Answer:
Explanation:
Let v₁ and v₂ be velocities at lowest and topmost position . Let r be the radius of the circle .
Let N₁ and N₂ be the normal reaction force .
At the top position
centripetal force = N₂ + mg ; so
N₂ + mg = m v₂² / r
At the bottom position
centripetal force = N₁ - mg ; so
N₁ - mg = m v₁² / r
subtracting these two equations
N₁ - mg - N₂ - mg = m v₁² / r - m v₂² / r
N₁ - N₂ - 2mg = 1/r (m v₁² - m v₂² )
N₁ - N₂ - 2mg = 1/r x mg x 2r ( loss of potential energy = gain of kinetic energy )
N₁ - N₂ = 2mg + 2mg
= 4 mg .
While watching a movie a spaceship explodes and there is a loud bang and flash of light. What is wrong with this scene? Explain how you know using evidence and scientific reasoning from the lesson.
Doubling the amplitude of a sound wave will ?
A) Double the speed of the sound and cause it to have a higher pitch
B) Havel the speed of sound and cause it to have a lower pitch
C) Leave the speed of sound unchanged, but cause the sound to be louder
D) Leave the speed of sound unchanged and cause the sound to have a lower pitch
Answer:
(C) is correct option.
Explanation:
The loudness of sound wave determines its amplitude. If the sound is loud, it will have more amplitude while if the sound is soft, it will have less amplitude.
On doubling the amplitude of sound wave, it will become more loud.
Doubling the amplitude of a sound wave will leave the speed of sound unchanged, but cause the sound to be louder. Hence, the correct option is (C).
This is a measure of quantity of matter
Answer:
Mass
Explanation:
Mass is the measure of amount of matter contained within any substance and hence mass determines the weight. Unit of mass is kilogram as per ISI system of units.
Mass is measured through a balance. The more is the mass of an object, the more the balance tilts towards the object side.
Weight is equal to product of mass and the gravitational constant i.e 9.8m/s^2
The space probe Deep Space 1 was launched on October 24th, 1998 and it used a type of engine called an ion propulsion drive. An ion propulsion drive generates only a weak force (or thrust), but can do so for long periods of time using only a small amount of fuel. Suppose the probe, which has a mass of 474 kg is travelling at an initial speed of 275 m/s. No forces act on it except the 5.60 x 10⁻² N thrust from the engine. This external force is directed PARALLEL to the displacement. The displacement has a magnitude of 2.42 x 10⁹ m. {PART A} Calculate the INITIAL kinetic energy of the probe [2 marks] {PART B} Find the work done BY THE ENGINE on the space probe [2 marks] {PART C} Calculate the FINAL KINETIC ENERGY of the probe (Hint W=∆E) [2 marks] {PART D} Determine the final speed of the probe, assuming that its mass remains constant [3 marks]
Answer:
Explanation:
mass of probe m = 474 Kg
initial speed u = 275 m /s
force acting on it F = 5.6 x 10⁻² N
displacement s = 2.42 x 10⁹ m
A )
initial kinetic energy = 1/2 m u² , m is mass of probe.
= .5 x 474 x 275²
= 17923125 J
B )
work done by engine
= force x displacement
= 5.6 x 10⁻² x 2.42 x 10⁹
= 13.55 x 10⁷ J
C ) Final kinetic energy
= Initial K E + work done by force on it
= 17923125 +13.55 x 10⁷
= 1.79 x 10⁷ + 13.55 x 10⁷
= 15.34 x 10⁷ J
D ) If v be its velocity
1/2 m v² = 15.34 x 10⁷
1/2 x 474 x v² = 15.34 x 10⁷
v² = 64.72 x 10⁴
v = 8.04 x 10² m /s
= 804 m /s
why can you see the path of light in a sunbeam?
Answer:
Sunbeams are seen because of light separated from water droplets and dust and smoke particles suspended in the air. If the cloud cover only has a few small holes in it, then separate rays of light will sprinkle light in every direction so you can see sunbeams.
If you apply a force of 130 N to the lever, how much force is applied to lift the
crate?
Answer:171 N
Explanation:
Answer:
171 N.
Explanation:
The universal law of gravitation states that the force of attraction between two objects depends on which quantities?
the masses of the objects and their densities
the distance between the objects and their shapes
the densities of the objects and their shapes
the masses of the objects and the distance between them
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Subm
Kandretum
Answer:depends on the masses of the objects and the distance between them
Explanation:
According to Newton's law of universal gravitation,the force of attraction between two objects depends on the masses of the objects and the distance between them
A person drives north 6 blocks, then turns west, and drives 6 blocks. The driver then turns south and drives 6 blocks. How could the driver have made the distance shorter while maintaining the same displacement?
Answer:
Considering that there is no obstructions, he could go west from the start.
Explanation:
Two wires, both with current out of the page, are next to one another. The wire on the left has a current of 1 A and the wire on the right has a current of 2 A. We can say that:
A. The left wire attracts the right wire and exerts twice the force as the right wire does.
B. The left wire attracts the right wire and exerts half the force as the right wire does.
C. The left wire attracts the right wire and exerts as much force as the right wire does.
D. The left wire repels the right wire and exerts twice the force as the right wire does.
E. The left wire repels the right wire and exerts half the force as the right wire does.
F. The left wire repels the right wire and exerts as much force as the right wire does.
Answer:
C. The left wire attracts the right wire and exerts as much force as the right wire does.
Explanation:
To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:
[tex]B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\[/tex]
Next, you use the formula for the magnetic force produced by the wires:
[tex]\vec{F_B}=I\vec{L}\ X \vec{B}[/tex]
if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:
[tex]\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}[/tex]
Hence, due to this result you have that:
C. The left wire attracts the right wire and exerts as much force as the right wire does.
Three packing crates of masses, M1 = 6 kg, M2 = 2 kg
and M3 = 8 kg are connected by a light string of
negligible mass that passes over the pulley as shown.
Masses M1 and M3 lies on a 30o
incline plane which
slides down the plane. The coefficient of kinetic friction
on the incline plane is 0.28.
Determine the acceleration of the system.
Answer:
a = 2.5 m / s²
Explanation:
This is an exercise of Newton's second law, in this case we fix a coordinate system with the x axis parallel to the plane with positive direction
Let's write the second law for bodies in the inclined plane
W₁ₓ + W₃ₓ - fr = (m₁ + m₃) a
N₁ - [tex]W_{1y}[/tex] + N₃- W_{3y} = 0
N₁ + N₃ = W_{1y} + W_{3y}
let's use trigonometry to find the weight components
sin 30 = Wₓ/ W
Wₓ = W sin 30
cos 30 = W_{y} / W
W_{y} = W cos 30
we substitute
N₁+ N₃ = W₁ cos 30 + W₃ cos 30
W₁ₓ + W₃ₓ - μ (m₁ + m₃) g cos30 = (m₁ + m₃) a
a = (m₁g sin 30 + m₃g sin 30 - μ (m₁ + m₃) g cos 30) / (m₁ + m₃)
a = g sin 30 - μ g cos30
let's calculate
a = 9.8 sin 30 - 0.28 9.8 cos 30
a = 4.9 - 2,376
a = 2.5 m / s²
Problem 2: A flat belt is used with a driving pulley (diameter 4 inches) and a driven pulley (diameter 18 inches) in a open configuration. The center distance between the pulleys is 48 inches. The friction coefficient between the belt and pulley is 0.6. Determine the following: a) If the belt is initially tensioned to 50 lbs, what is the force in the belt on the tight side just before slippage (neglect the centrifugal force of the belt). b) Find the maximum torque required at the driving pulley.
Answer:
Explanation:
The two pictures attached shows the solution to the problem
A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30 m/s. if the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10 m/s, how fast would the stone be traveling just before it hits the ground
Answer:
The velocity just before hitting the ground is [tex]v_f = 30 m/s[/tex]
Explanation:
From the question we are told that
The initial speed is [tex]u = 10 m/s[/tex]
The final speed is [tex]v = 30 \ m/s[/tex]
From the equations of motion we have that
[tex]v^2 =u^2 + 2as[/tex]
Where s is the distance travelled which is the height of the cliff
So making it the subject of the the formula we have that
[tex]s = \frac{v^2 - u^2 }{2a}[/tex]
Where a is the acceleration due to gravity with a value [tex]a = 9.8m/s^2[/tex]
So
[tex]s = \frac{30^2 - 10^2 }{2 * 9.8 }[/tex]
[tex]s = 40.8 \ m[/tex]
Now we are told that was through horizontally with a speed of
[tex]v_x =10 m/s[/tex]
Which implies that this would be its velocity horizontally through out the motion
Now it final velocity vertically can be mathematically evaluated as
[tex]v_y = \sqrt{2as}[/tex]
Substituting values
[tex]v_y = \sqrt{(2 * 9.8 * 40.8)}[/tex]
[tex]v_y = 28.3 \ m/s[/tex]
The resultant final velocity is mathematically evaluated as
[tex]v_f = \sqrt{v_x^2 + v_y^2}[/tex]
Substituting values
[tex]v_f = \sqrt{10^2 + 28.3^2}[/tex]
[tex]v_f = 30 m/s[/tex]
Potassium is a crucial element for the healthy operation of the human
body. Potassium occurs naturally in our environment (and thus our
bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-
41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical
human body contains about 3.0 grams of Potassium per kilogram of body
mass.
1. How much Potassium-40 is present in a person with a mass of 80
kg?
2. If, on average, the decay of Potassium-40 results in 1.10 MeV of
energy absorbed, determine the effective dose (in Sieverts) per year
due to Potassium-40 in an 80-kg body. Assume an RBE of 1.2. The
half-life of Potassium-40 is 1.28 x 10° years.
A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2300 kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.75, and the cars slide to a halt at a point 5.54 m west and 6.19 m south of the impact point. How fast was sedan traveling just before the collision? How fast was SUV traveling just before the collision?
Answer:
Explanation:
momentum of sedan of 1600 kg = 1600x v , where v is its velocity
momentum of suv of 2300 kg = 2300 x u where u is its velocity .
force of friction = ( 1600 + 2300 ) x 9.8 x .75 ( fiction = μ mg )
= 28665 N
distance by which friction acted = √ (5.54² + 6.19²)
= 8.3 m
work done by friction
= 28665 x 8.3
= 237919.5 J
Total kinetic energy of cars = work done by friction
1/2 x 1600 x v² + 1/2 x 2300 u² = 237919.5
16 v² + 23 u² = 4758.4
1600 x v / 2300 u = 6.19 / 5.54
v / u = 1.6
v = 1.6 u
putting this equation in fist equation
40.96 u² + 23 u² = 4758.4
= 63.96 u² = 4758.4
u² = 74.4
u = 8.62 m /s
v = 13.8 m /s
Which of the following statements are characteristics of magnetic fields? Select all that apply.
Magnetic fields point from the north pole to the south pole of a magnet.
The earth's magnetic field has no effect on the electron rays coming from the sun.
An example of the Biot-Savart law is the effect of the earth's maghytic field on the electron rays coming from the sun.
The north pole of a magnet will be attracted to the south pole of the earth.
If a bar magnet is cut in half two magnets with like poles will be created.
Answer:
Magnetic fields point from the north pole to the south pole of a magnet.
An example of the Biot-Savart law is the effect of the earth's maghytic field on the electron rays coming from the sun.
The north pole of a magnet will be attracted to the south pole of the earth.
If a bar magnet is cut in half two magnets with like poles will be created
Explanation:
The magnetic field of Earth is due to the presence of iron in the core of the Earth.
The metal emits the magnetic waves from it and the North and South pole of the planet.
Both the poles emit the magnetic rays which create magnetic sheet around it. The Earth acts like a magnet bar if which is cut into two half, the planet will act like two magnets. Also, Biot Savarts's law states that the magnetic field does not affect the electron rays coming from the Sun.
Thus, the selected options are correct.
Answer:
ACDE
Explanation:
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram-meter^3 and the density of silicon in other units: 2.33 gram-centimeter^3. You decide to convert the density of silicon into units of kilogram-meter^3 to perform the comparison.
By which combination of conversion factors will you multiply 2.33 gram-centimeter^3 to perform the unit conversion?
Answer:
Explanation:
To convert gram / centimeter³ to kg / m³
gram / centimeter³
= 10⁻³ kg / centimeter³
= 10⁻³ / (10⁻²)³ kg / m³
= 10⁻³ / 10⁻⁶ kg / m³
= 10⁻³⁺⁶ kg / m³
= 10³ kg / m³
So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³
2.33 gram / cm³
= 2.33 x 10³ kg / m³ .
You have two flashlights that operate on 1.50-V D batteries. The first flashlight uses two batteries in a chain, and the second uses five batteries in a chain. Each flashlight has a current of 2.50 A flowing through its circuit. What power is being transferred to the bulb in each flashlight
Answer:
The first flashlight,
P = 7.5 W
The second flashlight,
P = 18.75 W
Explanation:
P = VI.................. Equation 1
Where P = Power, V = Voltage, I = current.
For the first flashlight,
Given: I = 2.5 A, and V = 1.5×2 (The first uses two battery in chain) = 3 V
Substitute into equation 1
P = 3(2.5)
P = 7.5 W.
For the second flash light,
P = VI
Given: I = 2.5 A, V = 1.5×5 ( The second uses five batteries in a chain) = 7.5 V
Substitute into equation 1
P = 2.5(7.5)
P = 18.75 W.
Arrange the steps in order to describe what happens to a gas when it cools.
Answer:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
Explanation:
A bowler releases a 7.0kg bowling ball from rest to a final velocity of 8.0m/s. What is the magnitude of the change in momentum of the bowling ball?
Answer:
56 kg m/s
Explanation:
Δp = mΔv
Δp = (7.0 kg) (8.0 m/s − 0 m/s)
Δp = 56 kg m/s
What happens to the brightness of the lightbulb when its resistance is increased? *
Answer:
It will be dimmer than before
Answer:For parallel connection,the brightness would be dimmer, while for series connection it would be brighter
Explanation:
For parallel connection,resistance and brightness are inversely proportional.meaning as resistance increases, brightness decreases.
For series connection,resistance and brightness are directly proportional. Meaning as the resistance increases, brightness also increases.
Match these items.
1 . pls help
asteroids
between Mars and Jupiter
2 .
fission
ice, dust, frozen gases
3 .
energy
sun's atmosphere
4 .
fusion
ability to do work
5 .
corona
splitting atoms
6 .
comets
the combining of atomic nuclei to form one nucleus
Answer:
Here's your answer :
Asteroids - Between mars and JupiterFission - splitting atomsEnergy - Sun's atmosphereFusion - The combining of atomic nuclei to form one nucleusCorona - Ability to do workComets - Ice, dust, frozen gaseshope it helps!
Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with a uniform forward acceleration of 2.40 m/s2. How long after car A begins to accelerate does it take car A to overtake car B?
Answer:
The taken is [tex]t_A = 19.0 \ s[/tex]
Explanation:
Frm the question we are told that
The speed of car A is [tex]v_A = 22 \ m/s[/tex]
The speed of car B is [tex]v_B = 29.0 \ m/s[/tex]
The distance of car B from A is [tex]d = 300 \ m[/tex]
The acceleration of car A is [tex]a_A = 2.40 \ m/s^2[/tex]
For A to overtake B
The distance traveled by car B = The distance traveled by car A - 300m
Now the this distance traveled by car B before it is overtaken by A is
[tex]d = v_B * t_A[/tex]
Where [tex]t_B[/tex] is the time taken by car B
Now this can also be represented as using equation of motion as
[tex]d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300[/tex]
Now substituting values
[tex]d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]
Equating the both d
[tex]v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]
substituting values
[tex]29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]
[tex]7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]
[tex]7 t_A =1.2 t_A^2 - 300[/tex]
[tex]1.2 t_A^2 - 7 t_A - 300 = 0[/tex]
Solving this using quadratic formula we have that
[tex]t_A = 19.0 \ s[/tex]
What must x be so that the handle end of the bat remains at rest as the bat begins to move? (Hint: Consider the motion of the center of mass and the rotation about the center of mass. Find x so that these two motions combine to give v=0 for the end of the bat just after the collision. Also, note that integration of equation ∑τ⃗ =dL⃗ dt gives ΔL=∫t1t2(∑τ)dt. )
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told that
The mass of the bat is [tex]m_b = 0.800 \ kg[/tex]
The bat length is [tex]L_b = 0.900 \ m[/tex]
The distance of the bat's center of mass to the handle end is [tex]z_c = 0.600 \ m[/tex]
The moment of inertia of the bat is [tex]I = 0.0530 \ kg \cdot m^2[/tex]
The objective of the solution is to find x which is the distance from the handle of the bat to the point where the baseball hit the bat
Generally the velocity change at the end of the bat is mathematically represented as
[tex]\Delta v_e = \Delta v_c - \Delta w* z_c[/tex]
Where [tex]\Delta v_c[/tex] is the velocity change at the center of the bat which is mathematically represented as
[tex]\Delta v_c = \frac{Impulse}{m_b }[/tex]
We are told that the impulse is J so
[tex]\Delta v_c = \frac{J}{m_b }[/tex]
And [tex]\Delta w[/tex] is the change in angular velocity which is mathematically represented as
[tex]\Delta w = \frac{J (z -z_c)}{I}[/tex]
Now we have that
[tex]\Delta v_e = \frac{J}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]
Before a swing when the bat is at rest the velocity change a the end of the bat handle is zero and the impulse will be 1
So
[tex]0 = \frac{1}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]
=> [tex]x = \frac{I}{m_b z_c} + m_b[/tex]
substituting values
[tex]x = \frac{0.530}{0.800 * 0.600} + 0.600[/tex]
[tex]x = 0.710 \ m[/tex]
In the figure calculates the acceleration of the block friction not today
Answer:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
Explanation:
A"boat"is"moving"to"the"right"at"5"m/s"with"respect"to"the"water."A"wave"moving"to"the"left,"opposite"the"motion"of"the"boat."The"waves"have"2.0"m"between"the"top"of"the"crests"and"the"bottom"of"the"troughs."The"period"of"the"wave"is"8.3"s"and"their"wavelength"is"110"m."At"one"instant"the"boat"sits"on"a"crest"of"the"wave,"20"seconds"later,"what"is"the"vertical"displacement"of"the"boat
Answer:
0.99m
Explanation:
Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:
[tex]v=\lambda f=\lambda\frac{1}{T}=(110m)\frac{1}{8.3s}=13.25\frac{m}{s}[/tex]
the relative velocity is:
[tex]v'=13.25m/s-5m/s=8.25\frac{m}{s}[/tex]
This velocity is used to know which is the distance traveled by the boat after 20 seconds:
[tex]x'=v't=(8.25m/s)(20s)=165m[/tex]
Next, you use the general for of a wave:
[tex]f(x,t)=Acos(kx-\omega t)=Acos(\frac{2\pi}{\lambda}x-\omega t)[/tex]
you take the amplitude as 2.0/2 = 1.0m.
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{8.3s}=0.75\frac{rad}{s}[/tex]
by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:
[tex]f(165,20)=1.0m\ cos(\frac{2\pi}{110m}(165)-(0.75\frac{rad}{s})(20s))\\\\f(165,20)=0.99m[/tex]
Q1. What is the frequency of rotation of 1000 loop coil of area 20cm2 in a magnetic field of 5T to
generate an emf that has a maximum value of 15.7V?
Answer:
Explanation:
Emf e generated in a coil with no of turn n and area A rotating in a magnetic field B with angular speed of ω is given by the expression
e = e₀ sinωt
where e₀ = nωAB which is the maximum emf generated
Putting the given values
15.7 = 1000xω x 20 x 10⁻² x 5
ω = .0157
frequency of rotation
= ω / 2π
= .0157 / 2 x 3.14
= .0025 /s
9 rotation / hour .
By which process does the heat from the Sun reach the Earth? (AKS 4b DOK 1) *
Two charged particles are accelerated through a uniform electric field and zero magnetic field, then enter a region with zero electric field and a uniform magnetic field. The particles start at rest from the same position (but at different times; they do not interact with each other). They have identical charges, but different masses. Particle 2 has a cyclotron radius 1.5 times as large as that of particle 1. Find ratio m2/m1
Answer:
Explanation:
In magnetic field , charged particle will have circular path . Let the radius of their circular path be r₁ and r₂ . Let their velocity at the time of entering magnetic field be v₁ and v₂ .
The velocity with which they will come out of electric field can be measured from following equation
Eq = 1/2 m v² , E is electric field , q is charge on the particle , m is mass and v is velocity .
v² = 2Eq / m
radius of circular path can be measured by the following expression
m v² / r = Bqv
2Eq / r = Bqv
r = 2Eq / Bqv
= 2E / Bv
r² = 4E² / B²v²
= 4E²m / B²x 2Eq
since E , B and q are constant
r² = K . m
r₂² / r₁² = m₂ / m₁
1.5²
m₂ / m₁ = 1.5²
= 2.25
A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is what is the kinetic energy of the block?
Complete Question
A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 150 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 0.024 m, what is the kinetic energy of the block?
Answer:
The kinetic energy is [tex]KE = 0.4368\ J[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m= 0.025\ kg[/tex]
The spring constant is [tex]k = 150 N/m[/tex]
The length of first displacement is [tex]x_1 = 0.80 \ m[/tex]
The length of first displacement is [tex]x_2 = 0.024 \ m[/tex]
At the [tex]x_2[/tex] the kinetic energy is mathematically evaluated as
[tex]KE = \Delta E[/tex]
Where [tex]\Delta E[/tex] is the change in energy stored on the spring which is mathematically represented as
[tex]\Delta E = \frac{1}{2} k (x_1 ^2 - x_2^2)[/tex]
=> [tex]KE = \frac{1}{2} k (x_1 ^2 - x_2^2)[/tex]
Substituting value
[tex]KE = \frac{1}{2} * 150 * (0.08^2 - 0.024^2)[/tex]
[tex]KE = 0.4368\ J[/tex]
The figure shows a crane whose weight is 12.5 kN and center of gravity in G. (a) If the crane needs to suspend the 2.5kN drum, determine the reactions on the wheel in A and B when the boom is in the position shown.(b) Considering the same situation illustrated, what is the value of the maximum weight that the crane can suspend without tipping over?
Answer:
(a) Ra = 9.25 kN; Rb = 5.75 kN
(b) 26.7 kN
Explanation:
(a) Draw a free-body diagram of the crane. There are four forces:
Reaction Ra pushing up at A,
Reaction Rb pushing up at B,
Weight force 12.5 kN pulling down at G,
and weight force 2.5 kN pulling down at F.
Sum of moments about B in the counterclockwise direction:
∑τ = Iα
-Ra (0.66 m + 0.42 m + 2.52 m) + 12.5 kN (2.52 m + 0.42 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0
-Ra (3.6 m) + 12.5 kN (2.94 m) − 2.5 kN (1.38 m) = 0
Ra = 9.25 kN
Sum of moments about A in the counterclockwise direction:
∑τ = Iα
Rb (0.66 m + 0.42 m + 2.52 m) − 12.5 kN (0.66 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° + 0.66 m + 0.42 m) = 0
Rb (3.6 m) − 12.5 kN (0.66 m) − 2.5 kN (4.98 m) = 0
Rb = 5.75 kN
Alternatively, you can use sum of the forces in the y direction as your second equation.
∑F = ma
Ra + Rb − 12.5 kN − 2.5 kN = 0
Ra + Rb = 15 kN
9.25 kN + Rb = 15 kN
Rb = 5.75 kN
However, you must be careful. If you make a mistake in the first equation, it will carry over to this equation.
(b) At the maximum weight, Ra = 0.
Sum of the moments about B in the counterclockwise direction:
∑τ = Iα
12.5 kN (2.52 m + 0.42 m) − F ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0
12.5 kN (2.94 m) − F (1.38 m) = 0
F = 26.7 kN
