Given:
the current through motor is
[tex]I=210\text{ A}[/tex]The potential of the battery is
[tex]V=12\text{ V}[/tex]The time is
[tex]t=10\text{ s}[/tex]Required: the energy delivered to the motor in 10 sec is?
Explanation:
first, we calculate the power delivered to the motor.
The power is given by
[tex]P=VI[/tex]Plugging all the values in the above relation, we get:
[tex]\begin{gathered} P=12\text{ V}\times210\text{ A} \\ P=2520\text{ watt} \end{gathered}[/tex]The energy delivered to the motor is given by
[tex]E=Pt[/tex]Plugging all the values in the above relation, we get:
[tex]\begin{gathered} E=2520\text{ w}\times10\text{ s} \\ E=25200\text{ J} \\ \end{gathered}[/tex]Thus, the energy delivered to the motor is 25200 J.
A substance has a mass of 15,000 kg and a volume of 30 m 3 . What is the density of thesubstance?
ANSWER
500 kg/m³
EXPLANATION
Given:
• The mass of the substance, m = 15,000 kg
,• The volume of the substance, V = 30 m³
Find:
• The density of the substance, ρ
The density of a substance of mass m and volume V is,
[tex]\rho=\frac{m}{V}[/tex]Substitute the known values and solve,
[tex]\rho=\frac{15,000kg}{30m^3}=500kg/m^3[/tex]Hence, the density of the substance is 500 kg/m³.
A book of mass m = 10kg is balanced on a table Determine the weight of the book and the component of the reaction of the table in each of the following cases: 1)The book is simply placed on the table2) Someone puts their hand on the book by exerting a vertical force of 40N3)Someone is pulling this book vertically upwards with a force of 40N
Given,
The mass of the book, m=10 kg
The weight of an object is given by the product of its mass and the acceleration due to gravity.
Thus the weight of the book is given by,
[tex]W=mg[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} W=10\times9.8 \\ =98\text{ N} \end{gathered}[/tex]Therefore the weight of the book is 98 N
1)
When the book is placed on the table reaction of the table will be equal to the weight of the book. Thus, the reaction of the table, in this case, will be 98 N
2)
The vertically downward force applied on the book, F=40 N
The reaction of the table when there is a vertically downward force is equal to the sum of the weight and the applied force.
Thus the reaction of the table, in this case, is given by,
[tex]N=W+F[/tex]On substituting the known values,
[tex]\begin{gathered} N=98+40 \\ =138\text{ N} \end{gathered}[/tex]Thus, the reaction of the table when a vertically downward force is acting on it is 138 N
3)
The vertically upward force acting on the book, F=40 N
As the book is in equilibrium, the net force acting on the book must be equal to zero. That is total upward forces acting on the book must be equal to the total downward forces. Thus,
[tex]\begin{gathered} N+F=W \\ \Rightarrow N=W-F \end{gathered}[/tex]Where N is the normal force.
On substituting the known values,
[tex]\begin{gathered} N=98-40 \\ =58\text{ N} \end{gathered}[/tex]Thus the reaction of the table, in this case, is 58 N
Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 33.55 N when they are separated by 55.75 cm. What is the magnitude of the charges in microCoulombs ?
Given
Two equal, but oppositely charged particles are attracted to each other electrically.
Force of attraction is F=33.55 N
Distance between them, d=55.75 cm=0.5575 m
To find
What is the magnitude of the charges in microCoulombs ?
Explanation
Let the charge be q.
We know the force of attraction is given by
[tex]\begin{gathered} F=k\frac{q^2}{d^2} \\ \Rightarrow33.55=9\times10^9\times\frac{q^2}{(0.5575)^2} \\ \Rightarrow q=3.39\times(10)^{-5}C \\ \Rightarrow q=\pm33.9\mu C \end{gathered}[/tex]Conclusion
The charges are:
[tex]+33.9\mu C,-33.9\mu C[/tex]The magnitude of equal charges are:
[tex]\lvert{q}\rvert=33.9\mu C[/tex]Please help me I know it’s not the second choice
Which quantity is not required in order to calculate the mass of an ion using a mass spectrometer?O velocity vO magnetic field BO charge qO change in time t
To answer your question, lets think about the formulas needed
Centripetal force = mv^2/r
and force on a point charge = qvb
The force of the magnet is the same on the particle so you can set them equal to each other
qvb = mv^2/r
Set equal to mass
m = qbr/v
Change in time is not needed
A block has a velocity of 6 m/s to the East and 360 J of kinetic energy. The block is pushed West with a 30 N external force, while the block moves 3 m East. How much work is done by the force?
Work done will be 270 J
What is work energy theorem?
The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
according to work force theorem
Work done = Force x direction = FD Cosθ
Even if the force is applied to the opposite direction, the box will move in the direction of East. Firstly it was already 6m towards East and after applying force, the box moves further 3m towards same direction i.e East.
= 30 N x (9) cos 0⁰
= cos 0⁰ is 1
= 270 J
270 J work is done by the force.
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If I have a USB 2.0 port and it provides a maximum current of 0.5A at a voltage of 5 volts, if I connect a device with a resistance of 3 Ohms, how much current would flow through it and what will the dissipated power be?
1. The amount of current that will flow through it is 1.67 A
2. The power dissipated is 8.35 watts
1. How do I determine the current that will flow through it?
Ohm's law states as follow
Voltage (V) = Current (I) × resistance (R)
V = IR
With the above formula, we can obtain the current. This is shown below
Voltage (V) = 5 VoltsResistance (R) = 3 ohmsCurrent (I) =?Voltage (V) = Current (I) × resistance (R)
5 = Current × 3
Divide both sides by 3
Current = 5 / 3
Current = 1.67 A
2. How do I determine the power dissipated?
The power dissipated can be obtained as follow:
Voltage (V) = 5 VoltsCurrent (I) = 1.67 APower dissipated (P) = ?Power (P) = voltage (V) × current (I)
Power dissipated = 5 × 1.67
Power dissipated = 8.35 watts
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Please help me create a table and graph for the experiment below.Mass of bob: 25g, Time noted: 0.54 s, 0.54 s, 0.54 sMass of bob: 50g, Time noted: 0.54 s, 0.56 s, 0.49 sMass of bob: 100g: Time noted: 0.53 s, 0.53 s, 0.46 s
Given:
Mass of bob: 25g, Time noted: 0.54 s, 0.54 s, 0.54 s
Mass of bob: 50g, Time noted: 0.54 s, 0.56 s, 0.49 s
Mass of bob: 100g: Time noted: 0.53 s, 0.53 s, 0.46 s
To find:
The average time period for each mass of the pendulum bob.
Explanation:
As the sensor records the time it takes for the pendulum to complete one-half of the period, the time period of the pendulum can be calculated by multiplying the recorded value by 2.
For the bob of mass 25 g,
Period (s) Trial 1: 0.54 s × 2 = 1.08 s
Period (s) Trial 2: 0.54 s × 2 = 1.08 s
Period (s) Trial 3: 0.54 s × 2 = 1.08 s
Average Period (s): T = (1.08 s + 1.08 s + 1.08 s)/3 = 1.08 s
For the bob of mass 50 g,
Period (s) Trial 1: 0.54 s × 2 = 1.08 s
Period (s) Trial 2: 0.56 s × 2 = 1.12 s
Period (s) Trial 3: 0.49 s × 2 = 0.98 s
Average Period (s): T = (1.08 s + 1.12 s + 0.98 s)/3 = 1.06 s
For the bob of mass 100 g,
Period (s) Trial 1: 0.53 s × 2 = 1.06 s
Period (s) Trial 2: 0.53 s × 2 = 1.06 s
Period (s) Trial 3: 0.46 s × 2 = 0.92 s
Average Period (s): T = (1.06 s + 1.06 s + 0.92 s)/3 = 1.013 s
Final answer:
The average value of the period is
For bob of mass 25 g, Average Period = 1.08 s
For bob of mass 50 g, Average Period = 1.06 s
For bob of mass 100 g, Average Period = 1.013 s
A box is pulled a distance of 20 meters by a force of magnitude 40N, in the direction of the force ,over a period of 10 seconds Find the power generated by the force
The formula for the power is,
[tex]\begin{gathered} P=\frac{W}{t} \\ Where, \\ P=Power \\ W=Work \\ t=Time \end{gathered}[/tex]Work done is,
[tex]\begin{gathered} W=F\times S \\ Where, \\ W=Work \\ F=Force \\ S=Displacement \\ here, \\ F=40N;\text{ }and\text{ }S=20m; \\ So, \\ W=F\times S=40\times20=800J \end{gathered}[/tex]Here time taken to do the work is 10 sec
So,
[tex]P=\frac{W}{t}=\frac{800}{10}=80\text{ watt/sec}[/tex]So power generated by the force is 80 watt/sec.
A 15 V constant voltage battery is connected to 120 kΩ resistor. A 15 V constant voltage battery is connected to 120 kΩ resistor. 1) What is the current in the resistor in [mA]?Multiple Choice question2) How many electrons pass through the resistor each second?a) 7.8x1014 electronsb) 1.25x10-4 electronsc) 1.6x10-19 electrons
We will have the following:
1)
[tex]I=\frac{V}{R}\Rightarrow I=\frac{15V}{120000\Omega}\Rightarrow I=1.25\cdot10^{-4}A[/tex][tex]\Rightarrow I=0.125mA[/tex]2)
First we determine the quantity of coulombs:
[tex]Q=(1.25\cdot10^{-4}A)(1s)\Rightarrow Q=1.25\cdot10^{-4}C[/tex]Now, we know that 1 coulomb represents 6.24*10^18 electrons, so:
[tex]1.25\cdot10^{-4}C\cdot\frac{6.24\cdot10^{18}\text{electrons}}{1C}=7.8\cdot10^{14}\text{electrons}[/tex]So, approximately 7.8*10^14 electrons pass each second.
The half-life of tungsten 188 is 69.4 days. Initially, there are 0.725 kg of this isotope. How much of the isotope will remain after 147 days?(a) 0.104 kg(b) 0.167 kg(c) 0.237 kg(d) 0.255 kg
In order to determine the amount of tungsten after 147 days, use the following formula for the radioactive decay:
[tex]A=A_oe^{-\lambda t}[/tex]where
A: amount of tungsten after t days
Ao: initial amount of tungsten = 0.725 kg
t: time = 147 days
λ: decay constant
Then, it is necessary to find the value of λ. Use the following formula:
[tex]\lambda=\frac{\ln 2}{t_{\frac{1}{2}}}[/tex]where t1/2 is the half-life of tungsten (69.4 days)
[tex]\lambda=\frac{\ln 2}{69.4}=0.00998[/tex]next, replace the previous result and the values of the other parameters into the formula for A:
[tex]A=(0.725kg)e^{-(0.00998)(147)}=(0.725kg)(0.23)=0.167\operatorname{kg}[/tex]Hence, after 147 days, there are 0.167 kg of tungsten 188
hello can you help me with my AP physics assignment
1)
R is the required distance from the starting point. The right triangle representing this scenario is shown below
We would apply pythagorean theorem which is expressed as
hypotenuse^2 = one leg^2 + other leg^2
hypotenuse = R
one leg = 25
other leg = 18
Thus,
R^2 = 25^2 + 18^2 = 949
R = √949
R = 30.81
You're 30.81m from your starting point
2) We would find θ by applying the tangent trigonometric ratio which is expressed as
tanθ = opposite side/adjacent side
opposite side = 25
adjacent side = 18
tanθ = 25/18
θ = tan^-1(25/18)
θ = 54.25 degrees
Position = 54.25 degrees west of north
Alexander's hobby is dirt biking. On one occasion last weekend, he accelerated from rest to 12.8 m/s in 4.43 seconds. He then maintained this speed for 8.83 seconds. Seeing a coyote run cross the trail ahead of him, he abruptly stops in 6.36 seconds. Determine Alexander's average speed for this motion?
The average speed of Alexander, given that he accelerated from rest to 12.8 m/s in 4.43 seconds is 12.8 m/s
How do I determine the average speed of Alexander?First, we shall determine the distance travelled in 4.43 seconds. This is shown below:
Speed = 12.8 m/sTime = 4.43 secondDistance =?Distance = speed × time
Distance = 12.8 × 4.43
Distance = 56.704 m
Next, we shall determine the distance travelled in 8.83 seconds. This is shown below:
Speed = 12.8 m/sTime = 8.83 secondDistance =?Distance = speed × time
Distance = 12.8 × 8.83
Distance = 113.024 m
Next, we shall determine the distance travelled in 6.36 seconds. This is shown below:
Speed = 12.8 m/sTime = 6.36 secondDistance =?Distance = speed × time
Distance = 12.8 × 6.36
Distance = 81.408 m
Finally, we shall determine the average speed of Alexander. This is shown below:
Total distance = 56.704 + 113.024 + 81.408 = 251.136 mTotal time = 4.43 + 8.83 + 6.36 = 19.62Average speed =?Average speed = Total distance / total time
Average speed = 251.136 / 19.62
Average speed = 12.8 m/s
Thus, the average speed is 12.8 m/s
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1. Battery voltage = 12v. the current in the circuit is 0.5A. the resistance R must be 2. A 12 volt battery is connected across resistor, and a current 1.5A flows in the resistor. What is the value of the resistor?
To find the resistance, we have to use the following
[tex]R=\frac{V}{I}[/tex]Then, we replace the values
[tex]R=\frac{12}{1.5}=8[/tex]Hence, the answer is A.A student conducts an investigation into the motion of objects dropped from rest in the absence of air resistance. The student releases objects of various masses in a vacuum chamber and records the speed and distance fallen every 0.1 seconds for each object. What experimental evidence would lead the student to conclude that free-fall motion is constant acceleration? Select two answers. A. All objects released from rest at the same time fall together.B. The distance fallen is proportional to the square of the falling time.C. A graph of velocity vs. time is linear. D. Objects speed up as they free-fall. One of your answers can be explained using a basic equation of motion. Which answer is it and which equation is it? One of your answers can be explained using a rule about how acceleration appears on a certain graph. State the graph rule, then explain why “constant acceleration” looks the way that it does on this certain graph.
7. A 50 kg woman standing on a polished floor (with no friction) has a cat jump into her armsand she catches the cat. If the cat has a mass of 5 kg and jumps with a speed of 12 m/sa) How much momentum does the cat have before the woman catches it?b) What happens to the woman/cat combination after the cat is caught?c) What is the total momentum of the woman/cat combination after the cat is caught?d) What is the speed of the woman/cat combination after the cat is caught?
a) The momentum of the cat before the woman catches it = 60 kgm/s
b) After the cat is caught, the woman and the cat move with an equal velocity
c) Total momentum of the woman/cat combination after the cat is caught =
60 kgm/s
d) The speed of the woman/cat combination after the cat is caught = 1.09 m/s
Explanation:Mass of the woman, m₁ = 50 kg
The woman was standing before she catches the cat. This means that she was at rest and the initial velocity is 0
u₁ = 0 m/s
The mass of the cat, m₂ = 5 kg
The cat's speed, u₂ = 12 m/s
Momentum = Mass x Velocity
a) The momentum of the cat before the woman catches it = m₂u₂
The momentum of the cat before the woman catches it = 5(12)
The momentum of the cat before the woman catches it = 60 kgm/s
b) After the cat is caught, the woman and the cat move with an equal velocity
c)
Total momentum of the woman/cat combination before the cat is caught= Total momentum of the woman/cat combination after the cat is caught
The initial momentum of the woman = m₁u₁
The initial momentum of the woman = 50(0) = 0 kgm/s
Total momentum of the woman/cat combination after the cat is caught = 0 + 60
Total momentum of the woman/cat combination after the cat is caught =
60 kgm/s
d) The speed of the woman/cat combination after the cat is caught
m₁u₁ + m₂u₂ = (m₁ + m₂)v
50(0) + 5(12) = (50+5)v
60 = 55v
v = 60/55
v = 1.09 m/s
The speed of the woman/cat combination after the cat is caught = 1.09 m/s
Point p, to me, doesn't look like a conjunction point where I can apply Kirchhoff's first law, which is the sum of the current in is equal to the sum of the current out. And how do I explain using charges that flow through?
ANSWER
a) The current at P is 1.0A and it is moving towards the node.
b) A total of 1C charges flow through P in 1.0 seconds
EXPLANATION
a) To find this, we have to apply Kirchoff's node law which states that:
Therefore, we have that:
[tex]4.0+3.0-8.0+P=0[/tex]Solve for P:
[tex]\begin{gathered} 4.0+3.0-8.0+P=0 \\ -1.0+P=0 \\ \Rightarrow P=1.0A \end{gathered}[/tex]This implies that the current at point P is 1A and it is moving towards the node (in the left direction).
b) To find the electric charge that flows through P in 1.0s, we have to use the formula for current:
[tex]I=\frac{q}{t}[/tex]where I = current, q = electric charge and t = time
Therefore, we have to find q:
[tex]\begin{gathered} q=I\cdot t \\ \Rightarrow q=1.0\cdot1.0 \\ q=1.0C \end{gathered}[/tex]Therefore, a total of 1C charges flow through P in 1.0 seconds.
ball is launched with an initial speed of 30 m/s making an angle of 45° above the horizontal. How long does it take the ball to reach a vertical displacement Δy= +5 m for the first time?*1.12 s0.54s0.9s0.25s0.7 s
Answer: t = 0.25 s
Explanation:
To find the time it will take the ball to travel a vertical height of 5m, we would apply one of Newton's equation of motion shown below
h = ut + 1/2gt^2
where
h is the height or vertical displacement
u is the initial velocity
g is the acceleration due to gravity
t is the time taken to reach height h
From the information given,
h = 5
g = - 9.8 m/s^2 because the ball is decelerating while moving upwards.
Given that the initial velocity is 30m/s and it was launched at an angle of 45 degrees, we would find the y or vertical component of the velocity. Thus,
u = 30sin45
By substituting these values into the equation, we have
5 = 30sin45t - 1/2 x 9.8 x t^2
5 = 21.21t - 4.9t^2
4.9t^2 - 21.21t + 5 = 0
This is a quadratic equation. The standard form of a quadratic equation is expressed as
ax^2 + bx + c = 0
By comparing both equations,
a = 4.9
b = - 21.21
c = 5
We would solve the equation by using the quadratic formula which is expressed as
[tex]\begin{gathered} x\text{ = }\frac{-\text{ b }\pm\sqrt{b^2-4ac}}{2a} \\ x\text{ = }\frac{-\text{ - 21.21 }\pm\sqrt{-\text{ 21.21}^2-4(4.9\text{ }\times5}}{2\times4.9} \\ x\text{ = }\frac{21.21\pm\sqrt{449.8641\text{ - 98}}}{9.8} \\ x\text{ = }\frac{21.21\text{ }\pm18.758}{9.8} \\ x\text{ = }\frac{21.21\text{ + 18.758}}{9.8}\text{ or x = }\frac{21.21\text{ - 18.758}}{9.8} \\ x\text{ = 4.08 or x = 0.25} \end{gathered}[/tex]Replacing x with t, we have
t = 0.25 s
If you ride a bicycle 36 m in 12 seconds, what is your speed?O A. 0.3 m/sB. 2 m/sC. 3 m/sD. 0.09 m/s
We are given that the distance is 36 m and the time is 12 seconds.
We are asked to find the speed.
Recall that the time, speed, and distance formula is given by
[tex]v=\frac{x}{t}[/tex]Where v is the speed, x is the distance, and t is the time.
Let us substitute the given values into the above formula.
[tex]\begin{gathered} v=\frac{36}{12} \\ v=3\; \; \frac{m}{s} \end{gathered}[/tex]Therefore, the speed is 3 m/s
Option C is the correct answer.
A student pushes a heavy box that is originally stationary across the floor with a force 112 N. What is the coefficient of static friction between the classroom floor and the 140 N box?
Given data:
* The force applied on the heavy box is,
[tex]F_a=112\text{ N}[/tex]* The weight of the heavy box is,
[tex]w=140\text{ N}[/tex]Solution:
The frictional force acting on the box is,
[tex]\begin{gathered} F_r=\mu\times w \\ \mu\text{ is the coefficient of friction,} \end{gathered}[/tex]The frictional force acting on the heavy box is equal to the applied force as the heavy box is not moving under the action of applied force.
Thus,
[tex]\begin{gathered} F_r=F_a \\ \mu\times w=112 \\ \mu\times140=112 \\ \mu=\frac{112}{140} \end{gathered}[/tex]By simplifying,
[tex]\mu=0.8\text{ }[/tex]Thus, the coefficient of friction between the floor and the box is 0.8.
An astronaut drops two pieces of paper from the door of a lunar landing module. One piece of paper is
crumpled, and the other piece is folded into an airplane. Why do the two pieces of paper land on the Moon's
surface at the same time? (1 point)
O The Moon's gravity is much weaker than Earth's.
O The mass of the paper folded into an airplane must be greater than the mass of the crumpled paper.
O The pieces of paper were not dropped from a sufficient height for air resistance to affect their falls.
O The Moon has practically no atmosphere, so there is no air resistance.
The moon has practically no atmosphere so there is no air resistance.
What is Lunar Landing Module?Grumman created the two-stage Apollo Lunar Module (LM) to transport two men from lunar orbit to the lunar surface and back. The ascending rocket engine, equipment bays, and pressurized crew quarters made up the higher ascent stage.
The landing gear, descent rocket engine, and lunar surface experiments were all located on the lower descent stage.
For a second unmanned Earth-orbit test trip, LM 2 was constructed. A second unmanned LM test mission was deemed unnecessary because the LM 1 test flight, which was carried out as part of the Apollo 5 mission, was so successful.
Therefore, The moon has practically no atmosphere so there is no air resistance.
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how much work is required to stop an electron which is moving with a speed of 1.10x10^6m/s
Work required to stop an electron which is moving with a speed of 1.10x10^6m/s is [tex]-\frac{3.971}{5^{19}\cdot \:1048576}[/tex]
What is electron?There are three main types of particles in an atom: protons, neutrons, and an electron that is bonded to an atom. For a variety of reasons, electrons are different from other particles. They possess both wave-like and particle-like properties, exist outside of the nucleus, have a substantially lower mass than protons. A particle that is not composed of smaller parts is an electron, which is also an elementary particle. They are not elementary particles because quarks are assumed to make up protons and neutrons.
We can solve the problem by using the work-energy theorem: in fact, the work done on the electron must be equal to its change in kinetic energy, therefore
[tex]W=K_{f}-K_{i}[/tex]
[tex]W=\frac{1}{2} mv^{2} - \frac{1}{2} mu^{2}[/tex]
where,
m=9.11 × 10⁻³¹kg is the mass of the electron
v is the final velocity of the electron
u is the initial velocity of the electron
In the problem, we are told that the electron is initially moving at,
v = 1.90×10⁶ m/s
Therefore, the work required to stop it as
W = -1/2 mv²
W = -1/2 (1.10×10⁻³¹)(1.90×10⁶)²
[tex]W= -\frac{3.971}{5^{19}\cdot \:1048576}\\[/tex]
and the work is negative since it is in the opposite direction to the motion of the electron.
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convert 675 nm to meters
Given,
The number needed to be converted is 675 nm
A nanometer is equal to one billionth of a meter.
Mathematically,
[tex]1nm^{}=1\times10^{-9}\text{ m}[/tex]Thus to convert any number from nanometer to meter, we need to multiply the number by 10⁻⁹ m.
Thus,
[tex]675\text{ nm}=675\times10^{-9}\text{ m}[/tex]Therefore 675 nm is equal to 675×10⁻⁹ m
Which of the following choices correctly describes the orbital relationship between Earth and the sun?
a.
The sun orbits Earth in a perfect circle
c.
The sun orbits Earth in an ellipse, with Earth at one focus
b.
Earth orbits the sun in a perfect circle
d.
Earth orbits the sun in an ellipse, with the sun at one focus
The correct statement which describes the relationship between earth and the sun is that earth orbits the sun in an ellipse, with the sun at one focus.
What is an Orbit?An orbit in celestial mechanics is the curved path taken by an object, such as the path taken by a planet around with a star, a celestial body around a planet, or a manufactured satellite around an object or location in space, such as a planet, satellite, meteorite, or Lagrange point.
Ordinarily, the physics term "orbit" implies a trajectory that repeats itself with a regular basis, though it can also denote a non-repeating trajectory.
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Please help me solve From a previous question, the index of refraction of the liquid is 1.37
ANSWER
EXPLANATION
From the previous part, we have that the index of refraction of the liquid is 1.37, so we have to replace this in the equation and solve,
[tex]\sin\theta_c=\frac{n_{air}}{n_{liquid}}=\frac{1.00}{1.37}\approx0.73[/tex]And then, take the inverse of the sine to find the critical angle,
[tex]\theta_c=\sin^{-1}0.73\approx46.9\degree[/tex]Hence, the critical angle is 46.9°, rounded to the nearest tenth.
Jontell finds a giant spring that has a 350N/m spring constant. Jontell figures out how to compress is 1.5m. If Jontell and his spring propelled cart's mass is 85kg, how fast is he going after the push fromthe spring? PEspring1/2*k*x?^2
Given data
*The given mass of the propelled cart's is m = 85 kg
*The given spring constant is k = 350 N/m
*The spring compresses at a distance is x = 1.5 m
The formula for the speed is given by the conservation of energy as
[tex]\begin{gathered} U_{p.e}=U_k \\ \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v=\sqrt[]{\frac{kx^2}{m}} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=\sqrt[]{\frac{350\times(1.5)^2}{85}} \\ =3.04\text{ m/s} \end{gathered}[/tex]If the period of a certain wave (wavelength = 4.5 m) is 2 seconds, what is the speed of the wave?1) 0.44 m/s2) 1.1 m/s3) 9.0 m/s4) 2.3 m/s
Given data:
*The given wavelength of the wave is
[tex]\lambda=4.5\text{ m}[/tex]*The given time is t = 2 s
The formula for the speed of the wave is given as
[tex]v=\frac{\lambda}{t}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=\frac{4.5}{2} \\ =2.25\text{ m/s} \\ \approx2.3\text{ m/s} \end{gathered}[/tex]Hence, the speed of the wave is v = 2.3 m/s and the correct option is (4)
Why doesn't an object falling from an airplane continue to accelerate? (1 point)
O Gravity's force diminishes as the object nears the ground.
O Air resistance on the object will eventually equal the force of force of gravity.
O The object's weight varies as it nears the ground.
O Hitting the ground stops the object's acceleration.
A falling object accelerates as it descends. The quantity of air resistance rises in proportion to the speed. The pull of gravity eventually is balanced by the force of air resistance as it grows. The item will cease accelerating since there is no net force at this point in time (0 Newton).
Since the upward force of air resistance eventually equals the downward force of gravity, a falling item cannot continue to accelerate indefinitely before reaching its terminal velocity.In contrast to air resistance, which operates in the opposite direction and slows acceleration, gravity causes objects to accelerate downhill. Greater surface area falling objects encounter more air resistance. In the absence of air, or in a vacuum, all objects fall with the same precise rate of acceleration.To know more about gravity
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Answer: (B). Air resistance on the object will eventually equal the force of force of gravity.
Explanation:
Numerical problem on Coulomb's law 1. Find the magnitude of product of two charges kept 1 metre aparat when a active force of 9.10^9 newton is acting between them.
In order to calculate the product of charges, we can use the formula below for the force acting on two charges (Coulomb law):
[tex]|F|=k_e\frac{|q_1q_2|_{}}{r^2}[/tex]Where ke is the Coulomb constant (ke = 9 * 10^9), q1 and q2 are the charges and r is the distance between the charges.
So, for r = 1 and F = 9 * 10^9, we have:
[tex]\begin{gathered} 9\cdot10^9=9\cdot10^9\frac{|q_1q_2|}{1^2} \\ |q_1q_2|=1 \end{gathered}[/tex]Therefore the product of the charges is 1 Coulomb.
A 5000. kg elevator accelerates at 3.0m/s? as it begins to move upward. Calculate the tension in the cable
Given data;
* The mass of the elevator is 5000 kg.
* The acceleration of the elevator is,
[tex]a=3ms^{-2}[/tex]Solution:
The free body diagram of the elevator is,
The weight of the elevator is,
[tex]W=mg[/tex]where m is the mass of the elevator and g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} W=5000\times9.8 \\ W=49000\text{ N} \end{gathered}[/tex]The net force acting on the elevator is,
[tex]F_{\text{net}}=ma[/tex]where a is the acceleration of the elevator moving upwards,
Substituting the known values,
[tex]\begin{gathered} F_{\text{net}}=5000\times3 \\ F_{\text{net}}=15000\text{ N} \end{gathered}[/tex]From the free body diagram, the tension acting on the cable is,
[tex]\begin{gathered} T-W=F_{\text{net}} \\ T=F_{\text{net}}+W \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} T=15000+49000 \\ T=64000\text{ N} \\ T=64\text{ kN} \end{gathered}[/tex]Thus, the tension acting in the cable is 64 kN.