Answer:
the correct answer is A :)
Two objects (48.0 and 25.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find (a) the acceleration of the objects and (b) the tension in the string.
Answer:
a. 3.09 m/s^2
b. 322.52 N
Explanation:
The computation of the acceleration of the object and the tension of the string is as follows:
The acceleration of the system is a
= (m1 - m2) × g ÷ (m1 + m2)
= (48 - 25) × 9.81 ÷ (48 + 25)
= 3.09 m/s^2
b. The tension in the string is T
= 2 × m1 × m2 × g÷ (m1+m2)
= (2 × 48 × 25 × 9.81) ÷ (48 + 25)
= 322.52 N
In the following circuit (Fig.3), calculate the intensity I through the resistance 3 using the principle of superposition.
Answer:
time
Explanation:
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]
voltage needed to raise current to 3.75a using 20,20,200 resistor set
Answer: The voltage needed is 35.7 V
Explanation:
Assuming that the resistors are arranged in parallel combination.
For the resistors arranged in parallel combination:
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}[/tex]
We are given:
[tex]R_1=20\Omega\\R_2=20\Omega\\R_3=200\Omega[/tex]
Using above equation, we get:
[tex]\frac{1}{R}=\frac{1}{20}+\frac{1}{20}+\frac{1}{200}\\\\\frac{1}{R}=\frac{10+10+1}{200}\\\\R=\frac{200}{21}=9.52\Omega[/tex]
Calculating the voltage by using Ohm's law:
[tex]V=IR[/tex] .....(1)
where,
V = voltage applied
I = Current = 3.75 A
R = Resistance = [tex]9.52\Omega[/tex]
Putting values in equation 1, we get:
[tex]V=3.75\times 9.52\\\\V=35.7V[/tex]
Hence, the voltage needed is 35.7 V
There are six different ways to arrange those resistors.
Each way leads to a different answer.
You didn't tell us how you want them connected.
The attached drawing shows them all.
Por que nos dias de hoje o futebol é mais técnico, com mas estratégias e mais disputado?
Explanation:
[tex]which \: language \: is \: this[/tex]
[tex]pls \: write \: in \: english[/tex]
[tex]then \: only \: i \: can \: answer \: u[/tex]
[tex]otherwise \: sry[/tex]
what change occurs to the mass of an object when a unbalanced
Answer:
The mass decreases
Explanation:
Just smart
A wave has angular velocity 12 rad/sec and maximum displacement X cm (X= last two digit [1] of your student’s ID). Calculate the maximum acceleration of the wave.
Answer:
[tex]a=1440\ m/s^2[/tex]
Explanation:
Given that,
The angular velocity of a wave, [tex]\omega=12\ rad/s[/tex]
The maximum displacement of the wave, A = 10 cm (let)
The maximum acceleration of the wave is given by :
[tex]a=-A\omega^2[/tex]
Put all the values,
[tex]a=10\times 12^2\\\\a=1440\ m/s^2[/tex]
So, the maximum acceleration of the wave is equal to [tex]1440\ m/s^2[/tex].
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]
The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?
Answer:
62.8 μC
Explanation:
Here is the complete question
The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?
Solution
The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ
So, Q = ∫∫∫ρdV
Q = ∫∫∫ρr²sinθdθdrdΦ
Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ
Q = ∫∫∫0.2r⁴sinθdθdrdΦ
We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π
So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ
Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ
Q = ∫∫0.2r⁴[-cosθ]drdΦ
Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ
Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ
Q = ∫∫0.2r⁴-[- 2]drdΦ
Q = ∫∫0.2r⁴(2)drdΦ
Q = ∫∫0.4r⁴drdΦ
Q = ∫0.4r⁴dr∫dΦ
Q = ∫0.4r⁴dr[Φ]
Q = ∫0.4r⁴dr[2π - 0]
Q = ∫0.4r⁴dr[2π]
Q = ∫0.8πr⁴dr
Q = 0.8π∫r⁴dr
Q = 0.8π[r⁵/5]
Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]
Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]
Q = 0.8π[0.025 m⁵ - 0 m⁵]
Q = 0.8π[0.025 m⁵]
Q = (0.02π mC/m⁵) m⁵
Q = 0.0628 mC
Q = 0.0628 × 10⁻³ C
Q = 62.8 × 10⁻³ × 10⁻³ C
Q = 62.8 × 10⁻⁶ C
Q = 62.8 μC
A parallel-plate capacitor is connected to a battery of electric potential difference V. If the plate separation is decreased, do the following quantities increase, decrease, or remain the same: (a) the capacitor’s capacitance, (b) the potential difference across the capacitor, (c) the charge on the capacitor, (d) the energy stored by the capacitor, (e) the magnitude of the electric field between the plates, and (f ) the energy density of that electric field?
Answer:
a) increases.
b) remains the same.
c) increases.
d) increases.
e) increases.
f) increases.
Explanation:
a)
Since the capacitance of a parallel-plate depends only on geometric constants and the dielectric between the plates, we can use the following expression to asess the value of the capacitance:[tex]C = \epsilon_{0}*\frac{A}{d} (1)[/tex]
where ε₀ = permitivitty of free space
A = area of one of the plates
d= plate separation
As we can see, if the plate separation is decreased, the value of the capacitance must increase.b)
Per definition the capacitance explains the relationship between the charge on one of the conductors, and the potential difference between them, as follows:[tex]C = \frac{Q}{V} (2)[/tex]
Assuming that the capacitor remains connected to the battery when the plate separation is decreased, since the voltage can't change (as it must hold the same voltage than previously since it's directly connected to the battery) the potential difference between plates must remain the same.c)
From B, we know that V in (2) must remain constant. Since we know from (1) that C must increase, this means from (2) that Q must increase too.d)
The energy stored in the electric field between the plates can be expressed as follows in terms of the capacitance C and the potential difference V:[tex]U = \frac{1}{2}* C*V^{2} (3)[/tex]
From (1) in a) and from b) we know that the capacitance C must increase whilst V must remain the same, so U in (3) must increase also.e)
In the capacitor the magnitude of the Electric field between the plates is constant, and is related to the potential difference between them by the following linear relationship:[tex]V = E*d (4)[/tex]
Since we know that V must remain the same, if the distance d decreases, the electris field E must increase in the same ratio in order to keep the equation balanced.f)
The energy density of the electric field is defined as the energy stored between plates by unit volume, as follows:[tex]u = \frac{U}{v} = \frac{\frac{1}{2}* C*V^{2}}{A*d} =\frac{1}{2}* \epsilon_{0}*\frac{A*V^{2} }{A*d*d} = \frac{1}{2} *\epsilon_{0}*E^{2} (5)[/tex]
Since it's proportional to the square of the electric field, and we know from e) that the magnitude of the electric field must increase, u must increase too.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.
A physicist wants to estimate the rate of emissions of alpha particles from a certain source. He counts 400 emissions in 80 seconds. Estimate the rate, in emissions per second and find the uncertainty in this estimate.
Answer:
Emissions per second = 0.36
Explanation:
Please find the attached question
Solution
Given
Let X be the rate of background emission.
X = B/t
Where B = 36
And t = 100
X = 36/100 = 0.36
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:
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]
Q2/Deceleration of a particle is based on relation a=-3 v² m/s² where v in m/s. If it moves along a straight line and has velocity 10 m/s and position s = 8m when t=0, determine its velocity and position when t= 3 s. Where the velocity become zero. Discuss briefly.
Explanation:
Given: a = -3v^2
By definition, the acceleration is the time derivative of velocity v:
[tex]a = \frac{dv}{dt} = - 3 {v}^{2} [/tex]
Re-arranging the expression above, we get
[tex] \frac{dv}{ {v}^{2} } = - 3dt[/tex]
Integrating this expression, we get
[tex] \int \frac{dv}{ {v}^{2} } = \int {v}^{ - 2}dv = - 3\int dt[/tex]
[tex] - \frac{1}{v} = - 3t + k[/tex]
Since v = 10 when t = 0, that gives us k = -1/10. The expression for v can then be written as
[tex] - \frac{1}{v} = - 3t - \frac{1}{10} = - ( \frac{30 + 1}{10} )[/tex]
or
[tex]v = \frac{10}{30t +1 } [/tex]
We also know that
[tex]v = \frac{ds}{dt} [/tex]
or
[tex]ds = vdt = \frac{10 \: dt}{30t + 1} [/tex]
We can integrate this to get s:
[tex]s = \int v \: dt = \int ( \frac{10}{30t + 1}) \: dt = 10 \int \frac{dt}{30t + 1} [/tex]
Let u = 30t +1
du = 30dt
so
[tex] \int \frac{dt}{30t + 1} = \frac{1}{30} \int \frac{du}{u} = \frac{1}{30}\ln |u| + k[/tex]
[tex]= \frac{1}{30}\ln |30t + 1| + k[/tex]
So we can now write s as
[tex]s = \frac{1}{3}\ln |30t + 1| + k[/tex]
We know that when t = 0, s = 8 m, therefore k = 8 m.
[tex]s = \frac{1}{3}\ln |30t + 1| + 8[/tex]
Next, we need to find the position and velocity at t = 3 s. At t = 3 s,
[tex]v = \frac{10}{30(3) +1 } = \frac{10}{91}\frac{m}{s} = 0.11 \: \frac{m}{s} [/tex]
[tex]s = \frac{1}{3}\ln |30(3) + 1| + 8 = 9.5 \: m[/tex]
Note: velocity approaches zero as t --> [tex]\infty [/tex]
A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed (12 m/s), the total force on the driver has a magnitude of 146 N. What is the total vector force (in N) on the driver if the speed is 18 m/s instead?
Answer:
a1 = v1^2 / R
a2 = v2^2 / R
a2 = (v2 / v1)^2 = (3 / 2)^2 = 9/4
F2 = 9/4 * F1 = 9/4 * 146 = N 329 N since F = m * a
can anyone answer this fast pls
(a) Suppose that a NASCAR race car is moving to the right with a constant velocity of +96 m/s. What is the average acceleration of the car? (b) Twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed. What is the average acceleration of the car?
what is microeconomics
Answer:
Microeconomics is a part of economics and the study of decisions made by people and businesses regarding the allocation of resources, and prices at which they trade goods and services.
Microeconomics helps business planning i.e. helps the business community to plan their costs, production, etc. in anticipation of demand in order to maximize profits. Microeconomics is useful in explaining how market mechanism determines the price in a free market economy.
The wheels on a skateboard have a diameter of 2.93 inches. If a skateboarder is traveling downhill at 23 miles per hour, determine the angular velocity of the wheels in radians per second.
Answer:
The angular velocity of the wheels is 276.314 radians per second.
Explanation:
The wheels are rolling, which is a combination of rotation and translation, whose center of rotation is the point of contact between the wheel and the ground and the geometrical center of the skateboard experiments pure translation. Then, the angular velocity can be found by using the following kinematic expression:
[tex]\omega = \frac{v}{R}[/tex] (1)
Where:
[tex]\omega[/tex] - Angular velocity, in radians per second.
[tex]v[/tex] - Velocity of the wheel at its center, in inches per second.
[tex]R[/tex] - Radius of the wheel, in inches.
If we know that [tex]v = 404.8\,\frac{in}{s}[/tex] and [tex]R = 1.465\,in[/tex], then the angular velocity of the wheels are:
[tex]\omega = \frac{404.8\,\frac{in}{s} }{1.465\,in}[/tex]
[tex]\omega = 276.314\,\frac{rad}{s}[/tex]
The angular velocity of the wheels is 276.314 radians per second.
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.
turn this scentence to repirted speach.
i ate icecream
She said that..........
Answer:
dhfhffuththt9tr8tujtngigjtjrjrjrurur
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!
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 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
20. Using the picture, how many neutrons does lithium have?
Answer:
No. of Neutrons = 3
Explanation:
The atomic number of Lithium is given as 3 in the symbol while the mass number is given as 5.941 which is approximately equal to 6.
Mass Number = No. of Protons + No. of Neutrons = 6
Atomic Number = Number of Electrons = No. of Protons = 3
Therefore,
Mass Number - Atomic Number = (No. of Protons + No. of Neutrons) - No. of Protons
Mass Number - Atomic Number = No. of Neutrons
No. of Neutrons = 6 - 3
No. of Neutrons = 3
Answer:
3
Explanation:
Honestly, it is the only one in the picture and as an answer.
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]
Water at 200 C has a bulk modulus of 2.2109 Pa, and the speed of sound in water at this temperature is 1480m/s. For 1000Hz sound waves in water at 200 C, what displacement amplitude is produced if the pressure amplitude is 310-2 Pa?
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
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
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