Answer: 1.348 meters
Explanation: Although the sign is missing from the location of the 4.00 kg object, it is assumed to be positive. The net moment of all the objects about the center of mass must be zero. Let the center of mass be on the y axis at a point c . Adding the four moments together, we get:
(2.00)(3.00−c)+(3.00)(2.50−c)+(2.50)(0−c)+(4.00)(0.500−c)=0
6.00−2.00c+7.50−3.00c+0−2.50c+2.00−4.00c=0
11.5c=15.50
c= 1.348 metres
The center of mass is on the y axis at y = 1.348 metres.
You are designing a delivery ramp for crates containing exercise equipment. The 1390 N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22.0∘. The ramp exerts a 515 N kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 5.0 m along the ramp. Once stopped, a crate must not rebound back up the ramp.
The maximum force constant of the spring Kmax is 2337.9 N/m.
What is force constant of a spring?The force constant or spring constant is defined as the force required to stretch or compress a spring such that the displacement in the spring is 1 meter.
Force constant is denoted by K and its unit is N/m.
Force = K * xWhere;
K = spring constant
x = displacement
The work done by the spring is given below as follows:
Work done = Fx/2
Kinetic Energy = mv²/2
Force on an inclined plane = mgsinθ
Total force, F = mgsinθ + frictional force
F = 1390 * sin 22° + 515
F = 1035.7 N
Work done = change in KE
Fx/2 = mv²/2
Fx = mv²
m = 1390/9.81 = 141.692
Solving for x;
x = mv²/F
x = 141.692 * 1.8²/1035.7
x = 0.443 m
The maximum force constant of the spring Kmax = 1035.7/0.443
Kmax = 2337.9 N/m
In conclusion, the maximum force constant of the spring is the ratio of the total force and displacement.
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Note that the complete question is given below:
You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 2.0 m/s at the top of a ramp that slopes downward at an angle 21.0 ∘. The ramp will exert a 533 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring Kmax that can be used in order to meet the design criteria
Suppose that the math man, a super here that fights crime with math and physics, can decelerate the rate of gravity. During a recent fight with a diamond thief, Math man fell from the top of a 500 meter building. The equation D=t^2*.25. How long us he falling from the top of the building to the ground?
Answer:
44.7 seconds
Explanation:
D = 500 m
500 = .25 t^2
500/.25 = t^2
2000 = t^2
t = 44.7 seconds
A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and
400°C. The location of the stops corresponds to 40 percent of the initial volume. Now the
steam is cooled. Determine the compression work if the final state is
a) 1.0 MPa and 250°C and
b) 500 kPa.
c) Also determine the temperature at the final state in part b.
A piston-work cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops corresponds to 40 percent of the initial volume. Now the steam is cooled. compression work is 44.32 KJ,
The amount of labor put into the piston to cause its reciprocating motion is known as the piston work. It is calculated by multiplying the piston's displacement by the net force.
An expanding gas cylinder's force output is transferred by pistons to the crankshaft, which then drives the flywheel's rotation. A reciprocating engine is a device like this.
Piston work is the effort made by the piston to make its reciprocating motion. The piston's displacement is calculated by multiplying it by the net force.
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A 92-kg fullback is running at 3.6 m/s to the east and is stopped in 0.85 s by a head-on tackle by a tackler running due west.
a)Calculate the original momentum of the fullback.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the momentum is to the east, and negative value if the direction of the momentum is to the west.
b)Calculate the impulse exerted on the fullback.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the impulse is to the east, and negative value if the direction of the impulse is to the west.
c)Calculate the impulse exerted on the tackler.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the impulse is to the east, and negative value if the direction of the impulse is to the west.
d)Calculate the average force exerted on the tackler.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the force is to the east, and negative value if the direction of the force is to the west.
a) The original momentum of the fullback is 331.2 kgm/s.
b) The impulse exerted on the fullback is - 331.2 kgm/s
c) The impulse exerted on the tackler is 331.2 kgm/s
d) The average force exerted on the tackler is 389.64 N
Given:Mass of the fullback, m = 92 kg
Initial velocity of the fullback , u = 3.6 m/s
Time of the motion of the fullback , t = 0.85s
The original momentum of the fullback ;
[tex]P_{i} = mv\\P_{i} = (92) (3.6)\\P_{i} = 331.2 kg m/s[/tex]
The impulse exerted on the full-back;
J = ΔP = [tex]m v_f - m v_i[/tex]
J = [tex]m ( v_f - v_i)[/tex]
J = 92 ( 0- 3.6)
J = - 331.2 kgm/s
The impulse exerted on the tackler;
[tex]J _1 = - J _2\\J_2 = - J_1\\J_2 = -( - 331.2)\\J_2 = 331.2 kgm/s[/tex]
The average force exerted on the tackler;
F= Δmv/t
[tex]F =\frac{331.2}{0.85} \\F = 389.64 N[/tex]
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Please help with this question, step by step
Thank you!
From the calculation, the force constant is 192 N. Also, friction would decrease the extension.
What is the force constant?We know that the force constant can be obtained by the use of the relation;
F = Ke
F = applied force
K = force constant
e = extension
We know from Hooks law that the force applied is directly proportional to the extension.
We can write;
F = mgcosθ
F = 43 Kg * 9.8 m/s^2 * sin31°
F = 217 N
K = 217 N/1.13 m
K = 192 N/m
If there is friction between the incline and the crate, it will stretch less because some work will be lost due to friction causing only some fraction of the elastic potential energy to be converted to kinetic energy.
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1. The diagram shows a satellite traveling in uniform circular motion around the Earth.
(a) Give the relation between radius of the orbit and the velocity of the satellite.
(b ) The satellite is kept in orbit by a force. On the diagram draw an arrow to show the
direction of this force.
Answer:
M V R = constant angular momentum is constant because no forces act in the direction of V
Since M (mass) = constant
V R = constant
The force is directed along the gravitational force vector (towards the center of rotation)
define universal vibrations ?
Answer:
The second universal law, the law of vibration, posits that everything (every atom, object, and living thing) is in constant motion, vibrating at a specific frequency.
Explanation:
Please mark brainly
The second universal law defines this.
The second universal law, also known as the Law of Vibration, The Law of Vibration states that everything in the universe is in a constant state of movement. We refer to these movements as vibration, and the speed or rate at which something vibrates is called its frequency. The only difference between one object and another is its vibration rate.
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The figure illustrates flow through a pipe with diameters of 1.0 mm and 2.0 mm and with different elevations. Px is the pressure in the pipe, and Vx is the speed of a non-viscous incompressible fluid at locations x = Q,R,S,T, or U. Options are: Greater than, Less than, Equal to
PU is ... PQ
VU is ... 2VT
PR is ... PU.
VR is ... VS
VQ is ... VU
PR is ... PS
a.
i. PU is greater than PQ.ii. VU is Greater than 2VTb.
i. PR is Equal to PU.ii. VR is Equal to VSc.
i. VQ is Equal to VUii. PR is Greater than PS.What is pressure?Pressure is the force per unit area on a surface.
What is speed?Speed is the distance moved per unit time.
Pressure
Since pressure, P = hρg where
h = depth, ρ = density of liquid and g = acceleration due to gravity.Since ρ and g are constant
P ∝ h
So, we see that pressure is directly proportional to depth.
a. i. Pressure between R and USince U is lower than Q, Pressure at U is greater than pressure at Q.
So,PU is greater than PQ.
ii. Speed between U and TUsing the continuity equation
VUAU = VTAT where
VU = speed at U, AU = cross-sectional area at U = π(dU)² where dU = diameter at U = 1.0 mmVUT= speed at T, AT = cross-sectional area at T = π(dT)² where dT = diameter at T = 2.0 mmSo, VUAU = VTAT
VUπ(dU)² = VTπ(dT)²
VU = VT(dT)²/(dU)²
VU = VT(2.0)²/(1.0)²
VU = VT(4)
VU = 4VT
Since VU = 4VT,VU is Greater than 2VT
b i. Pressure between R and USince R is at the same depth as U, Pressure at R is equal to pressure at U.
So,PR is Equal to PU.
ii. Speed between R and SUsing the continuity equation
VRAR = VSAS where
VR = speed at R, AR = cross-sectional area at R = π(dR)² where dR = diameter at R = 2.0 mmVS= speed at S, AS = cross-sectional area at S = π(dS)² where dS = diameter at S = 2.0 mmSo, VRAR = VSAS
VRπ(dR)² = VSπ(dS)²
VR = VS(dS)²/(dS)²
VR = VS(2.0)²/(2.0)²
VR = VS(1)
VR = VS
Since VR = VS,VR is Equal to VS
c. i. Speed between Q and UUsing the continuity equation
VQAQ = VUAU where
VQ = speed at Q, AQ = cross-sectional area at Q = π(dQ)² where dQ = diameter at Q = 1.0 mmVU = speed at U, AU = cross-sectional area at U = π(dU)² where dU = diameter at U = 1.0 mmSo, VQAQ = VUAU
VQπ(dQ)² = VUπ(dU)²
VQ = VU(dU)²/(dQ)²
VQ = VU(1.0)²/(1.0)²
VQ = VU(1)
VQ = VU
Since VQ = VU, VQ is Equal to VU
Ii. Pressure between R and SSince R is lower than S, Pressure at R is greater than pressure at S.
So,PR is Greater than PS.
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a guage is connected to tank in which the pressure of the fluid is 305kpa above atmospheric. if the absolute presssure of the fluid remains unchanged but the gage is in chamber where the air pressure is reduce to a vaccum of 648mmHg what reading in psi will then be observed
The pressure reading in psi observed after the gauge is placed in the chamber with a vacuum of 648 mmHg will be approximately 31.67 psi.
When the gauge is placed in a chamber with a vacuum of 648 mmHg we need to convert the pressure reading to a compatible unit to determine the pressure measurement in PSI.
First, convert the initial pressure of the fluid from kPa to mmHg.
7.5 mmHg is equal to about 1 kPa. So, 305 kPa * 7.5 mmHg/kPa = 2287.5 mmHg is the initial pressure of the fluid in mmHg.
Reducing the original pressure by vacuum pressure.
1639.5 mmHg = new pressure = 2287.5 mmHg - 648 mmHg
Converting the mmHg of the new pressure to the psi unit.
0.01934 psi is equal to approximately 1 mmHg. As a result, the new pressure in psi is equal to 31.67 psi: 1639.5 mmHg * 0.01934 psi/mmHg
Therefore, the pressure reading in psi observed after the gauge is placed in the chamber with a vacuum of 648 mmHg will be approximately 31.67 psi.
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An Airbus A380-800 passenger airplane is cruising at constant altitude on a straight line with a constant speed. The total surface area of the two wings is 395 m^2. The average speed of the air just below the wings is 259 m/s, and it is 288 m/s just above the surface of the wings.
What is the mass of the airplane? (The average density of the air around the airplane is ρair = 1.21 kg/m^3.)
The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.
We need to be aware of the Bernoulli principle in order to determine the solution.
How can I determine an airplane's mass?According to the Bernoulli's principle, the total amount of pressure energy, kinetic energy, and potential energy in a streamlined flow of an incompressible, non-viscous fluid is constant.It can be stated as follows:[tex]P+\frac{1}{2}dv^2+ dgh = constant.[/tex] We substitute d for to represent density.
We've done that,[tex]V_1=259m/s\\V_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]
We compare the governing idea for the wing's bottom and upper surfaces to:[tex]P_1+\frac{1}{2}dV_1^2+dgh=P_2+ \frac{1}{2}dV_2^2+dgh\\P_1-P_2=\frac{1}{2}d(V_2^2-V_1^2)\\\frac{F}{A}= \frac{1}{2}d(V_2^2-V_1^2)\\[/tex]
Consequently, using the aforementioned equation, the airplane's mass will be,[tex]m=\frac{\frac{1}{2}d(V_2^2-V_1^2)A\\}{g} \\m=387*10^3kg.[/tex]
Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.
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Suppose you need your silicon circuit element to run continuously for 3 minutes before it shuts off long enough to cool back down to its initial temperature. If the circuit element can withstand a temperature change of 5.1 ∘C without being damaged, what is the maximum rate at which energy can be added to the circuit element?
The maximum rate at which energy can be added to the circuit element mathematically given as
[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]
What is the maximum rate at which energy can be added to the circuit element?Generally, the equation for P is mathematically given as
[tex]P=\ln s \frac{\Delta T}{\Delta t}[/tex]
Therefore
[tex]Rate\ of\ Change\ of\ Temp =\frac{p}{lnS}[/tex]
[tex]\frac{p}{lnS}=\frac{7.4 \times 10^{-3}}{23 \times 10^{-6} \times 705}[/tex]
[tex]\frac{p}{lnS}=0.456^{\circ \mathrm{c}} / \mathrm{sec}[/tex]
Max temp Change
[tex]MaxT=5.6^{\circ} \mathrm{C}[/tex]
[tex]\text { time }=3 \times 60[/tex]
t=180s
In conclusion, Max Energy Rate
[tex]MER =23 \times 10^{-6} \times \frac{301 \times 5.6}{180}[/tex]
[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]
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The resistivity of a 1.0 m long wire is 1.72 × 10-8 ωm and its cross sectional area is 2.0 × 10-6 m2. if the wire carries a current of 0.20 a, what is the voltage across the wire?
The voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V
What does the Resistance of a wire depend on ?The resistance of a wire is the opposition to the flow of current. It depends on the following;
TemperatureLength of the wireCross sectional areaResistivity of the wireGiven that the resistivity of a 1.0 m long wire is 1.72 × 10-8 ωm and its cross sectional area is 2.0 × 10-6 m2. if the wire carries a current of 0.20A
The given parameters are;
Resistivity ρ = 1.72 × 10-8 ωm Length L = 1.0 mCross sectional area A = 2.0 × [tex]10^{-6}[/tex] m²Current I = 0.2 AResistance R = ?Voltage V = ?The formula to use to get R will be
R = ρL / A
Substitute all the necessary parameters into the formula
R = 1.72 x [tex]10^{-8}[/tex] x 1 / 2 x [tex]10^{-6}[/tex]
R = 8.6 x [tex]10^{-3}[/tex] Ω
From Ohm's law, V = IR
Substitute all the necessary parameters into the formula
V = 0.2 x 8.6 × [tex]10^{-3}[/tex]
V = 1.72 x [tex]10^{-3}[/tex] V
Therefore, the voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V
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A thin-walled hollow sphere has a radius 4cm from the center
of the sphere, the eletric field points radially inward and
has a magnitude of 1.5 × 10^4NC^-1 How much charge is on the surface
A thin-walled hollow sphere has a radius of 4 cm from the center
of the sphere, the electric field points radially inward and
has a magnitude of 1.5 × 10⁴ N/C then the charge on the surface would be 2.6690×10⁻⁹ C
What is an electric charge?Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge.
The electric field inside a spherical shell is given by the formula
E = q/4πεr²
where q is the charge on the surface
r is the radius of the sphere
ε is the electrical permeability
By substituting the respective values in the given formula
1.5 × 10⁴ = q/4π(8.85✕ 10⁻⁻¹²)(.04²)
q = 2.6690×10⁻⁹ C
Thus, the charge on the surface would be2.6690×10⁻⁹ C
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A simple pendulum suspended in a rocket ship has a period [tex]T_{0}[/tex]. Assume that the rocket ship is near the earth in a uniform gravitational field.
True or false:
1. If the ship moves upward with a constant velocity, the period increases.
2. If the length of the pendulum is doubled, the new period will be: square root of 2 times [tex]T_{0}[/tex].
3. If the ship accelerates upward, the period increases.
4. If the ship accelerates downward at 9.81 [tex]m/s^{2}[/tex], the pendulum will no longer oscillate.
5. If the mass of the pendulum is halved, the period decreases.
(1) If the ship moves upward with a constant velocity, the period increases, false.
(2) If the length of the pendulum is doubled, the new period will be: square root of 2 times T₀. true.
(3) If the ship accelerates upward, the period increases, true.
(4) If the ship accelerates downward at 9.81 m/s² , the pendulum will no longer oscillate, false.
(5) If the mass of the pendulum is halved, the period decreases, false.
Period of a simple pendulum
The period of a simple pendulum is given as;
T = 2π√(L/g) ------- (1)
where;
L is length of the pendulumg is acceleration due to gravityWhen the ship moves upward with a constant velocity, the period will not change.
When the length of the pendulum is doubledT ∝√L
when, L = 2L₀
T = T₀√2
Thus, if the length of the pendulum is doubled, the new period will be: square root of 2 times T₀.
when ship accelerates upwardt = √2h/g
when the height increases, time or period increaseswhen mass of the pendulum is halvedT = √I/mgd
where;
m is mass of the pendulumwhen the mass is halved, the period increases.
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Consider the f(x) = Acos(x) function shown in the figure in blue color. What is the value of amplitude A for this function?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.
The amplitude of the red colored wave is 1 unit and the amplitude of the red colored wave is 2.1 unit.
What is amplitude of a wave?
The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
Amplitude of the red colored waveFrom the graph, the amplitude of the red colored wave is 1 unit.
Amplitude of the blue colored waveFrom the graph, the amplitude of the red colored wave is 2.1 unit.
Thus, the amplitude of the red colored wave is 1 unit and the amplitude of the red colored wave is 2.1 unit.
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Write a properly formatted hypothesis statement to answer this question: How does the amount of salt added to ice affect the rate at which the ice will melt?
Specify how you plan to change the independent variable by using terms such as increase or decrease. Also, specify how the dependent variable will change in response by using terms such as increase, decrease, or stays the same.
Criteria pts
Correct placement of IV 5
Correct placement of DV 5
If, then format 5
IV indicates either "increases" or "decreases" 5
DV indicates either "increases", "decreases", or "stays the same" 5
The hypothesis will be:
H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.
H₁ = The amount of salt added to ice will affect the rate at which the ice will melt.
The independent variable which is can be changed by increasing the rate of salt added to the equation.
The dependent variable which is ice will change or melt in response as it will decrease if the rate of the salt added increases.
What is the effect of salt on the melting temperature of ice?Salt does not really lower the temperature of an ice cubes, it is known to just lowers their freezing point, that is lowers their melting point.
Note that if salt is around, ice cubes are known to be colder to be solid, and they tend to melt at a temperature that is said to be lower than the freezing point of pure water.
If the ionic compound salt is known to be added, it tends to lowers the freezing point of the water, which implies that the ice on the ground is not able to freeze that layer of water at all.
Therefore, The hypothesis will be:
H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.
H₁ = The amount of salt added to ice will affect the rate at which the ice will melt.
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a trolley and a sandbag have a combined mass of 4 kg. a bullet with a mass of 150 gram is fired towards the trolley and is lodged in the sand bag immediately after the Collision, on the trolley and sandbag in combination with the bullet, move backwards at 53 metre per second calculate the kinetic energy of the trolley , with the sandbag and bullet ,directly after the Collision
The correct answer is 5828.675 J.
Given combined mass 4kg and mass of bullet 150gm=0.150kg.
Total mass= 4+0.150=4.150kg
Velocity=53 m/s
Kinetic energy = [tex]\frac{1}{2} *m*v^{2}[/tex] =0.5*4.150*[tex]53^{2}[/tex] =5828.675 J
Kinetic energyKinetic energy is a type of power that an item or particle possesses as a result of motion. When an item undergoes work—the transfer of energy—by being subjected to a net force, it accelerates and consequently obtains kinetic energy. A moving object or particle's kinetic energy, which depends on both mass and speed, is one of its characteristics. Any combination of motions, including translation (or travel along a path from one location to another), rotation about an axis, and vibration, may be used as the type of motion.
A body's translational kinetic energy is equal to [tex]\frac{1}{2} *m*v^{2}[/tex] , or one-half of the product of its mass, m, and square of its velocity, v.
a trolley and a sandbag have a combined mass of 4 kg. a bullet with a mass of 150 gram is fired towards the trolley and is lodged in the sand bag immediately after the Collision, on the trolley and sandbag in combination with the bullet, move backwards at 53 metre per second calculate the kinetic energy of the trolley , with the sandbag and bullet ,directly after the Collision
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Two uncharged spheres are separated by 1.70 m. If 2.40 ✕ 10¹² electrons are removed from one sphere and placed on the other, determine the magnitude of the Coulomb force (in N) on one of the spheres, treating the spheres as point charges.
_______N
**Hint** Find the net charge on each sphere and substitute values into Coulomb's law.
The magnitude of the Coulomb force (in N) on one of the spheres, given the data is 4.59×10⁻⁴ N
How to determine the charge on each spheresSphere 1 losses 2.40×10¹² electrons
But
1 electron = 1.6x10¯¹⁹ C
Thus,
Charge on sphere 1 = +1.6x10¯¹⁹ × 2.40×10¹² = +3.84×10¯⁷ C
Sphere 2 gains 2.40×10¹² electrons
But
1 electron = 1.6x10¯¹⁹ C
Thus,
Charge on sphere 2 = -1.6x10¯¹⁹ × 2.40×10¹² = -3.84×10¯⁷ C
How to determine the coulomb forceCharge on sphere 1 (q₁) = +3.84×10¯⁷ CCharge on sphere 2 (q₂) = 3.60 mC = -3.84×10¯⁷ CElectric constant (K) = 9×10⁹ Nm²/C²Distance apart (r) = 1.7 mForce (F) =?
Using the Coulomb's law equation, the force can be obtained as illustrated below:
F = Kq₁q₂ / r²
F = (9×10⁹ × 3.84×10¯⁷ × 3.84×10¯⁷) / (1.7)²
F = 4.59×10⁻⁴ N
Thus, the magnitude of the Coulomb's force is 4.59×10⁻⁴ N
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A uniform plank of length 2.00 m and mass 30.23 kg is supported by three ropes. A 700 N person is a distance, of 0.55 m from the left end. A) find the magnitude of the tension in the vertical rope on the left end. Give your answers in newtons. B) find the magnitude of the tension in the rope in the right end. Give your answers in newtons C) find the magnitude of the tension in the horizontal rope on the left end. Give your answers in newtons
a. The magnitude of the tension, T₂ in the vertical rope on the left end is T₂ = 655.62 N
b. The magnitude of the tension in the rope in the right end is T₁ = 530.4 N
c. The magnitude of the tension in the horizontal rope on the left end, T₃ is T₃ = 406.3 N
What is tension force?Tension force refers to a pulling force that is exerted by a string or cable about an axis.
a. The magnitude of the tension, T₂ in the vertical rope on the left end is given as follows:
Taking moment about the vertical axis
T₂ = 30.23 * 9.81 + 700 - T₁ * Sin40°
Solving for T₁ by taking the left end as the pivot;
T₁ Sin 40° * 2.00 = 700 * 0.55 + (30.23 * 9.81) * 1.0
T₁ * 1.285 = 681.5563
T₁ = 530.4 N
Therefore;
T₂ = 30.23 * 9.81 + 700 - 530.4 * Sin 40°
T₂ = 655.62 N
b. From calculation, the magnitude of the tension in the rope in the right end is T₁.
T₁ = 530.4 N
c. The magnitude of the tension in the horizontal rope on the left end, T₃ is determined thus:
Taking moments about the left end in the horizontal direction;
T₃ = T₁ * cos 40°
T₃ = 530.4 N * cos 40°
T₃ = 406.3 N
In conclusion, the tension at the rope in the various ends is determined by taking moments about the left end.
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Only using dimensional analysis, answer the following question:
Convert 65 miles per hour to kilometers per second. Show your work and explain how you got it please!
Brainliest goes to most detailed response
Answer:
[tex]2.9055\times 10^-2 km/s \approx 0.03 km/s[/tex]
Explanation:
One (statute) mile is defined as [tex]1\ 609.344 m = 1.609\ 334 km[/tex] or, cutting the last significant digits, [tex]1.6 km[/tex]. An annoying European would drop some kind of remark here like "but what about nautical miles?", but let's assume it's a statute mile.
One hour is 3600 seconds (60 minutes in a hour, times 60 seconds in a minute), no complains across the pond here.
Now we're ready to crunch numbers.
[tex]65\frac{mi}h = 65\frac{1.6km}{3600s} \approx 0.03 \frac {km}s[/tex]
You can use the exact definition of mile and you will get the exact value of [tex]0.029055 \frac{km}s = 2.9055\times10^-^2 km/s[/tex] with an error of about a meter.
Answer:
See below
Explanation:
Since you want detailed dimensional analysis here is a lot of details:
mile / hr / (60 min / hr * 60 sec / min) * 1760 yds/mile * 36 in/yd * 1m /39.37 in * 1 km / 1000 m
After cancelling everything out (dimensional analysis) , you are left with:
= mile/ hr * 1/3600 * 1760 * 36 /39.37 * 1/1000
= mile /hr * .00044704089 km -hr/ sec-mile = km/s
the underlined portion is the conversion factor for miles/hr to km/s
65 * .00044704089 = .029057 km / sec
Consider the f(x) = cos(x) + D function shown in the figure in blue color. What is the value of parameter D?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.
The value of parameter D is -3.
What is angular frequency of a wave?Angular frequency is the angular displacement of any element of the wave per unit time.
f(x) = Acos(x) + D
where;
A is amplitude of the waveB is phase difference of the waveWhat is amplitude of a wave?The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
From the blue colored graph, y = -2 at x = 0
-2 = cos(0) + D
-2 = 1 + D
-2 - 1 = D
-3 = D
Thus, the value of parameter D is -3.
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Standing 38.6 m away from a rock wall, you yell. How much time in seconds will it take you to hear your echo to two significant digits? Make sure to account for the travel from you to the wall and from the wall back to you.
It will take you 0.23 seconds to hear your echo
What is Echo ?Echo can be simply defined as the reflection of sound wave.
Given that you are 38.6 m away from a rock wall, you yell. To know how much time in seconds it will take you to hear your echo to two significant digits, You must make sure to account for the travel from you to the wall and from the wall back to you.
The speed of sound V = 340 m/sThe distance D to and fro = 2 x 38.6 = 77.2 mThe time taken T = ?Speed V is the distance per time
V = D / T
Substitute all the parameters into the formula
340 = 77.2 / T
Make T the subject of formula
T = 77.2 / 340
T = 0.2270
T = 0.23 s to two significant digits
Therefore, it will take you 0.23 seconds to hear your echo.
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A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface. Neglect any effects due to air resistance.
a) What is the kinetic energy of the projectile when it reaches the highest point in its trajectory?
b) How much work was done in firing the projectile?
Answer:
Kinetic energy of the projectile at the vertex of the trajectory: [tex]900\; {\rm J}[/tex].
Work done when firing this projectile: [tex]2500\; {\rm J}[/tex].
Explanation:
Since the drag on this projectile is negligible, the horizontal velocity [tex]v_{x}[/tex] of this projectile would stay the same (at [tex]30\; {\rm m\cdot s^{-1}}[/tex]) throughout the flight.
The vertical velocity [tex]v_{y}[/tex] of this projectile would be [tex]0\; {\rm m\cdot s^{-1}}[/tex] at the vertex (highest point) of its trajectory. (Otherwise, if [tex]v_{y} > 0[/tex], this projectile would continue moving up and reach an even higher point. If [tex]v_{y} < 0[/tex], the projectile would be moving downwards, meaning that its previous location was higher than the current one.)
Overall, the velocity of this projectile would be [tex]v = 30\; {\rm m\cdot s^{-1}}\![/tex] when it is at the top of the trajectory. The kinetic energy [tex]\text{KE}[/tex] of this projectile (mass [tex]m = 2.0\; {\rm kg}[/tex]) at the vertex of its trajectory would be:
[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}[/tex].
Apply the Pythagorean Theorem to find the initial speed of this projectile:
[tex]\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Hence, the initial kinetic energy [tex]\text{KE}[/tex] of this projectile would be:
[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}[/tex].
All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be [tex]2500\; {\rm J}[/tex].
It's your first day on the job without your mentor as a member of mission control.
Your shift begins and your counterpart, who happens to be in their 10th year in this
position, speaks to you for about 45 minutes as they end their shift and you begin
yours. What were they talking to you about?
getting you up to date with recent mission developments
sanitation protocol that all employees must memorize
connecting you with each of the astronauts in space at the time
how to change settings for each system that's in use
Explanation:
getting me up to date with recent mission development
PLEASE HELP!!!!
Compared to driving at 20 mph, about how much longer will it take for you to stop at 60 mph?
about nine times as far
about six times as far
about twice as far
Answer:
The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)
Explanation:
Let [tex]u[/tex] denote the initial velocity of the vehicle ([tex]20\; \text{mph}[/tex] or [tex]60\; \text{mph}[/tex]) and let [tex]v[/tex] denote the velocity of the vehicle after braking ([tex]0\; \text{mph}[/tex]). Let [tex]x[/tex] denote the braking distance.
Assume that the acceleration during braking are both constantly [tex]a[/tex] in both scenarios. The SUVAT equations would apply. In particular:
[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}[/tex].
Since [tex]v = 0[/tex] (the vehicle has completely stopped), the equation becomes [tex]x = (-u^{2}) / (2\, a)[/tex].
Assuming that [tex]a[/tex] (braking acceleration) stays the same, the braking distance [tex]x[/tex] would be proportional to [tex]u^{2}[/tex], the square of the initial velocity.
Hence, increasing the initial speed from [tex]20\; \text{mph}[/tex] to [tex]60\; \text{mph}[/tex] would increase the braking distance by a factor of [tex]3^{2} = 9[/tex].
Answer:
9 times
Explanation:
Can we take water instead of clock oil in Milikan oil drop experiment. Explain.
Answer:
No you can't cuz,if you put water instead of clock oil in Millikan oil drop your experiment will fail and it won't turn out the way you wanted it to be
A ball of mass 0.200 kg has a velocity of 150m/s; a ball of mass 0.300 kg has a velocity of - 0.4m/s. They meet in a head-on elastic collision. (a) Find their velocities after the collision. (b) Find the velocity of their center of mass before and after the collision.
(a) The velocity of 0.2 kg after the collision is 30.48 m/s and the velocity of the 0.3 kg mass is 119.92 m/s.
(b) The velocity of their center of mass before collision is 60.24 m/s and after the collision is 84.14 m/s.
Velocity of the balls after collision
Apply the principle of conservation of linear momentum to determine the velocity of the balls;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.2(150) + (0.3)(-0.4) = 0.2v₁ + 0.3v₂
29.88 = 0.2v₁ + 0.3v₂
Apply one directional linear velocity
u₁ + v₁ = u₂ + v₂
v₁ = u₂ + v₂ - u₁
v₁ = -0.4 + v₂ - 150
v₁ = v₂ - 150.4
Substitute the value of v₁ into the first equation;
29.88 = 0.2(v₂ - 150.4) + 0.3v₂
29.88 = 0.2v₂ - 30.08 + 0.3v₂
59.96 = 0.5v₂
v₂ = 59.96/0.5
v₂ = 119.92 m/s
v₁ = 119.92 - 150.4
v₁ = -30.48 m/s
Velocity of their center mass before collisionV(cm) = (0.2 x 150 + 0.3 x 0.4) / (0.2 + 0.3)
V(cm) = 60.24 m/s
Velocity of their center mass after collisionV(cm) = (0.2 x 30.48 + 0.3 x 119.92) / (0.2 + 0.3)
V(cm) = 84.14 m/s
Thus, the velocity of 0.2 kg after the collision is 30.48 m/s and the velocity of the 0.3 kg mass is 119.92 m/s.
The velocity of their center of mass before collision is 60.24 m/s and after the collision is 84.14 m/s.
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According to current evidence, what happens when gays or lesbians rear a child?
a. the child is more likely than average to develop depression or anxiety disorders.
O b. the child develops about the same as other children.
O c. the child is more likely than average to become gay or lesbian.
O d. the child does poorly in school.
Answer: The answer is A.
Explanation:
Just because u have gay parents doesnt mean theyre bad lol XD
As the resistance increases, what happens to the slope on the current vs voltage graph?
A. It gets steeper
B. It gets flatter
C. It keeps a constant slope
D. Not enough info
B. As the resistance increases, the slope on the current vs voltage graph gets flatter.
What is resistance?Resistance is the opposition to the follow of current in electrical material.
The mathematical relationship between resistance, current and voltage is given by Ohm's law.
R = V/I
where;
R is resistance, measured in ohmsV is voltage, measured in voltsI is current, measured in ampereFrom the equation above, increase in resistance means an increase in voltage or decrease in current.
In Current vs voltage graph, the following will occur;
The current decreases - the vertical axis decreasesThe voltage increases - the horizontal axis becomes largerThus, as the resistance increases, the slope on the current vs voltage graph gets flatter.
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An object is thrown upward with initial velocity of 30m/s at angle of 30 degree to the horizontal. calculate the components of the initial velocity.
Answer:
Vo (vertical) = Vo sin 30 = 30 m/s / 2 = 15 m/s
Vo (horizontal) = Vo cos 30 = .866 30 m/s = 26 m/s
