600 = 5.2 × 10^-3 m/s.
Also, Bowser Jr.'s speeds in m/s = ( 30 × 0.47) ÷ 3600 = 3.92 × 10^3 m/s.
(2). So, the next thing to do now.is to determine or calculate how long it took for Mario to catch Bowser Jr.
Thus, the time it took for Mario to catch Bowser Jr. Can be related as below;
[ ( 5.2 × 10^-3 m/s) - (3.92 × 10^-3 m/s) × (time,t taken for Mario to catch Bowser Jr.) = 75 × 0.47.
Therefore, the time it took for Mario to catch Bowser Jr. = 27001.2 seconds.
(3). Now, we calculate How far down the track Mario from the point at which he reaches his top speed.
The distance = 5.2 × 10^-3 m/s × 27001.2m = 141m
A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?
Answers
Answer:
The answer is 45 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distanceFrom the question
distance = 3 meters
force = 15 newtons
We have
workdone = 15 × 3
We have the final answer as
45 JHope this helps you
An electron moving in the direction of the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the
Answers
Answer:
-z
Explanation:
The force on a moving charge due to a magnetic field follows the right hand rule, so a positive charge, experiencing a magnetic deflection in the -y direction, while it moves in the direction of the x-axis, will do it due to a magnetic field pointing in the +z direction.
As the electron has a negative charge, the magnetic field will point in the opposite direction, i.e., in the -z direction.
If a power utility were able to replace an existing 500 kV transmission line with one operating at 1 MV, it would change the amount of heat produced in the transmission line to
Answers
Answer:
It would change the amount of heat produced in the transmission line to four times the previous value.
Explanation:
Given;
initial voltage in the transmission line, V₁ = 500 kV = 500,000 V
Final voltage in the transmission line, V₂ = 1 MV = 1,000,000
The power lost in the transmission line due to heat is given by;
[tex]P = \frac{V^2}{R}[/tex]
Power lost in the first wire;
[tex]P_1 = \frac{V_1^2}{R}[/tex]
[tex]R = \frac{V_1^2}{P_1}[/tex]
Power lost in the second wire
[tex]P_2 = \frac{V_2^2}{R}\\\\ R = \frac{V_2^2}{P_2}[/tex]
Keeping the resistance constant, we will have the following equation;
[tex]\frac{V_2^2}{P_2} = \frac{V_1^2}{P_1} \\\\P_2 = \frac{V_2^2P_1}{V_1^2}\\\\[/tex]
[tex]P_2 = \frac{(1,000,000)^2P_1}{(500,000)^2}\\\\P_2 =4P_1[/tex]
Therefore, it would change the amount of heat produced in the transmission line to four times the previous value.
A soccer ball is kicked with a velocity of 8 m/s at an angle of 23°. What is the
ball's acceleration in the vertical direction as it flies through the air?
A. -7.4 m/s2
B. O m/s2
C. 3.1 m/s2
D. -9.8 m/s2
Answers
Answer: -9.8 m/s2
Explanation: