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.
What does it mean when work is positive?
O Velocity is greater than kinetic energy.
O Kinetic energy is greater than velocity.
O The environment did work on an object.
O An object did work on the environment.
Answer:
O The environment did work on an object
Explanation:
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
Kulsum’s TV uses 45 W. How much does it cost her to watch TV for one month (30 days). She watches TV for 4 hours/day during mid-peak time (10.4 cents/kWh).
Answer:
Total cost = 56.16 cents
Explanation:
Given the following data;
Power = 45 Watts
Time = 4 hours
Number of days = 30 days
Cost = 10.4 cents
To find how much does it cost her to watch TV for one month;
First of all, we would determine the energy consumption of the TV;
Energy = power * time
Energy = 45 * 4
Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).
Energy consumption = 0.18 Kwh
Next, we find the total cost;
Total cost = energy * number of days * cost
Total cost = 0.18 * 30 * 10.4
Total cost = 56.16 cents
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]
A camera lens with focal length f = 50 mm and maximum aperture f>2
forms an image of an object 9.0 m away. (a) If the resolution is limited
by diffraction, what is the minimum distance between two points on the
object that are barely resolved? What is the corresponding distance
between image points? (b) How does the situation change if the lens is
“stopped down” to f>16? Use λ= 500 nm in both cases
Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
what change occurs to the mass of an object when a unbalanced
Answer:
The mass decreases
Explanation:
Just smart
Una pelota de basket es soltada desde 2.5 m de altura y rebota con una velocidad igual a 3/4 partes de la velocidad que llego. ¿ a qué altura alcanza la bola en el rebote ? ¿ cuánto tiempo transcurre desde que rebota ?
Answer:
Tenemos dos problemas a resolver acá:
Primero, debemos encontrar la velocidad con la que la pelota impacta el suelo.
Acá podemos usar la conservación de la energía.
E = U + K
U = energía potencial = m*g*H
m = masa
g = aceleración gravitatoria = 9.8m/s^2
H = altura
K = energía cinética = (m/2)*V^2
donde V es la velocidad.
Inicialmente, cuando la pelota es soltada, su velocidad es cero, entonces solo tenemos energía potencial:
Ei = U = m*(9.8m/s^2)*2.5m
Al final, cuando la pelota esta por impactar el suelo, la altura tiende a cero, entonces ya no hay energía potencial, solo hay energía cinética:
Ef = (m/2)*V^2
Y como la energía se conserva, la energía final es igual a la inicial, entonces:
m*(9.8m/s^2)*2.5m = (m/2)*V^2
Podemos resolver esto para V, y asi obtener la velocidad con la que la pelota impacta el suelo.
V = √(2*(9.8m/s^2)*2.5m) = 7m/s
Ahora respondamos la segunda parte.
Una vez la pelota rebota, su aceleración va a estar dada solamente por la aceleración gravitatoria, entonces tenemos:
A(t) = -9.8m/s^2
Para obtener su velocidad integramos:
V(t) = (-9.8m/s^2)*t + V0
donde V0 es la velocidad con la que la pelota reboto, que sabemos que es 3/4 de 7m/s
V0 = (3/4)*7m/s = (21/4) m/s
Así, la ecuación de la velocidad es:
V(t) = (-9.8m/s^2)*t + (21/4) m/s
Sabemos que la altura máxima se da cuando la velocidad es igual a cero, entonces primero calculemos el valor de t tal que esto ocurra:
V(t) = 0 = (-9.8m/s^2)*t + (21/4) m/s
t = (21/4) m/s/9.8m/s^2 = 0.54 s
Ahora debemos encontrar la ecuación de la posición y evaluarlo en este tiempo.
Para ello integramos de vuelta:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t + P0
donde P0 es la posición inicial, como la pelota rebota en el suelo, la posición inicial es el suelo, el cual representamos con 0, entonces la ecuación de la posición es:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t
La altura máxima estará dada por esta ecuación evaluada en t = 0.54 s
P(0.54s) = (1/2)(-9.8m/s^2)*(0.54s)^2 + (21/4 m/s)*0.54s = 1.81 m
La altura máxima es 1.81 metros.
Y entre que rebota y llega a esta altura máxima, transcurren 0.54 segundos.
Which describes farsightedness? O Distant objects are blurry. O Concave lenses can correct it. O Objects appear larger when wearing corrective glasses. O Corrective glasses do not change apparent the size of objects.
Answer:
O Distant objects are blurry. describes farsightedness.
Explanation:
Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.
Determine the tension in the string that connects M2 and M3.
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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
Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if the system it is rotated about (a) one end of the rod and (b) the center of the rod?
Explanation:
a) [tex]I=\displaystyle \sum_{i}m_ir_i^2[/tex]
where [tex]r_i[/tex] is the distance of the mass [tex]m_i[/tex] from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,
[tex]I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2[/tex]
b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,
[tex]\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2[/tex]
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]
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]
Ahmed is pushing a 4 Kg box to the right and Rashid is Pushing it to the right as well with a force of 12 N , the box accelerates by 5 m/s^2. What is the Force that is applied by Ahmed
Answer:
8 N
Explanation:
Applying,
(F'+F) = ma............... Equation 1
Where F' = Amhed's force, F = Rashid's force, m = mass of the box, a = acceleration of the box.
From the question,
Given: F = 12 N, m = 4 kg, a = 5 m/s²
Substitute these values into equation 1
(F'+12) = 4×5
(F'+12) = 20
F' = 20-12
F' = 8 N.
Hence Ahmed's force is 8 N
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]
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. We assume the upward direction to be positive, and the downward direction to be negative.
(a) How long are her feet in the air?(b) What is her highest point above the board?(c) What is her velocity when her feet hit the water?
Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
[tex]2gh = v_f^2 - v_i^2[/tex]
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
[tex](2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\[/tex]
h = 0.82 m
Now, for the time in air during upward motion we use first equation of motion:
[tex]v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s[/tex]
(c)
Now we will consider the downward motion and use the third equation of motion:
[tex]2gh = v_f^2-v_i^2[/tex]
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
[tex]2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\[/tex]
vf = 7.17 m/s
Now, for the time in air during downward motion we use the first equation of motion:
[tex]v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s[/tex]
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
t = 1.14 s
Select the correct answer.
What are the directions of an object's velocity and acceleration vectors when the object moves in a circular path with a constant speed?
OA. The question is meanimgless, since the acceleration is zero.
ов.
The vectors point in opposite directions.
Oc.
Both vectors point in the same direction.
OD
The vectors are perpendicular,
Answer:
A
Explanation:
If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.
Answer: C.
Explanation: plato users
plz answer the question
Answer:
Ray A - incident ray
Ray B - reflected ray
12. What type of circuit is the diagram below?
series circuit
parallel circuit
Answer:
parallel circuit
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
Basically, the components of an electric circuit can be connected or arranged in two forms and these includes;
I. Series circuit
II. Parallel circuit: it's an electrical circuit that has the same potential difference (voltage) across its terminals or ends. Thus, its components are connected within the same common points so that only a portion of current flows through each branch.
Hence, the type of circuit that the above diagram above represents is a parallel circuit.
Answer:
parallel circuit
Explanation:
I got it right on my exam
Who stated that man is an animal
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.
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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 pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will Group of answer choices be behind the package. be over the package. be in front of the package depend of the speed of the plane when the package was released.
Answer:
The location of helicopter is behind the packet.
Explanation:
As the packet also have same horizontal velocity as same as the helicopter, and also it has some vertical velocity as it hits the ground.
The horizontal velocity remains same as there is no force in the horizontal direction. The vertical velocity goes on increasing as acceleration due to gravity acts.
So, the helicopter is behind the packet.
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.
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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.
A 5.0-kg mass is placed at (3.0, 4.0) m, and a 6.0-kg mass is placed at (3.0, -4.0) m. What is the moment of inertia of this system of masses about the y-axis?
Answer:
the moment of inertia of this system of masses about the y-axis is 99 kgm²
Explanation:
Given the data in the question;
mass m₁ = 5.0 kg at point ( 3.0, 4.0 )
mass m₂ = 6.0 kg at point ( 3.0, -4.0 )
Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;
Moment of inertia [tex]I[/tex]ₓ = mixi²
Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²
we substitute
Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ] + [ 6.0 × ( 3 )² ]
Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ] + [ 6.0 × 9 ]
Moment of inertia [tex]I[/tex] = 45 + 54
Moment of inertia [tex]I[/tex] = 99 kgm²
Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²
who is the biggest man in the world
Answer:
Sultan Kösen
here is a pic
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]
why are you teachers regarded as professionals
Answer:
coz teaching is their profession.
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.
