This question is incomplete, the complete question is;
The displacement of a string carrying a traveling sinusoidal wave is given by y(x,t)=ymsin(kx - ωt -φ) .
At time t = 0, the point at x = 0 has velocity v₀ and displacement y₀.
The phase constant φ is given by tanφ =:
A) ωv₀ /y₀
B) ωv₀ y₀
C) v₀ /ωy₀
D) y₀ /ωv₀
E) ωy₀ /v₀
Answer:
E) ωy₀ /v₀
Explanation:
Given that;
displacement of a wave is; y(x,t) = ym sin (kx - ωt - φ)
we differentiate the given equation with respect to time
d/dt (y(x,t)) = d/dt(ym sin(kx - ωt - φ) )
v(0,0)) = -ym ωcos (k(0) - ω(0) - φ) )
v₀ = -ym ωcos (-φ) ......... lets leave thisas equ 1
At t = 0, x = 0
the displacement of the wave is
y(0,0) = ym sin (k(0) - ω(0) - φ)
y₀ = ym sin(-φ) ..............let this be equ 2
y₀/v₀ = (ym sin(-φ)) / (-ym ωcos (-φ)) = ( -ym sin(φ)) / (-ym ωcos (φ))
(tanφ)/ω = y₀/v₀
tanφ = y₀ω/v₀
therefore the required value is y₀ω/v₀
option (E).
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
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
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
Answer: -9.8 m/s2
Explanation:
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
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.
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
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 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
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
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
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.
What is the answer to this ?
how far will a brick starting from rest fall freely in 3.0 seconds?
Answer: It will be about 44.1m
Explanation:
What kind of variation is there in the mechanical energy as the cart rolls down the ramp? Does this agree with your prediction? Explain.
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.
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?
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 pair of glasses are dropped from the top of a 32.0 m high stadium. A pen is dropped 2.00 s later. How high above the ground is the pen when the glasses hit the ground? (Disregard air resistance. a = -g = -9.81 m/s2.)
Answer:
Explanation:
the distance have the following relation:
d = (1/2)gt2
D=32.0 m
t =√ (2D/g) = √(2*32.0m/9.8m/s2) = 2.56s
it take 2.56s from the glasses to hit the ground
when the glasses hit the ground, the pen only travel Δt =2.56s - 2.00s = 0.56s
x = (1/2)g(Δt)2 = 0.5*9.8m/s2*(0.56s)2 = 1.54 m
the pen only travel 1.54m
so the pen is above the ground 32.0m - 1.54m = 30.46m
The pen is 30.46m above the ground. when the glasses hit the ground. It is represented by x.
What is the height?Height is a numerical representation of the distance between two objects or locations on the vertical axis.
The height can refer to a physical length or an estimate based on other factors in physics or common use. |
The given data in the problem is;
h is the height from the top of a stadium = 32.0 m
t is the time period when the pen is dropped later = 2.00 s
x is the height above the ground
a is the air resistance. a = -g = -9.81 m/s²
From the second equation of motion;
[tex]\rm H =ut+\frac{1}{2} gt^2 \\\\ \rm H =\frac{1}{2} gt^2 \\\\ \rm t = \sqrt{\frac{2H}{g} } \\\\ \rm t = \sqrt{\frac{2 \times32.0 }{9.81} } \\\\ \rm t =2.56\ sec[/tex]
When the glasses fall to the ground, the pen only travels a short distance;
[tex]\rm \triangle t = 2.56 -2.00 \\\\ \rm \triangle t = 0.56 \ sec[/tex]
So the pen travel the distance;
[tex]\rm h= \frac{1}{2} g \triangle t^2 \\\\ \rm h= \frac{1}{2} \times 9.81 (0.56)^2 \\\\ h=1.54 \ m[/tex]
The pen above the ground is found as;
[tex]\rm x = H-h \\\\ \rm x = 32.0-1.54 \\\\ \rm x =30.46 \ m[/tex]
Hence the pen is 30.46m above the ground. when the glasses hit the ground.
To learn more about the height refer to the link;
https://brainly.com/question/10726356
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? (68.15 m/s)
v² - u² = 2 a ∆x
where u = initial velocity, v = final velocity, a = acceleration, and ∆x = distance traveled.
So
v² - (15 m/s)² = 2 (6.5 m/s²) (340 m)
v² = 4645 m²/s²
v ≈ 68.15 m/s
The particles of a more dense substance are closer together
than the particles of a less dense substance.
TRUE
FALSE
The particles of a more dense substance are closer together than the particles of a less dense substance. Thus, the given statement is true.
What is density of particles?Density of the particles is the substance's mass per unit of volume. The symbol which is most often used for the density is ρ (rho), although the Latin letter D can also be used to denote density.
Density is the mass of a unit volume of a material substance or particle. The formula for density is d = M/V (mass per unit volume), where d is density, M is the mass of particle, and V is the volume. Density is commonly expressed in the units of grams per cubic centimeters.
The S.I. unit of density is made up of the mass of the particle which is kg and that of volume is meter cube. Hence, the S.I. unit of density is kg/m³.
Learn more about Density here:
https://brainly.com/question/29775886
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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.
A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?
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
Which of the filling is a fossil fuel?
A ray is incident at at 50 degrees angle on a plane mirror. What will be the deviation after reflection from the mirror?
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°.
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?
Answer:
Explanation:
Have y’all seen steeleflag19 at all on here?
Matching type. Send help please. ASAP!
Answer:
46-D
47-C
48-F
49-A
50-B
I am not very sure I am right about those answers though.
What type of force holds atoms together in a crystal?
Answer:
Covalent Bond
Explanation:
i took the test , mark me brainliest.
Answer: Electrical
Explanation: Atoms are tied together by electrical bonding forces.
first correct answer gets brainliest
Answer:
electrical energy transforming into sound energy in speaker
Answer:
the first one. Electrical energy transforming into sound energy in a speaker
what is the summary for Electrons and protons
Explanation:
the link enjoy
calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun
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
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?
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
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?
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
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
Answer:
5
Explanation:
Bextra in bf x vi d sj by
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
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.
correct me if im wrong
What are the significant transitions middle adulthood?
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:
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:
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