When the reflecting wave flips upside-down on a stretchedstring, which of the following is correct?a The stretched string is with a fixed boundaryb The stretched string is with a free boundaryсThe stretched string may have a fixed boundary ora free.d The given information is not enough.e None of the above is correct.
We are given a reflecting wave on a string. This can be exemplified in the following diagram:
Calculate the depth in the ocean at which thepressure is three times atmospheric pressure.Atmospheric pressure is 1.013 x 10^5 Pa. Theacceleration of gravity is 9.81 m/s^2and them/sdensity of sea water is 1025 kg/m^3Answer in units of m.
In order to determine the depth in the ocean, use the following equation:
[tex]h=\frac{P-P_o}{\rho g}[/tex]h: depth
P: pressure = 3*Po
Po: atmospheric pressure = 1.013*10^5Pa
g: gravitational acceleration constant = 9.8m/s^2
p: density of water = 1025 kg/m^3
Replace the previoua values into the formula for h and simplify:
[tex]\begin{gathered} h=\frac{3P_o-P_o}{\rho g}=\frac{2P_o}{\rho g} \\ h=\frac{2(1.013\cdot10^5Pa)}{(1025\frac{kg}{m^3})(9.8\frac{m}{s^2})}\approx20.17m \end{gathered}[/tex]Hence, the depth in the ocean is approximately 20.17m
b. Andrea has a particularly old bike, and she has not lubricated the chain orgears. This causes friction inside the gears as she rides. She begins coasting onflat ground at a speed of 10 m/s, and after a few seconds is only going 8 m/s.How much energy has been converted to heat? (1 point) (The combined mass is 50kg)c. Andrea is going 10 m/s toward a hill. She coasts up the hill without pedaling. Iffriction causes 10% of her energy to be converted to heat inside the gears, howhigh up the hill will she be able to coast? (2 points)
b.
The energy converted in heat is the change in kinetic energy, that is, the kinetic energy lost is the same as the heat.
The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]The change in kinetic energy is given by:
[tex]K_f-K_0=\frac{1}{2}mv^2_f-\frac{1}{2}mv^2_0[/tex]Plugging the values given we have that:
[tex]\begin{gathered} K_f-K_0=\frac{1}{2}(50)(8)^2-\frac{1}{2}(50)(10)^2 \\ =-900 \end{gathered}[/tex]Hence Andrea has lost 900 J of energy and therefore 900 J is converted to heat.
c.
We know that the energy is conserved, in this case this means that the initial kinetic energy has to be equal to the energy lost by heat in the gears and the potential energy gained by climbing up the hill, that is:
[tex]K=Q+U[/tex]where K is the kinetic energy, Q is the heat and U is the potential energy.
We know that 10% of her energy is converted in heat, this means that:
[tex]Q=0.1K[/tex]and hece we have:
[tex]\begin{gathered} K=0.1K+U \\ U=0.9K \end{gathered}[/tex]the potential energy is given by:
[tex]U=mgh[/tex]then we have that:
[tex]mgh=0.9(\frac{1}{2}mv^2)[/tex]Plugging the values given and solving for h we have:
[tex]\begin{gathered} (50)(9.8)h=0.9(\frac{1}{2}(50)(10^2)) \\ 490h=2250 \\ h=\frac{2250}{490} \\ h=4.59 \end{gathered}[/tex]Therefore Andrea will be able to climp up to 4.59 meters
Forces that are equal in magnitude
but opposite in direction will:
Answer:
The two forces equal in magnitude but acting opposite in direction on a body are called balanced forces
Explanation:
Three bulbs of resistance 100. Ω, 200, Ω and 300 Ω are connected in parallel to a 120. V DC power supply. Draw the diagram and find thea) current in each bulb b) current drawn from the power supplyc) total power drawn power supply d) the net resistance of all bulbs
Let's use the formula for electric current.
[tex]I=\frac{V}{R}[/tex]Where V is the power supply 120 V, and R is the resistance. Let's find the current in each bulb.
[tex]\begin{gathered} I=\frac{120V}{100\Omega}=1.20A \\ I=\frac{120V}{200\Omega}=0.6A \\ I=\frac{120V}{300\Omega}=0.4A \end{gathered}[/tex](a) The current in each bulb is 1.20A, 0.6A, and 0.4A, respectively.(b) (c) The diagram of the circuit isTo find the net resistance, we use the following formula.
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}[/tex]Let's use the given magnitudes.
[tex]\begin{gathered} \frac{1}{R}=\frac{1}{100\Omega}+\frac{1}{200\Omega}+\frac{1}{300\Omega} \\ \frac{1}{R}=\frac{6+3+2}{600\Omega} \\ \frac{1}{R}=\frac{11}{600\Omega} \\ R=\frac{600}{11}\Omega \\ R\approx54.55\Omega \end{gathered}[/tex]Therefore, the net resistance of all bulbs is 54.55 ohms.How much work is done when a 25.0 kg object is lifted 3.00 m?
The amount of work done is 735J.
We all know that the potential energy U is equal to the work we must do against the force for moving an object from the reference point to the exact position.
The work done is equal to the potential energy through the work energy theorem.
W = mgh
here,
m = mass
g = gravitational acceleration
h = height
W = work done
On solving the above equation
W = 25x9.8x3
W = 75x9.8
W = 735 J
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A hammer is used to hit a nail into a block of wood. The hammer hits the nail with a speed of 8.0 m/s and then stops. The hammer is in contact with the nail for 0.0015 s.,hammer has mass 0.15 kg.Calculate the average force between the hammer and the nail.
800 Newtons
Explanation
The average force is the force exerted by a body moving at a defined rate of speed (velocity) for a defined period of time.
the average force is given by:
[tex]F=ma[/tex]and
[tex]a=\frac{\Delta v}{\Delta t}[/tex][tex]\begin{gathered} F_{average}=m\frac{\Delta v}{\Delta t} \\ where\text{ m is the mass of the objectt} \\ \Delta v\text{ is the change in velocity} \\ \Delta t=\text{ time} \end{gathered}[/tex]Step 1
a) Let
[tex]\begin{gathered} m=0.0015\text{ kg} \\ \Delta v=0.0015s \\ \Delta v=8\text{ }\frac{m}{s} \end{gathered}[/tex]now, replace
[tex]\begin{gathered} F=0.15\text{ kg}\frac{0-8\frac{m}{s}}{0.0015\text{ s}} \\ F=-800\text{ N} \end{gathered}[/tex]the negative sign indicates the force is in the opposite way ( the force is exerted by the nail to the hammer), so the force is opposite to the direction of the movement
so, the answer is
800 Newtons
I hope this helps you
Which of the following is an example of Newton's third law of motion?A. A skydiver slows down when her parachute opens.B. A grocery cart moves forward when it is pushed.C. A cannon recoils backwards when it is fired.D. A rolling rock slows down due to friction.
Explanation:
The third law of Newton says that when an object exerts a force on a second object, the first object experiences an equal and opposite force that is exerted by the second object.
So, the example that shows this law is:
C. A cannon recoils backward when it is fired.
Because the cannon e
S=3-2t+3t^2
What is the instantaneous velocity and it’s acceleration at t=3s
At what time is the particle at rest
Answer:
Explanation:
Given:
X(t) = 3 - 2*t + 3*t²
t = 3 s
_______________
V(t) - ?
a(t) - ?
Speed is the first derivative of the coordinate, acceleration is the second.
1)
V(t) = X' = (3 - 2*t + 3*t²)' = 0 - 2 +6*t = 6*t - 2
V(3) = 6*3 - 2 = 16 m/s
2)
a(t) = X'' = V' = (6*t - 2)' = 6 m/s²
a(3) = 6 m/s²
3)
The body will stop (V = 0 ) in (t) seconds:
V(t) = 6*t - 2
0 = 6*t - 2
6*t = 2
t = 2/6 = 1/3 ≈ 0,33 s
It’s an assessment Jane multiplied 825x22 and got 3,300. Flynn multiples the same numbers and got 18,150 Which student is correct?
In order to determine which student is correct, multiply the given numbers, as follow:
8 2 5
x 2 2
1 6 5 0
1 6 5 0
1 8 1 5 0
As you can notice, the result of the multiplcation is 18,150, hence, Flynn is right.
A light, inextensible cord passes over alight, frictionless pulley with a radius of15 cm. It has a(n) 18 kg mass on the left and a(n) 2.6 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 1.5 m apart.The acceleration of gravity is 9.8 m/s².
At what rate are the two masses accelerating when they pass each other answer in units of m/s^2 boo
Answer:
have you heard of quizlet
Explanation:
they are very helpful to these things
Three vectors are shown in this figure. Their respective moduli are A = 4.00m.B = 3, 20m and C = 2.70mCalculate 2.00 A - B + 1.30 CExpress your answer according toa) Unit vectorsb) The modulus and orientation with respect to the positive part of the x-axis
Given that,
Modulus of vector A=4.00
The angle made by the vector A with the y axis, θ₁=33.0°
The modulus of vector B=3.20 m
The angle made by the vector B with the x-axis is θ₂=40.0+90.0=130°
The modulus of the vector C=2.70 m
The angle made by the vector C with x-axis θ₃=-90°
The x and y components of the vector can be written as
[tex]\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \end{gathered}[/tex]Where r is the magnitude (or modulus) of the vector and θ is the angle made by the vector.
Or a vector, in cartesian coordinates, can be written as,
[tex]R=r\cos \theta\hat{\text{i}}+r\sin \theta\hat{j}[/tex]Therefore, vector A is cartesian coordinates is
[tex]\begin{gathered} \vec{A}=4.00\cos 33^{\circ}\hat{i}+4.00\sin 33.0^{\circ}\hat{j} \\ =3.35\hat{i}+2.18\hat{j} \end{gathered}[/tex]And the vector B is
[tex]\begin{gathered} \vec{B}=3.20\cos (130^{\circ})\hat{i}+3.20\sin (130^{\circ})\hat{j} \\ =-2.06\hat{i}+2.45\hat{j} \end{gathered}[/tex]And vector C is given by,
[tex]\begin{gathered} \vec{C}=2.70\cos (-90^{\circ})\hat{i}+2.70\sin (-90^{\circ})\hat{j} \\ =-2.7\hat{j} \end{gathered}[/tex]The given equation is
[tex]2.00\vec{A}-\vec{B}+1.30\vec{C}[/tex]Let this represents a vector V
On substituting the known values,
[tex]\begin{gathered} \vec{V}=2.00\vec{A}-\vec{B}+1.30\vec{C} \\ =2.00\times(3.35\hat{i}+2.18\hat{j})-(-2.06\hat{i}+2.45\hat{j)}+1.30(-2.7\hat{j}) \\ =8.76\hat{i}-1.6\hat{j} \end{gathered}[/tex](a) This is the representation with the unit vectors, where i and j are the unit vectors along the x-axis and y-axis respectively.
[tex]\vec{V}=8.76\hat{i}-1.6\hat{j}[/tex]b) The modulus of any vector is the square root of the sum of the squares of its components.
That is, the magnitude of the vector V is
[tex]\begin{gathered} V=\sqrt[]{8.76^2+(-1.6)^2} \\ =8.90\text{ m} \end{gathered}[/tex]The angle of this vector with the x-axis is given by
[tex]\begin{gathered} \phi=\tan ^{-1}(\frac{-1.6}{8.75}) \\ =-10.36^{\circ}^{} \end{gathered}[/tex]The negative sign indicates that the vector is below the positive x-axis
Therefore the modulus of the resultant of the above equation is 8.90 m and its angle with the positive x-axis is -10.36°
For each letter, write a word to describe its role:A + B = C D x E = F
A is an addend
B is an addend
C is a sum
D is a factor
E is a factor
F is a product
Explanations:Note:Numbers(or characters) that are added together with the addition operator are called addends
The result of an addition operation is called sum
Numbers that are multiplied together are called factors
The result of a multiplication operator is called product
Considering the definitions above:In A + B = C
A is an addend
B is an addend
C is a sum
In D x E = F
D is a factor
E is a factor
F is a product
Point charges create equipotential lines that are circular around the charge (in the plane of the paper). What is the potential energy, in nJ, of a 1 nC charge located 1.99 m from a 2 nC charge ?
The potential energy between two charges can be written as:
[tex]U_e=\frac{kq_1q_2}{r}[/tex]In our case, it'll be equal to:
[tex]U_e=\frac{9*10^9*1*10^{-9}*2*10^{-9}}{1.99}=9.045nJ[/tex]Then, our answer is PE=9.045nJ
Hunter pushed a couch across the room. He did 800 J of work in 20 seconds.The couch weighed 500 N. How much power did he have?A. 16,000 WB. 1.6WC. 800 WD. 40 W
Power = 40 W
Option D
Explanation:The workdone by the Hunter = 800 Joules
The weight of the couch = 500 N
Time, t = 20 seconds
Power = Workdone/Time
Power = 800/20
Power = 40 W
A golf ball is initially on a tee when it is
struck by a golfer. The ball is given an
initial velocity of 50 m/s at a 37° angle. The
ball hits the side of a building that is 200
meters horizontally away from the golfer.
(a) What are the horizontal and vertical
components of the ball's initial
velocity?
(b) How much time elapses before the
ball strikes the side of the building?
(c) How far from the ground does the ball
strike the building?
Answer:
a.
[tex]horizontal=39.9[/tex] m/s
[tex]vertical=30.1[/tex] m/s
b.
[tex]t=5.009[/tex]
c.
[tex]y=27.7[/tex]
Explanation:
Lets write down what we were given.
Angle = 37°
Initial Velocity = 50 m/s
Displacement in x direction = 200 m
Take note:
I am having some trouble with the theta symbol so let theta = [tex]N[/tex]
Lets do question C first.
We know that time is equal to [tex]\frac{displacement}{velocity}[/tex] aka [tex]t=\frac{x}{v}[/tex].
[tex]x=v[/tex]₀ₓ [tex]t[/tex] ⇒ [tex]\frac{x}{v_{0x} }[/tex] ⇒ [tex]\frac{x}{v_{0} *cos(N)}[/tex]
Now substitute the expression for t into the equation for the position.
[tex]y=(v_{0}sin(N))*(\frac{x}{v_{0}cos(N) })-\frac{1}{2}g(\frac{x}{v_{0}cos(N) }) ^{2}[/tex]
Rearranging terms, we have
[tex]y=(tan(N)*x)-[\frac{g}{2(v_{0}cos(N))^{2} } ]x^{2}[/tex]
Now lets substitute our numbers in for the variables. Then simplify.
[tex]y=(tan37*200)-[\frac{9.81}{2(50*cos37)^{2} } ]200^{2}[/tex]
[tex]y=150.7108-[\frac{9.81}{2(50*cos37)^{2} } ]200^{2}[/tex]
[tex]y=150.7108-[0.0030761]200^{2}[/tex]
[tex]y=150.7108-(0.0030761*40000)[/tex]
[tex]y=150.7108-123.0444[/tex]
[tex]y=27.7[/tex]
Now lets do question B.
Lets steal this from the last question.
We know that time is equal to [tex]\frac{displacement}{velocity}[/tex] aka [tex]t=\frac{x}{v}[/tex].
[tex]x=v[/tex]₀ₓ [tex]t[/tex] ⇒ [tex]\frac{x}{v_{0x} }[/tex] ⇒ [tex]\frac{x}{v_{0} *cos(N)}[/tex]
Now substitute the expression for t into the equation for the position.
[tex]y=(v_{0}sin(N))*(\frac{x}{v_{0}cos(N) })-\frac{1}{2}g(\frac{x}{v_{0}cos(N) }) ^{2}[/tex]
We can substitute [tex]t[/tex] for [tex]\frac{x}{v_{0}cos(N) }[/tex]
[tex]y=(v_{0}sin(N))*(t)-\frac{1}{2}g(t) ^{2}[/tex]
We can rewrite the equation as
[tex](v_{0}sin(N)(t)-\frac{1}{2}*(g(t)^{2})=y[/tex]
Now lets substitute our numbers in for the variables.
[tex](50sin(37)(t)-\frac{1}{2}*(9.81(t)^{2})=27.7[/tex]
After some painful algebra and factoring we get
[tex]30.09075115t-4.905t^{2}=27.6664[/tex]
Subtract [tex]27.6664[/tex] from both sides.
[tex]30.09075115t-4.905t^{2}-27.6664=0[/tex]
Use the quadratic formula to find the solutions.
[tex]\frac{-b+-\sqrt{b^{2}-4ac } }{2a}[/tex]
After some more painful algebra we get
[tex]t=5.00854263, 1.12616708[/tex]
1.126 does not make any sense so.
[tex]t=5.009[/tex]
Finally lets do question A.
Lets draw a triangle. We have the velocity which is the hypotenuse and we have the angle. From there we can solve for the opposite and adjacent sides.
Let [tex]A=horizontal[/tex] and [tex]O=vertical[/tex]
[tex]cos(37)=\frac{A}{50}[/tex]
[tex]A=39.9[/tex]
[tex]sin37=\frac{O}{50}[/tex]
[tex]O=30.1[/tex]
How much power is created when you perform 55 Joule of work with a time 20 sec?
Answer:
2.75 watts
Explanation:
The power is equal to the work divided by time, so
P = W/t
Then, replacing W = 55 J and t = 20 sec, we get:
P = 55 J / 20 s = 2.75 Watts
Therefore, the power created is 2.75 watts
Identify the kinematic equation which relates the velocity and time.
The kinematic equation which relates velocity and time is
[tex]v=v_0+at[/tex]As when the acceleratio
Use Newton’s Law of Universal Gravitation and Newton’s Second Law to
Find g = acceleration due to gravity
Show g is independent of mass
By equating the two forces, acceleration due to gravity g is obtained which is independent of mass
What is Newton’s Second Law ?The law state that the rate of change of momentum is directly proportional to the force applied.
From Newton’s Second Law, Force F = mg. And from Newton’s Law of Universal Gravitation, Force F = GMm/r²
Where m is the mass of the satellite or the body revolving round the earth.
Equate the two forces.
mg = GMm/r²
The two m cancelled out leaving
g = GM/r²
Where
g = Acceleration due to gravityM = Mass of the earthG = Universal gravitational constantr = Distance between them.Therefore, since the mass of the satellite m has cancelled out, acceleration due to gravity g is independent of mass.
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According to Newton’s second law of motion,how much force will be required to accelerate an object at the same rate if it mass is reduced by half ?
According to Newton's second law of motion,
[tex]\text{Force = mass}\times acceleration[/tex]Let the initial force be F, acceleration be a and the initial mass be m.
The acceleration is the same but now the mass is reduced by half.
So, the force will be
[tex]\begin{gathered} F^{\prime}=\frac{m}{2}\times a \\ =\frac{F}{2} \end{gathered}[/tex]Thus, the force will also be half of the initial force if the mass is reduced by half.
Tsunami waves generally carry a mass (m) of 770 kg of water, travel at a velocity (v) of approximately 10 m/s and have a height (h) of 10 m at landfall. The colony structures can withstand a total energy (TE) 135,000 J before catastrophic damage occurs.ANSWER (a) AND (b)(a) Using your answers from #4 and #5 calculate the total energy (TE) of a tsunami wave. TE = KE + PE (b) Using your calculations and the provided data, explain to the colonizing council whether this crash site can be used to start a colony.
ANSWER:
(a)
Potential energy = 75460 J
Kinetic energy = 38500 J
Total energy = 113960 J
(b)
The site can be used to start a colony.
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 770 kg
Velocity (v) = 10m/s
Height (h) = 10 m
(a)
We calculate in each case the kinetic and potential energy by means of their formulas
[tex]\begin{gathered} E_k=\frac{1}{2}m\cdot v^2=\frac{1}{2}\cdot770\cdot10^2=38500\text{ J} \\ E_p=m\cdot g\cdot h=770\cdot9.8\cdot10=75460\text{ J} \end{gathered}[/tex]The total energy is the sum of both calculated energies:
[tex]\begin{gathered} E_T=38500+75460 \\ E_T=113960\text{ J} \end{gathered}[/tex](b)
Since the tsunami energy is less than the energy that can destroy the colony, then the site can support a permanent colony.
what do fusion and fission have in common
Answer:
Explanation:
they both involve nuclear reactions that produce energy, but the application are not the same
How much work does Scott do to push a 74 kg sofa 2.1 m across the floor at a constant speed? The coefficient of kinetic friction between the sofa and the floor is 0.23.
Work does Scott do to push a 74 kg sofa 2.1 m across the floor at a constant speed. The coefficient of kinetic friction between the sofa and the floor is 0.23 is 349 Nm.
given that :
mass = 74 kg
distance d = 2.1 m
coefficient of kinetic friction , μk = 0.23
work done is given as :
w = fd
f = μk m g
f = 0.23 × 74 × 9.8
f = 166 N
therefore ,
work = fd
w = 168 × 2.1
w = 349 Nm
Work does Scott do to push a 74 kg sofa 2.1 m across the floor at a constant speed. The coefficient of kinetic friction between the sofa and the floor is 0.23 is 349 Nm.
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johns mass is 92.0kg on the earth, what is his mass on mars where g= 3.72m/s^2
The mass is a fundamental property of all matter.
Mass is not the same as weight, the weight depends on the gravity while the mass does not. Therefore the John's mass is the same in mars.
Which has more kinetic energy: a 0.0014-kg bullet traveling at 397 m/s or a 5.9 107-kg ocean liner traveling at 13 m/s (25 knots)?
the bullet has greater kinetic energy
the ocean liner has greater kinetic energy
Justify your answer.
Ek-bullet =
J
Ek-ocean liner =
J
Answer:
Ocean Liner because its mass is 5000 times more and the bullet's velocity squared is only 1600 times more. This means that the kinetic energy of the ocean liner will be roughly 3 times greater.
Explanation:
The kinetic energy of an object is dependent on mass and velocity.
E = ( 1 / 2 )mv²;
Energy is measured in Joules which is equal to kilogram metres squared per second squared.
1J = ( kg )( m² )( s⁻² ) or 1J = ( kg )( m² ) / ( s² )
We can substitute the mass and velocity directly into the equation because the question gives the values in metres and kilograms.
BULLET:
E = ( 1 / 2 )( 0.0014kg )( 397m / s )²;
E = ( 0.0007kg )( 157609m² )( s⁻² );
E = 110.3263kgm²s⁻²;
E = 110.3263J;
I will just skip some steps because you get the idea.
OCEAN LINER:
E = ( 1 / 2 )( 5.9107kg )( 13m / s )²;
E = 499.45415J;
An object of mass m moves a circular path with a constant speed v. The centripetal force of the object is F. If the objects speed were halved in the mass was tripled, what would happen to the centripetal force?
An object of mass m moves a circular path with a constant speed v. The centripetal force of the object is F. If the object's speed were halved in the mass was tripled, then the centripetal force would be 0.75 times the original centripetal force.
What is a uniform circular motion?It is defined as motion when the object is moving in a circle with a constant speed and its velocity is changing with every moment because of the change of direction but the speed of the object is constant in a uniform circular motion.
A mass m object travels in a circle at a constant speed v. The object's centripetal force is F. The centripetal force would be 0.75 times greater if the object's mass were tripled and its speed was cut in half.
Centripetal force = m × v²/r
=3m × (0.5v)² / r
= 0.75 mv² / r
Thus, the centripetal force would become 0.75 times the original centripetal force.
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You push a 2.1 kg object on a table with 42N of force. The box then slowly skids to a stop over 2.2 m. What is the force of friction?
If you push a 2.1 kg object on a table with 42N of force. The box then slowly skids to a stop over 2.2 meters, then the force of the friction would be 42 Newtons as per the concept of limiting friction.
What is friction?Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.
As given in the problem if push a 2.1 kg object on a table with 42N of force. The box then slowly skids to a stop over 2.2 m, then we have to find the force of the friction.
The force of the friction = The limiting friction force
If you apply 42N of force to a 2.1-kilogram item on a table. According to the theory of limiting friction, when the box gently slides to a halt over a distance of 2.2 meters, the force of friction would be 42 Newtons.
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A convex spherical mirror has a radius of curvatureof 9 40 cm. A) Calculate the location of the image formed by an 7.75mm tall object whose distance from the mirror is 17.5 cmCalculate the size of the imageC) Calculate the location of the image formed by an 7.75mm tall object whose distance from the mirror is 10.0cmE) Calculate the location of the image formed by an 7.75mm tall object whose distance from the mirror is 2.65cmG) Calculate the location of the image formed by an 7.75mm tall object whose distance from the mirror is 9.60m
0.We are asked to determine the location of an image formed by an 7.75mm tall object that is located a distance of 17.5 cm from a convex mirror.
First, we will calculate the focal length using the following formula:
[tex]f=-\frac{R}{2}[/tex]Where:
[tex]\begin{gathered} f=\text{ focal length} \\ R=\text{ radius} \end{gathered}[/tex]Substituting the values we get:
[tex]f=-\frac{9.40cm}{2}[/tex]Solving the operations:
[tex]f=-4.7cm[/tex]Now, we use the following formula:
[tex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/tex]Where:
[tex]\begin{gathered} d_0=\text{ distance of the object} \\ d_i=\text{ distance of the image} \end{gathered}[/tex]Now, we substitute the known values:
[tex]\frac{1}{17.5cm}+\frac{1}{d_i}=-\frac{1}{4.7cm}[/tex]Now, we solve for the distance of the image. First, we subtract 1/17.5 from both sides:
[tex]\frac{1}{d_i}=-\frac{1}{4.7cm}-\frac{1}{17.5cm}[/tex]Solving the operation:
[tex]\frac{1}{d_i}=-0.27\frac{1}{cm}[/tex]Now, we invert both sides:
[tex]d_i=\frac{1}{-0.27}cm=-3.7cm[/tex]Therefore. the location of the image is -3.7 centimeters.
The other parts are solved using the same procedure.
Part B. To calculate the size of the image we will use the following relationship:
[tex]\frac{h_i}{h_o}=-\frac{d_i}{d_0}[/tex]Where:
[tex]h_i,h_0=\text{ height of the image and height of the object}[/tex]Substituting we get:
[tex]\frac{h_i}{7.75mm}=-\frac{-3.7cm}{17.5cm}[/tex]Solving the operations on the right side:
[tex]\frac{h_i}{7.75mm}=0.21[/tex]Now, we multiply both sides by 7.75:
[tex]h_i=(7.75mm)(0.21)[/tex]Solving the operations:
[tex]h_i=1.64mm[/tex]Therefore, the height of the iamge is 1.64 mm.
Philip jumps up with to a height of 3 m above the ground. What was Philip's initial velocity? round to the tenth.
The initial velocity of Philip was 7.66 m/s
Given data:
The vertical height is h=3 m.
Considering ground as the reference, then the initial potential energy of Philip is zero, i.e., PEi=0
The formula for the kinetic energy is given by,
[tex]KE_i=\frac{1}{2}mv^2[/tex]Here, m is mass and v is the velocity.
After reaching the height of 3 m Philip comes to a stop. It means the final kinetic energy is zero, i.e. KEf=0.
The final potential energy is given by,
[tex]PE_f=mgh[/tex]Here, g is the gravitational acceleration.
Applying the conservation of energy between initial position and final position.
[tex]\begin{gathered} KE_i+PE_i=KE_f+PE_f \\ \frac{1}{2}mv^2+0=0+mgh \\ v=\sqrt[]{2gh} \\ v=\sqrt[]{2\times9.8\times3} \\ v=7.66\text{ m/s} \end{gathered}[/tex]Thus, the initial velocity of Philip was 7.66 m/s.
A flywheel with a moment of inertia of 3.45 kg·m2is initially rotating. In order to stopits rotation, a braking torque of -9.40 N·m is applied to the flywheel. Calculate the initialangular speed of the flywheel if it makes 1 complete revolution from the time the brake isapplied until it comes to rest
Given data
*The given moment of inertia is I = 3.45 kg.m^2
*The given braking torque is T = -9.40 N.m
*The angular distance traveled is
[tex]\theta=(1\times2\pi)rad_{}[/tex]*The final angular speed is
[tex]\omega=0\text{ rad/s}[/tex]The angular acceleration of the flywheel is calculated by using the torque and moment of inertia relation as
[tex]\begin{gathered} T=I\alpha \\ \alpha=\frac{T}{I} \\ =\frac{-9.4}{3.45} \\ =-2.72rad/s^2 \end{gathered}[/tex]The formula for the initial angular speed of the flywheel is given by the rotational equation of motion as
[tex]\omega^2-\omega^2_0=2a\theta[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} (0)^2-\omega^2_0=2\times(-2.72)(2\pi) \\ \omega_0=\sqrt[]{2\times2.72\times2\pi} \\ =5.88\text{ rad/s} \end{gathered}[/tex]Hence, the initial angular speed of the flywheel is 5.88 rad/s
