Answer:
The correct option is (b).
Explanation:
The relation between the wavelength and frequency is given by :
[tex]\lambda=\dfrac{v}{f}[/tex]
Where
v is the wave speed
f is the frequency of a wave
It is clear from the above equation that the wavelengths and frequency can vary inversely to produce the same wave speed.
A golf ball is dropped from rest from a height of 8.40 m. It hits the pavement, then bounces back up, rising just 5.60 m before falling back down again. A boy then catches the ball when it is 1.40 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.
Answer:
t1= 8.40/10 =.84 s
t2 = 5.60/10 = .56s
t3= 1.4/10 = .14s
total time = 1.54 sec
g you hang an object of mass m on a spring with spring constant k and find that it has a period of T. If you change the spring to one that has a spring constant of 2 k, the new period is
Answer:
a) T = 2π [tex]\sqrt{\frac{m}{k} }[/tex], b) T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]
Explanation:
a) A system formed by a mass and a spring has a simple harmonic motion with angular velocity
w² = k / m
angular velocity and period are related
w = 2π /T
we substitute
4π²/ T² = k / m
T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]
b) We change the spring for another with k ’= 2 k, let's find the period
T ’= 2π [tex]\sqrt{\frac{m}{k'} }[/tex]
T ’= 2π [tex]\sqrt{ \frac{m}{2k} }[/tex]
T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]
Determine the magnitude of the minimum acceleration at which the thief can descend using the rope. Express your answer to two significant figures and include the appropriate units.
Answer: hello your question is incomplete below is the missing part
A 69-kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg.
answer:
To 2 significant Figures = 1.6 m/s^2
Explanation:
Calculate the magnitude of minimum acceleration at which the thief can descend
we apply the relation below
Mg - T = Ma --- ( 1 )
M = 69kg
g = 9.81
T = 58 * 9.81
a = ? ( magnitude of minimum acceleration)
From equation 1
a = [ ( 69 * 9.81 ) - ( 58 * 9.81 ) ] / 69
= 1.5639 m/s^2
To 2 significant Figures = 1.6 m/s^2
ou charge a piece of aluminum foil (mass = 4.99 g) by touching it to a charged rod. The charged rod gives the aluminum foil a charge of 13 µC. Your professor brings a charged plate over and tells you to put the aluminum foil on top of the plate. To your surprise the aluminum foil hovers motionless in the air above it! Calculate the value of the electric field from the charged plate (assume it is a uniform field and the aluminum foil is a point charge).
Answer:
The appropriate answer is "3761.69 N/C".
Explanation:
Given that:
Mass,
m = 4.99 g
or,
= [tex]4.99\times 10^{-3} \ kg[/tex]
Charge,
q = 13 µC
or,
= [tex]13\times 10^{-6} \ C[/tex]
As we know,
⇒ [tex]F=mg=Eq[/tex]
then,
⇒ [tex]E=\frac{mg}{q}[/tex]
By putting the values, we get
[tex]=\frac{4.99\times 10^{-3}\times 9.8}{13\times 10^{-6}}[/tex]
[tex]=3761.69 \ N/C[/tex]
A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30.08 as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp.
Answer:
2.55 m/s
Explanation:
A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.
Solution:
The work done by friction is given as:
[tex]W_f=F_f\Delta S\\\\Where\ F_f\ is\ the \ frictional\ force=-5N(the\ negative \ sign\ because\ it\\acts\ opposite\ to \ direction\ of\ motion),\Delta S=slope\ length=1\ m\\\\W_f=F_f\Delta S=-5\ N*1\ m=-5J[/tex]
The work done by gravity is:
[tex]W_g=F_g*s*cos(\theta)\\\\F_g=force\ due\ to\ gravity=mass*acceleration\ due\ to\ gravity=3\ kg*9.81\\m/s^2, s=1\ m, \theta=angle\ between\ force\ and\ displacement=90-30=60^o\\\\W_g=3\ kg*9.81\ m/s^2*1\ m*cos(60)=14.72\ J\\\\The\ Kinetic\ energy(KE)=W_f+W_g=14.72\ J-5\ J=9.72\ J\\\\Also, KE=\frac{1}{2} mv^2\\\\9.72=\frac{1}{2} (3)v^2\\\\v=\sqrt{\frac{2*9.72}{3} } =2.55\ m/s[/tex]
Wind is caused by ___. the earth's tilt the Coriolis effect temperature differences humidity
I am guessing wind is caused by climate change in the atmosphere
Explanation:
wind is cause by climate change in the atmosphere that depends weather is is breezy really cold or rain and cold
Answer:
caused by the uneven heating of the Earth by the sun and the own rotation.
Increasing the surfactant concentration above the critical micellar concentration
will result in: Select one:
1.An increase in surface tension
2. A decrease in surface tension
3. No change in surface tension
4.None of the above
Answer:
Explanation:no change in surface tension
An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.
In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.
Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.
The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.
As the concentration of the surfactant increases before critical micellar concentration, the surface tension changes strongly with an increase in the concentration of the surfactant. After reaching the critical micellar concentration, any further increase in the concentration will result in no change of the surface tension, that is the surface tension will be constant.Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.
Learn more here: https://brainly.com/question/15785205
Your car breaks down in the middle of nowhere. A tow truck weighing 4000 lbs. comes along and agrees to tow your car, which weighs 2000 lbs., to the nearest town. The driver of the truck attaches his cable to your car at an angle of 20 degrees to horizontal. He tells you that his cable has a strength of 500 lbs. He plans to take 10 secs to tow your car at a constant acceleration from rest in a straight line along a flat road until he reaches the maximum speed of 45 m.p.h. Can the driver carry out the plan
Answer:
F = 1010 Lb
the tension on the cable is greater than its resistance, which is why the plan is not viable
Explanation:
For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.
v = v₀ + a t
how the car comes out of rest v₀ = 0
a = v / t
let's reduce to the english system
v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s
let's calculate
a = 66/10
a = 6.6 ft / s²
now let's write Newton's second law
X axis
Fₓ = ma
with trigonometry
cos 20 = Fₓ / F
Fₓ = F cos 20
we substitute
F cos 20 = m a
F = m a / cos20
W = mg
F = [tex]\frac{W}{g} \ \frac{a}{cos 20}[/tex]
let's calculate
F = [tex]\frac{2000}{32} \ \frac{6.6 }{cos20}[/tex](2000/32) 6.6 / cos 20
F = 1010 Lb
Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.
A 1 500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 12.0 m/s. What minimum coefficient of friction must exist between the road and tires to prevent the car from slipping
Explanation:
The centripetal force [tex]F_c[/tex] on the car must equal the frictional force f in order to avoid slipping off the road. Let's apply Newton's 2nd law to the y- and x-axes.
[tex]y:\:\:\:\:N - mg = 0[/tex]
[tex]x:\:\:F_c = f \Rightarrow \:\:\:m \dfrac{v^2}{r} = \mu N[/tex]
or
[tex]m \dfrac{v^2}{r} = \mu mg[/tex]
Solving for [tex]\mu[/tex],
[tex]\mu = \dfrac{v^2}{gr} = \dfrac{(12.0\:\frac{m}{s})^2}{(9.8\:\frac{m}{s^2})(52\:m)} = 0.28[/tex]
A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5500 Hz. When the siren is receding away from you, the frequency of the sound is 4500 Hz. Use the Doppler formula to determine the velocity of the police car. Use vsound=330 m/s.
What is the velocity v of the police car ?
When a police car in hot pursuit goes speeding past you, the velocity v of the police car is 33 m/s.
What is the Doppler formula?The formula is used when there exists a Doppler shift. The Doppler shift is due to the relative motion of sound waves between the source and observer.
The frequency increase by the Doppler effect is represented by the formula
f' = [tex]\dfrac{v-v_{o} }{v-v_{s} }[/tex]× f
Given the frequency of source f' is 5500 Hz . Velocity of the observer v₀ is 0.
Substituting the value into the equation will give us the velocity of the police car.
[tex]5500 = \dfrac{330}{330-v} \times f[/tex]...........(1)
When the car is receding, the frequency of the receiving signal f = 4500 Hz.
[tex]4500 = \dfrac{330}{330+v} \times f[/tex]..........(2)
Solving both equation, we get the velocity of a police car.
v = 33 m/s
Therefore, the velocity v of the police car is 33 m/s.
Learn more about Doppler equation.
https://brainly.com/question/15318474
#SPJ2
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5
Answer:
A. Power generated by meteor = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Workdone = 981000 J
Power required = 19620 Watts
Note: The question is incomplete. A similar complete question is given below:
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?
Explanation:
A. Power = workdone / time taken
Workdone = Kinetic energy of the meteor
Kinetic energy = mass × velocity² / 2
Mass of meteor = 1.5 g = 0.0015 kg;
Velocity of meteor = 50 km/s = 50000 m/s
Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J
Power generated = 1875000/2.1 = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Work done by elevator against gravity = mass × acceleration due to gravity × height
Work done = 1000 kg × 9.81 m/s² × 100 m
Workdone = 981000 J
Power required = workdone / time
Power = 981000 J / 50 s
Power required = 19620 Watts
Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.
help me with this question
Explanation:
Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:
x-axis:
[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]
[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]
y-axis:
[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]
Use Eqn 1 to solve for T,
[tex]T = m_1(g \sin \theta - a)[/tex]
Substitute this expression for T into Eqn 2,
[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]
Collecting all similar terms, we get
[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]
or
[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]
A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a forward velocity of The train then moves at a constant velocity of for 420 s. The train then slows down uniformly at until it is brought to a halt. The acceleration during the first 5.6 km of travel is closest to which of the following?
a. 0.19 m/s^2
b. 0.14 m/s^2
c. 0.16 m/s^2
d. 0.20 m/s^2
e. 0.17 m/s^2
Answer:
The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²
Option c) 0.16 m/s² is the correct answer.
Explanation:
Given the data in the question;
since the train starts from rest,
Initial velocity; u = 0 m/s
final velocity; v = 42 m/s
distance covered S = 5.6 km = ( 5.6 × 1000 )m = 5600 m
acceleration a = ?
From the third equation of motion;
v² = u² + 2as
we substitute in our values
( 42 )² = ( 0 )² + [ 2 × a × 5600 ]
1764 = 0 + [ 11200 × a ]
1764 = 11200 × a
a = 1764 / 11200
a = 0.1575 ≈ 0.16 m/s² { two decimal place }
Therefore, The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²
Option c) 0.16 m/s² is the correct answer.
the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during the impact with the bat, how many impules of importance are used to find the final velocity of the bat
Solution :
Given :
Mass of the baseball, m = 200 g
Velocity of the baseball, u = -30 m/s
Mass of the baseball after struck by the bat, M = 900 g
Velocity of the baseball after struck by the bat, v = 47 m/s
According to the conservation of momentum,
[tex]Mv+mu=Mv_1+mv_2[/tex]
(900 x 47) + (200 x -30) = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])
36300 = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])
[tex]9v_1 + 2v_2 = 363[/tex] ..............(i)
[tex]9v_1 = 363 - 2v_2[/tex]
[tex]v_1=\frac{363 - 2v_2}{9}[/tex]
The mathematical expression for the conservation of kinetic energy is
[tex]\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2[/tex]
[tex]\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2[/tex] ................(ii)
[tex]$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$[/tex]
[tex]21681 = 9v_1^2+2v_2^2[/tex]
Substituting (i) in (ii)
[tex]21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2[/tex]
[tex](363-2v_2)^2+18v_2^2=195129[/tex]
[tex](363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0[/tex]
[tex]22v_2^2-145v_2-63360=0[/tex]
Solving the equation, we get
[tex]v_2=96 \ m/s, -30 \ m/s[/tex]
The negative velocity is neglected.
Therefore, substituting 96 m/s for [tex]v_2[/tex] in (i), we get
[tex]v_1=\frac{363-(2 \times 96)}{9}[/tex]
= 19
Thus, only impulse of importance is used to find final velocity.
why are you teachers regarded as professionals
Answer:
coz teaching is their profession.
A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight but catches it as it is falling back to earth. If the friend catches the ball 3.0 s after it is thrown, at what time did it pass him on its upward flight
Answer:
[tex]t=1.9 sec[/tex]
Explanation:
From the question we are told that:
Height [tex]h=28m[/tex]
Time [tex]t=3s[/tex]
Generally the Newton's equation for Initial velocity upward is mathematically given by
[tex]s=ut+\frtac{1}{2}at^2[/tex]
[tex]28=3u-\frac{1}{2}*9.8*3^2[/tex]
[tex]u=24.03m/s[/tex]
Generally the velocity at elevation and depression occurs as ball arrives and passes through S=28
[tex]v=\sqrt{24.03-2*9.8*28}[/tex]
[tex]v=5.35m/s and -5.35m/s[/tex]
Generally the Newton's equation for time to reach initial velocity is mathematically given by
[tex]v=u+at[/tex]
[tex]5.35=24.03-9.8t[/tex]
[tex]t=\frac{28.03-5.35}{9.8}[/tex]
[tex]t=1.9 sec[/tex]
Who stated that man is an animal
A 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
Answer:
3.6 KJ
Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
The workdone = the energy.
There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )
P.E = mgh
P.E = 70 × 9.8 × 1.6
P.E = 1097.6 J
P.E = 1.098 KJ
K.E = 1/2mv^2
K.E = 1/2 × 70 × 8.5^2
K.E = 2528.75 J
K.E = 2.529 KJ
The non conservative workdone = K.E + P.E
Work done = 1.098 + 2.529
Work done = 3.63 KJ
Therefore, the non conservative workdone is 3.6 KJ approximately
1000 grams of water is heated from 0 degree to 200 degree . The specific heat of water is 4186 j/kg.°C. Estimate the change in entropy of the water.
Answer:2
Explanation:
turn this scentence to repirted speach.
i ate icecream
She said that..........
Answer:
dhfhffuththt9tr8tujtngigjtjrjrjrurur
A helicopter is ascending vertically witha speed of 5.40 m/s. At a height of 105 m above the earth a package is dropped from the helicopter. How much time does is take for the package to reach the ground
Answer: 5.21 s
Explanation:
Given
Helicopter ascends vertically with [tex]u=5.4\ m/s[/tex]
Height of helicopter [tex]h=105\ m[/tex]
When the package leaves the helicopter, it will have the same vertical velocity
Using equation of motion
[tex]\Rightarrow h=ut+\dfrac{1}{2}at^2\\\\\Rightarrow 105=-5.4t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-5.4t-105=0\\\\\Rightarrow t=\dfrac{5.4\pm \sqrt{5.4^2+4\times 4.9\times 105}}{2\times 4.9}\\\\\Rightarrow t=\dfrac{5.4\pm 45.68}{9.8}\\\\\Rightarrow t=5.21\ s\quad \text{Neglect negative value}[/tex]
So, package will take 5.21 s to reach the ground
An object with mass m = 0.56 kg is attached to a string of length r = 0.72 m and is rotating with an angular velocity ω = 1.155 rad/s. What is the centripetal force acting in the object?
Answer:
The centripetal force is 0.54 N.
Explanation:
mass, m = 0.56 kg
radius, r = 0.72 m
angular speed, w = 1.155 rad/s
The centripetal force is given by
[tex]F = m r w^2\\\\F =0.56\times 0.72\times 1.155\times 1.155\\\\F = 0.54 N[/tex]
Determine the tension in the string that connects M2 and M3.
☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️✌️✌️✌️✌️✌️✌️✌️✌️❤️❤️❤️
LC-circuit of the radio receiver consists of variable capacitor (Cmin= 1 pF, Cmax=10 pF) and inductor
with inductance 1 µH. Determine the wavelength range of this radio receiver.
Answer:
the radio can tune wavelengths between 1.88 and 5.97 m
Explanation:
The signal that can be received is the one that is in resonance as the impedance of the LC circuit.
X = X_c - X_L
X = 1 / wC - w L
at the point of resonance the two impedance are equal so their sum is zero
X_c = X_L
1 / wC = w L
w² = 1 / CL
w = [tex]\sqrt{\frac{1}{CL} }[/tex]
let's look for the extreme values
C = 1 10⁻¹² F
w = [tex]\sqrt{\frac{1}{ 1 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]
w = [tex]\sqrt{1 \ 10^{18}}[/tex]
w = 10⁹ rad / s
C = 10 10⁻¹² F
w = [tex]\sqrt{\frac{1}{10 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]Ra 1/10 10-12 1 10-6
w = [tex]\sqrt{0.1 \ 10^{18}}[/tex]Ra 0.1 1018
w = 0.316 10⁹ rad / s
Now the angular velocity and the frequency are related
w = 2π f
f = w / 2π
the light velocity is
c = λ f
λ = c / f
we substitute
λ = c 2π/w
we calculate the two values
C = 1 pF
λ₁ = 3 10⁸ 2π / 10⁹
λ₁= 18.849 10⁻¹ m
λ₁ = 1.88 m
C = 10 pF
λ₂ = 3 10⁸ 2π / 0.316 10⁹
λ₂ = 59.65 10⁻¹ m
λ₂ = 5.97 m
so the radio can tune wavelengths between 1.88 and 5.97 m
Page
E QON
1 What is force ? Write its unit and mention
any
three effects of the force.
Force is a push or a pull that changes or trends to change the state of rest or uniform motion of an object or changes the direction or shape of an object. It causes objects to accelerate. SI unit is Newton.
1) Can change the state of an object : For example, pushing a heavy stone in order to move it.
2) May change the speed of an object if it is already moving. For example, catching a ball hit by a batsman.
3) May change the direction of motion of an object.
Human vision cuts off on the red side of the spectrum at about 675 nm. What is the energy of a photon (in J) of this wavelength?
Answer:
The energy of a photon is 2.94x10⁻¹⁹ J.
Explanation:
The energy of the photon is given by:
[tex] E = \frac{hc}{\lambda} [/tex]
Where:
h: is Planck's constant = 6.62x10⁻³⁴ J.s
c: is the speed of light = 3.00x10⁸ m/s
λ: is the wavelength = 675 nm
Hence, the energy is:
[tex] E = \frac{hc}{\lambda} = \frac{6.62 \ccdot 10^{-34} J.s*3.00 \cdot 10^{8} m/s}{675 \cdot 10^{-9} m} = 2.94 \cdot 10^{-19} J [/tex]
Therefore, the energy of a photon is 2.94x10⁻¹⁹ J.
I hope it helps you!
A 0.0780 kg lemming runs off a
5.36 m high cliff at 4.84 m/s. What
is its kinetic energy (KE) when it
is 2.00 m above the ground?
Answer:
0.913
Explanation:
k.e=1/2mv square
k.e=1/2×0.078g×23.4256m/s square
k.e=0.913J
The kinetic energy when the lemming is 2.00 m above the ground is approximately 2.56 J (Joules).
To calculate the kinetic energy (KE) of the lemming when it is 2.00 m above the ground, we need to consider the change in its potential energy (PE) as it falls.
The potential energy at a height h is given by:
PE = m g h
Where:
m is the mass of the lemming (0.0780 kg)
g is the acceleration due to gravity (9.8 m/s²)
h is the height above the ground
Given:
Height of the cliff (h) = 5.36 m
Velocity of the lemming (v) = 4.84 m/s
Height above the ground (h') = 2.00 m
The lemming will lose potential energy as it falls from the cliff, which is converted into kinetic energy. Therefore, the kinetic energy when it is 2.00 m above the ground is equal to the difference between its total initial kinetic energy and the potential energy at that height.
Initial potential energy at the top of the cliff:
PE_initial = m g h
Potential energy when it is 2.00 m above the ground:
PE_final = m * g * h'
The change in potential energy is given by:
ΔPE = PE_final - PE_initial
The kinetic energy (KE) when it is 2.00 m above the ground:
KE = ΔPE = -ΔPE (due to energy conservation)
Let's calculate the potential energy at the top of the cliff and when it is 2.00 m above the ground:
PE_initial = m ×g × h
= 0.0780 kg × 9.8 m/s² × 5.36 m
PE_initial ≈ 4.09 J
PE_final = m ×g × h'
= 0.0780 kg ×9.8 m/s² ×2.00 m
PE_final ≈ 1.53 J
The change in potential energy (ΔPE) is:
ΔPE = PE_final - PE_initial = 1.53 J - 4.09 J
ΔPE ≈ -2.56 J
Since the change in potential energy is equal to the kinetic energy, the kinetic energy when the lemming is 2.00 m above the ground is approximately 2.56 J (Joules).
To know more about kinetic energy
https://brainly.com/question/999862
#SPJ2
An 8.50 kg point mass and a 14.5 kg point mass are held in place 50.0 cm apart. A particle of mass (m) is released from a point between the two masses 12.0 cm from the 8.50 kg mass along the line connecting the two fixed masses.Find the magnitude of the acceleration of the particle.
Answer:
[tex]a=2.8*10^{-9}m/s[/tex]
Explanation:
From the question we are told that:
First Mass [tex]m=8.50kg[/tex]
2nd Mass [tex]m=14.5kg[/tex]
Distance
[tex]d_1=50=>0.50m\\\\d_2=>12cm=>0.12m[/tex]
Generally the Newtons equation for Gravitational force is mathematically given by
[tex]F_n=\frac{Gm_nm}{(r_n)^2}[/tex]
Therefore
Initial force on m
[tex]F_1=\frac{Gm_1m}{(r_1)^2}[/tex]
Final force on m
[tex]F_2=\frac{Gm_2m}{(r_2)^2}\\\\F=\frac{Gm_1m}{(r_1)^2}-\frac{Gm_2m}{(r_2)^2}[/tex]
Acceleration of m
[tex]a=\frac{F}{m}\\\\a=\frac{Gm_1}{r_1^2}-\frac{Gm_2}{r_2^2}[/tex]
[tex]a=6,67*10^{-11}{\frac{8.5}{0.12}}-\frac{14.5}{0.50}[/tex]
[tex]a=2.8*10^{-9}m/s[/tex]
13. What type of lens bends light outwards and away from a point?
concave
Answer:
No,it isn't concave. The correct answer is convex lens.
Explanation:
A lens is a piece of transparent material bound by two surfaces of which at least one is curved. A lens bound by two spherical surfaces bulging outwards is called a bi-convex lens or simply a convex lens. A single piece of glass that curves outward and converges the light incident on it is also called a convex lens.
Convex lens is the answer.
See the attached diagram.
can anyone answer this fast pls
