Answer:
1- Period is the time for which one full rotation is completed. Regardless of their positions on the platform, periods of all three are the same. T2 = T.
2- Similarly, T3 = T.
3- The platform is making rotational motion. So, the relation between the angular velocity and the linear velocity is
[tex]v = \omega R[/tex]
For all the people, angular velocity is the same. Their linear velocities are different.
[tex]\omega = \frac{v_1}{R} = \frac{v_2}{3R/5}\\v_2 = \frac{3v}{5}[/tex]
4- Similarly,
[tex]\omega = \frac{v_1}{R} = \frac{v_2}{R/2}\\v_2 = \frac{v}{2}[/tex]
5- Radial acceleration in constant circular motion is
[tex]a_{rad} = \frac{v^2}{R}[/tex]
For the second person:
[tex]a_2 = \frac{v_2^2}{3R/5} = \frac{(3v/5)^2}{3R/5} = \frac{9v^2/25}{3R/5} = \frac{3v^2}{5R} = 3a/5[/tex]
6- Similarly,
[tex]a_3 = \frac{v_3^2}{R/2} = \frac{(v/2)^2}{R/2} = \frac{v^2/4}{R/2} = \frac{v^2}{2R} = a/2[/tex]
Explanation:
As a result, the period is same for every object on the rotating platform, as they all complete their revolutions at the same time. Their speed and radial acceleration is different according to their distance to the center.
The rotational kinematics relations allow to find the results for the questions about the movement of the three people on the turntable are:
1 and 2) All periods are equal, T₂ = T and T₃ = T.
3) The linear velocity of the 2nd person is: v₂ = [tex]\frac{3}{5} \ v[/tex]
4) The linear velocity of the 3rd person is: v₃ = ½ v
5) The linear acceleration of the 2nd person is: a₂ = [tex]\frac{3}{5} \ a[/tex]
6) The linear acceleration of the 3rd person: a₃ = ½ a
Rotational kinematics studies the rotational motion of bodies looking for relationships between angular position, angular velocity, and angular acceleration.
In the case where the angular accleration is zero, the expression for the velocity is:
[tex]w = \frac{\Delta \theta }{\Delta t}[/tex]
Where w is the angular velocity and Δw and Δt are the variation in angle t over time.
1 and 2)
Indicates that people are on a turntable, the period is when we have a complete rotation θ = 2π rad in time, therefore the period and the angular velocity are related.
[tex]w= \frac{2\pi }{T} \\T = \frac{2\pi }{w}[/tex]
In the apparatus of parks the angular velocity is constant and we see that it does not depend on the radius, therefore the period for all the people is the same.
T = T₁ = T₂
3) They indicate that the speed of the 1 person who is in the position r=R on the plate is v, let's calculate the speed for the 2 person who is in the position r = [tex]\frac{3}{5} \ R[/tex]
Linear and angular variables are related.
v = w r
Let's substitute for the 1st person.
v = w R
For the 2nd person.
v₂= w ( [tex]\frac{3}{5} R[/tex])
We solve these two equations.
[tex]v_2 = \frac{3}{5} \ v[/tex]
4) We carry out the same calculation for the 3rd person.
v₃ = w ½ R
We solve the two equations.
v₃ = ½ v
5) Ask for radial acceleration.
The relationship between radial and angular acceleration is.
a = α R
We substitute for the 1st person.
a = α R
For the second person.
a₂ = α ( [tex]\frac{3}{5} R[/tex])
We solve the two equations
a₂ = [tex]\frac{3}{5} \ a[/tex]
6) Ask the radial acceleration of the 3rd person.
We substitute.
a₃ = α (½ R)
We solve.
a₃ = ½ a
In conclusion, using the rotational kinematics relations we can find the results for the questions about the movement of the three people on the turntable are
1 and 2) All periods are equal, T₂ = T and T₃ = T.
3) The linear velocity of the 2nd person is: v₂ = [tex]\frac{3}{5} \ v[/tex]
4) The linear velocity of the 3rd person is: v₃ = ½ v
5) The linear acceleration of the 2nd person is: a₂ = [tex]\frac{3}{5} \ a[/tex]
6) The linear acceleration of the 3rd person: a₃ = ½ a
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Calculate the rotational inertia of a meter stick, with mass 0.71 kg, about an axis perpendicular to the stick and located at the 18 cm mark. (Treat the stick as a thin rod.)
To solve this problem we will use the parallel axis theorem for which the inertia of a point of an object can be found through the mathematical relation:
[tex]I = I_{cm} +mx^2[/tex]
Where
[tex]I_{cm}[/tex] = Inertia at center of mass
m = mass
x = Displacement of axis.
Our mass is given as 0.71kg,
m = 0.71kg
Para a Stick with length (L) the Moment of Inertia of the stick about and axis passing through the center and perpendicular to stick is
[tex]I_{cm} = \frac{1}{12} mL^2[/tex]
[tex]I_{cm} = \frac{1}{12} (0.71)(1)^2[/tex]
[tex]I_{cm} = 0.05916Kg\cdot m^2[/tex]
The distance between center of mass to the specific location is
[tex]x = 50cm - 18cm[/tex]
[tex]x = 38cm = 0.38m[/tex]
So, from parallel axis theorem ,
[tex]I = I_{cm} + mx^2[/tex]
[tex]I =0.05916Kg\cdot m^2+ (0.71kg)(0.38m)^2[/tex]
[tex]I = 0.161684Kg\cdot m^2[/tex]
Therefore the rotational inertia is [tex]0.161684Kg\cdot m^2[/tex]
The magnetic field in a plane monochromatic electromagnetic wave with wavelength λ = 598 nm, propagating in a vacuum in the z-direction is described by B =(B1sin(kz−ωt))(i^+j^) where B1 = 8.7 X 10-6 T, and i-hat and j-hat are the unit vectors in the +x and +y directions, respectively. What is k, the wavenumber of this wave?
Answer:
For this given plane monochromatic electromagnetic wave with wavelength λ=598 nm, the wavenumber is [tex]k=0,0105\ x\ 10^{-9}\ m^{-1}[/tex] .
Explanation:
For a plane electromagnetic wave we have that the electrical and magnetic field are:
[tex]E(r,t)=E_{0}\ cos ( wt-kr)\\\ B(r,t)=B_{0}\ cos(wt-kr)[/tex]
In this case we have the data for the magnetic field. We are told that the magnetic field in a plane electromagnetic wave with wavelength λ=598 nm, propagating in a vacuum in the z direction ([tex]\hat k[/tex]) is described by
[tex]B=8.7\ x\ 10^{-6}\ T sin(kz-wt) (\hat i+\hat j)[/tex]
([tex]\hat i,\hat j, \hat k[/tex] are the unit vectors in the x,y,z directions respectively)
The wavenumber k is a measure of the spatial frequency of the wave, is defined as the number of radians per unit distance:
[tex]k=\frac{2\pi}{\lambda}[/tex]
where λ is the wavelength
So we get that
[tex]k=\frac{2\pi}{\lambda} \rightarrow k=\frac{2\pi}{598 nm} \rightarrow k=0,0105\ x\ 10^{9}\ m^{-1}[/tex]
The wavenumber is
[tex]k=0,0105\ x\ 10^{9}\ m^{-1}[/tex] .
The headlights of a car are 1.6 m apart and produce light of wavelength 575 nm in vacuum. The pupil of the eye of the observer has a diameter of 4.0 mm and a refractive index of 1.4. What is the maximum distance from the observer that the two headlights can be distinguished?
To solve this problem it is necessary to apply the concepts related to angular resolution, for which it is necessary that the angle is
[tex]\theta = 1.22\frac{\lambda}{nd}[/tex]
Where
d = Diameter of the eye
n = Index of refraction
D = Distance between head lights
[tex]\lambda[/tex]= Wavelength
Replacing with our values we have that
[tex]\theta = 1.22 \frac{(1.22)(575*10{-9})}{1.4(4*10^{-3})}[/tex]
[tex]\theta = 1.252*10^{-4}rad[/tex]
Using the proportion of the arc length we have to
[tex]L = \frac{D}{\theta}[/tex]
Where L is the maximum distance, therefore
[tex]L = \frac{1.6}{1.252*10^{-4}}[/tex]
[tex]L = 12.77km[/tex]
Therefore the maximum distance from the observer that the two headlights can be distinguished is 12.77km
Which of the following statements correctly describes the law of conservation of energy? Group of answer choicesa. Mass cannot be created but it can be destroyed under extreme pressures.b. Mass cannot be conserved during a chemical reaction; a little bit of mass is always lost.c. The mass of a closed system cannot change over time; mass cannot be created nor destroyed.d. When added to a system, energy can destroy mass.
To solve this problem we will also apply the concept related to the conservation of the mass, which announces that: "In an isolated system, during any ordinary chemical reaction, the total mass in the system remains constant, that is, the mass consumed by the reagents is equal to the mass of the products obtained. "
If the mass is in a closed system, it cannot change. This assessment should not be confused with the transformation of the matter within it, for which it is possible that over time the matter will change from one form to another. For example during a chemical reaction, there is a rupture of links to reorganize into another, but said mass in the closed system is maintained.
The correct answer is:
C. "The mass of a closed system cannot change over time; mass cannot be created or destroyed."
The following statements correctly describe the law of conservation of energy - c. The mass of a closed system cannot change over time; mass cannot be created nor destroyed
The law of conservation of mass states that the mass is an isolated system that can not be created nor destroyed.
conserved means saved, so according to the law of conservation of mass refers to the "saving" of mass.
Thus, The following statements correctly describe the law of conservation of energy - c. The mass of a closed system cannot change over time; mass cannot be created nor destroyed
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Car drag racing takes place over a distance of a mile (402 m) from a standing start. If a car (mass 1600 kg) could be propelled forward with a pulling force equal to that of gravity, what would be the change in kinetic energy and the terminal speed of the car (in mph) at the end of the race be? (For comparison, a modern, high-performance sports car may reach a terminal speed of just over 100 mph = 44.7 m/s.)
Answer:
v = 88.76 m / s , K = 6.30 10⁶ J
Explanation:
For this exercise the force that is applied is that necessary for the acceleration of the car to be the acceleration of gravity, they do not indicate that there is friction, we look for the final speed
v² = v₀² + 2 a x
Since the car starts from rest, the initial speed is zero, vo = 0
v = √ 2 a x
v = √ (2 9.8 402)
v = 88.76 m / s
Let's look for kinetic energy
K = ½ m v²
K = ½ 160kg 88.76²
K = 6.30 10⁶ J
Why is it impossible for an astronaut inside an orbiting space station to go from one end to the other by walking normally?A. In an orbiting station, the gravitational force is too large and the astronaut can't take his feet off the floor.B. It is impossible to walk inside an orbiting space station because its rotation is too fast.C. In an orbiting station, after one foot pushes off there isn't a friction force to move forward. The astronaut "jumps" on the same place.D. In an orbiting station, after one foot pushes off there isn't a force to bring the astronaut back to the "floor" for the next step.
Final answer:
An astronaut cannot walk normally in a space station because there's no frictional force to move forward in the near-weightless environment. To move, astronauts use handholds and walls, pushing against them to create a reaction force.
Explanation:
It is impossible for an astronaut inside an orbiting space station to go from one end to the other by walking normally because C. In an orbiting station, after one foot pushes off there isn't a friction force to move forward. The astronaut would indeed "jump" in place due to the lack of friction between their feet and the floor of the space station, which is a result of the near-weightlessness they experience. In space, normal walking is ineffective because walking relies on gravity to pull the body back down to the floor after each step, which isn't present in the same way on a space station in orbit.
In order to move in such an environment, an astronaut must push against a solid object, creating a reaction force in the opposite direction according to Newton's third law of motion. This principle allows the astronaut to propel and steer themselves around the space station using handholds and walls. The environment inside the ISS is similar to that inside a freely falling box where gravity still exists, but occupants do not feel its effects because they are in free fall around Earth, which creates the sensation of weightlessness.
Final answer:
Astronauts cannot walk normally in an orbiting space station due to the lack of gravity and friction. They are in a state of free fall, creating a sensation of weightlessness. Movement can be achieved by utilizing the conservation of momentum and Newton's third law of motion. Therefore option C is the correct answer.
Explanation:
The reason it is impossible for an astronaut inside an orbiting space station to walk from one end to the other by walking normally is C. In an orbiting station, after one foot pushes off there isn't a friction force to move forward. The astronaut cannot walk from one end to the other by walking normally because, in the microgravity environment of an orbiting spacecraft, traditional walking, which relies on the force of gravity and friction between the feet and the ground, does not work. Instead, astronauts move about by pushing off surfaces or floating through the air.
In orbit, the International Space Station (ISS) and everything inside it, including the astronauts, are in a state of free fall. They are falling around Earth at the same rate as the space station, creating a sensation of weightlessness. This is akin to the sensation of temporary weightlessness one experiences at the topmost point of a roller coaster ride or when an elevator suddenly descends.
Achieving locomotion for an astronaut stranded in the center of the station without contact with any solid surface would necessitate a method that does not rely on gravity or friction. The astronaut would have to utilize the principle of conservation of momentum. For instance, by throwing an object in one direction, the astronaut would move in the opposite direction, as described by Newton's third law of motion: for every action, there is an equal and opposite reaction.
A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 4.10 L ? (The temperature was held constant.)
Answer:
0.358g
Explanation:
Density of Helium = 0.179g/L
ρ=m/v
m=ρv
when the volume was 2L
m1= 0.179*2
m1=0.358g
when the volume increased to 4L
m2= 0.179*4
m2=0.716g
gram of helium added = 0.716g-0.358g
=0.358g
Twist-on connectors without the spring-steel coils (plastic threads only) are suitable for making branch-circuit connections.
A. TrueB. False
Answer:
if it is a plastic connector it wont work but if there is metal or steel it will work
Explanation:
A rock of mass m is thrown straight up into the air with initial speed |v0 | and initial position y = 0 and it rises up to a maximum height of y = h. A second rock with mass 2m (twice the mass of the original) is thrown straight up with an initial speed of 2|v0 |. What maximum height does the second rock reach?
Answer:
Explanation:
Case 1:
mass = m
initial velocity = vo
final velocity = 0
height = y
Use third equation of motion
v² = u² - 2as
0 = vo² - 2 g y
y = vo² / 2g ... (1)
Case 2:
mass = 2m
initial velocity = 2vo
final velocity = 0
height = y '
Use third equation of motion
v² = u² - 2as
0 = 4vo² - 2 g y'
y ' = 4vo² / 2g
y' = 4 y
Thus, the second rock reaches the 4 times the distance traveled by the first rock.
The maximum height the second rock reach is :
-4 times the distance traveled by the first rock.
"Mass"Case 1:
mass = m
initial velocity = vo
final velocity = 0
height = y
using Third equation of motion
v² = u² - 2as
0 = vo² - 2 g y
y = vo² / 2g ... (1)
Case 2:
mass = 2m
initial velocity = 2vo
final velocity = 0
height = y '
Use third equation of motion
v² = u² - 2as
0 = 4vo² - 2 g y'
y ' = 4vo² / 2g
y' = 4 y
Therefore, the second rock reaches the 4 times the distance traveled by the first rock.
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8–4. The tank of the air compressor is subjected to an internal pressure of 90 psi. If the internal diameter of the tank is 22 in., and the wall thickness is 0.25 in., determine the stress components acting at point A. Draw a volume element of the material at this point, and show the results on the element.
Answer:
The stress S = 1935 [Psi]
Explanation:
This kind of problem belongs to the mechanical of materials field in the branch of the mechanical engineering.
The initial data:
P = internal pressure [Psi] = 90 [Psi]
Di= internal diameter [in] = 22 [in]
t = wall thickness [in] = 0.25 [in]
S = stress = [Psi]
Therefore
ri = internal radius = (Di)/2 - t = (22/2) - 0.25 = 10.75 [in]
And using the expression to find the stress:
[tex]S=\frac{P*D_{i} }{2*t} \\replacing:\\S=\frac{90*10.75 }{2*0.25} \\S=1935[Psi][/tex]
In the attached image we can see the stress σ1 & σ2 = S acting over the point A.
A projectile of mass m is fired straight upward from the surface of an airless planet of radius R and mass M with an initial speed equal to the escape speed vesc (meaning the projectile will just barely escape the planet's gravity -- it will asymptotically approach infinite distance and zero speed.) What is the correct expression for the projectile's kinetic energy when it is a distance 9R from the planet's center (8R from the surface). Ignore the gravity of the Sun and other astronomical bodies. KE (at r = 9R) is:a. GMm/9Rb. GMm/8Rc. 1/2mvesc^2d. -GMm/8Re. None of these
Answer:
K = G Mm / 9R
Explanation:
Expression for escape velocity V_e = [tex]\sqrt{\frac{2GM}{R} }[/tex]
Kinetic energy at the surface = 1/2 m V_e ²
= 1/2 x m x 2GM/R
GMm/R
Potential energy at the surface
= - GMm/R
Total energy = 0
At height 9R ( 8R from the surface )
potential energy
= - G Mm / 9R
Kinetic energy = K
Total energy will be zero according to law of conservation of mechanical energy
so
K - G Mm / 9R = 0
K = G Mm / 9R
A forward-biased silicon diode is connected to a 12.0-V battery through a resistor. If the current is 12 mA and the diode potential difference is 0.70 V, what is the resistance?
To solve this problem we will use the concepts related to Ohm's law for which voltage, intensity and resistance are related.
Mathematically this relationship is given as
[tex]V = IR \rightarrow R= \frac{V}{I}[/tex]
Where,
V= Voltage
I = Current
R = Resistance
The value of the given voltage is 12V, while the current is 12mA, therefore the resistance would be
[tex]R = \frac{12}{12*10^{-3}}[/tex]
[tex]R = 1000 \Omega[/tex]
Therefore the resistance is [tex]1000\Omega[/tex]
Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.
Answer:
115 ⁰C
Explanation:
Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies
[tex]q_{1} +q_{2} =-q_{3}[/tex] -----eqution 1
where,
[tex]q_{1}[/tex] is the heat absorbed by the solid at 0⁰C
[tex]q_{2}[/tex] is the heat absorbed by the liquid at 0⁰C
[tex]q_{3}[/tex] the heat lost by the warmer water sample
Important equations to be used in solving this problem
[tex]q=m *c*\delta {T}[/tex], where -----equation 2
q is heat absorbed/lost
m is mass of the sample
c is specific heat of water, = 4.18 J/0⁰C
[tex]\delta {T}[/tex] is change in temperature
Again,
[tex]q=n*\delta {_f_u_s}[/tex] -------equation 3
where,
q is heat absorbed
n is the number of moles of water
tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol
Step 2: calculate how many moles of water you have in the 100.0-g sample
[tex]=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O[/tex]
Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C
[tex]q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ[/tex]
This means that equation (1) becomes
79.13 KJ + [tex]q_{2} = -q_{3}[/tex]
Step 4: calculate the final temperature of the water
[tex]79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}[/tex]
Substitute in the values; we will have,
[tex]79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})[/tex]
79.13 kJ + 990.66J* [tex](T_{f}-218})[/tex] = -1463J*[tex](T_{f}-100})[/tex]
Convert the joules to kilo-joules to get
79.13 kJ + 0.99066KJ* [tex](T_{f}-218})[/tex] = -1.463KJ*[tex](T_{f}-100})[/tex]
[tex]79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3[/tex]
collect like terms,
2.45366[tex]T_{f}[/tex] = 283.133
∴[tex]T_{f} =[/tex] = 115.4 ⁰C
Approximately the final temperature of the mixture is 115 ⁰C
A continuous and aligned fiber-reinforced composite having a cross-sectional area of 1130 mm2 is subjected to an external tensile load. If the stresses sustained by the fiber and matrix phases are 156 MPa and 2.75 MPa, respectively, the force sustained by the fiber phase is 74,000 N and the total longitudinal strain is 1.25 x 10-3, what is the value of the modulus of elasticity of the composite material in the longitudinal direction?
Answer:
Ec=53.7×10⁹N/m² =53.7Gpa
Explanation:
To calculate the modulus of elasticity in the longitudinal direction. This is possible realizing Ec=σ/ε where σ=(Fm+Ff)/Ac
[tex]Ec=Sigma/E\\Ec=\frac{(Fm+Ff)/E}{Ac}\\ Ec=\frac{1802+74,000}{(1.25*10^{-3})(1130)(1/1000)^{2} }\\ Ec=53.7*10^{9}N/m^{2}\\or\\Ec=53.7GPa[/tex]
Final answer:
The modulus of elasticity of the composite material in the longitudinal direction is 124,800 MPa.
Explanation:
To find the modulus of elasticity of the composite material in the longitudinal direction, we can use the formula:
E = (stress sustained by the fiber phase)/(longitudinal strain)
Given that the stress sustained by the fiber phase is 156 MPa and the total longitudinal strain is 1.25 x 10^-3, we can plug in these values to calculate the modulus of elasticity:
E = 156 MPa / (1.25 x 10^-3) = 124,800 MPa
Therefore, the modulus of elasticity of the composite material in the longitudinal direction is 124,800 MPa.
If a nucleus decays by gamma decay to a daughter nucleus, which of the following statements about this decay are correct? (There may be more than one correct choice.)
a)The daughter nucleus has fewer protons than the original nucleus.
b)The daughter nucleus has the same number of nucleons as the original nucleus.
c)The daughter nucleus has more protons than the original nucleus.
d)The daughter nucleus has fewer neutrons than the original nucleus. The daughter nucleus has more neutrons than the original nucleus
Answer: Option (b) is the correct answer.
Explanation:
A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.
Symbol of a gamma particle is [tex]^{0}_{0}\gamma[/tex]. Hence, charge on a gamma particle is also 0.
For example, [tex]^{234}_{91}Pa \rightarrow ^{234}_{91}Pa + ^{0}_{0}\gamma + Energy[/tex]
So, when a nucleus decays by gamma decay to a daughter nucleus then there will occur no change in the number of protons and neutrons of the parent atom but there will be loss of energy as a nuclear reaction has occurred.
Thus, we can conclude that the statement daughter nucleus has the same number of nucleons as the original nucleus., is correct about if a nucleus decays by gamma decay to a daughter nucleus.
Answer: Option (b) is the correct answer.
Explanation:
A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.
Recall that force is a change in momentum over a change in time, the force due to radiation pressure reflected off of a solar sail can be calculated as 2 times the radiative momentum striking the sail per second. What is the approximate magnitude of the pressure on the sail in the vicinity of Earth’s Orbit?
Answer:
magnitude of the pressure on the sail in the vicinity of Earth’s Orbit= [tex]\frac{2I}{c}[/tex]
Explanation:
The momentum of a photon is:
p = E/c
E = the photon energy
c = the speed of light.
take the time derivative (gives the force)
F = dp/dt = (dE/dt)/c
F = 2(dE/dt)/c (is doubled for complete reflection of the light)
Intensity has the units of energy per unit time per unit area
= I
then,
Force/unit area = 2I/c
magnitude of the pressure on the sail in the vicinity of Earth’s Orbit= [tex]\frac{2I}{c}[/tex]
Did you think about this over Christmas? I did (-: Before Christmas a 65kg student consumes 2500 Cal each day and stays at the same weight. For three days in a row while visiting her parents she eats 3500 Cal and, wanting to keep from gaining weight decides to "work off" the excess by jumping up and down at the Christmas tree. With each jump she accelerates to a speed of 3.2 m/s before leaving the ground. a) How high will she jump each jump? b) How many jumps must she do to keep her weight? Assume that the efficiency of the body in using energy is 25%. c) Do you suggest that is a reasonable way for the student not to gain weight over Christmas? d) Possible enhancement: What other way/ways would you suggest for the student to keep her weight?
Answer:
a) Em = 332.8 J , b) # jump = 13, c) It is reasonable since there are not too many jumps , d) lower the calories consumed
Explanation:
a) Let's use energy conservation
Initial. On the floor
Em₀ = K = ½ m v²
Final. The highest point
Emf = U = m g h
Energy is conserved
Em₀ = Emf
½ m v² = m g h
h = ½ v² / g
h = ½ 3.2² /9.8
h = 0.52 m
b) When he was at home he maintained his weight with 2500 cal / day. In his parents' house he consumes 3500 cal / day, the excess of calories is
Q = 3500 -2500 = 1000cal / day
Let's reduce this value to the SI system
Q = 1000 cal (4,184 J / 1 cal) = 4186 J / day
Now the energy in each jump is
Em = K = ½ m v²
Em = ½ 65 3.2²
Em = 332.8 J
They indicate that the body can only use 25% of this energy
Em effec = 0.25 332.8 J
Em effec = 83.2 J
This is the energy that burns the body
Let's use a Proportion Rule (rule of three), if a jump spends 83.2J how much jump it needs to spend 1046 J
# jump = 1046 J (1 jump / 83.2 J)
# jump = 12.6 jumps / day
# jump = 13
c) It is reasonable since there are not too many jumps
d) That some days consume more vegetables to lower the calories consumed
Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1012 W) pulses of electromagnetic waves that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk 4.6 μm in diameter, with the pulse lasting for 3.4 ns with an average power of 2.46×1012 W . We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse.
Part A
How much energy is given to the cell during the pulse?
Express your answer to two significant figures.
The energy given to each cell during the pulse can be calculated by multiplying the power of the pulse by its duration, and then dividing by the number of cells.
Explanation:The energy supplied to the cell during the pulse is determined by the power multiplied by the duration of the pulse. In this scenario, the power is 2.46×1012 W and the duration is 3.4 ns (which is 3.4x10-9 s when converted to seconds for mathematical calculations).
We use the formula:
E = P * t
Where,
E is the Energy
P is the Power
t is the time (duration of the pulse)
Substituting the given values into the formula:
E = 2.46x1012 W * 3.4x10-9 s
This gives the total energy supplied. We know the energy is spread uniformly over the faces of 100 cells, so each cell will get 1/100 of the total energy. Using these calculations, we can determine the amount of energy given to each cell during the pulse.
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The inductance in the drawing has a value of L = 9.4 mH. What is the resonant frequency f0 of this circuit?
Answer:
The resonant frequency of this circuit is 1190.91 Hz.
Explanation:
Given that,
Inductance, [tex]L=9.4\ mH=9.4\times 10^{-3}\ H[/tex]
Resistance, R = 150 ohms
Capacitance, [tex]C=1.9\ \mu F=1.9\times 10^{-6}\ C[/tex]
At resonance, the capacitive reactance is equal to the inductive reactance such that,
[tex]X_C=X_L[/tex]
[tex]2\pi f_o L=\dfrac{1}{2\pi f_oC}[/tex]
f is the resonant frequency of this circuit
[tex]f_o=\dfrac{1}{2\pi \sqrt{LC}}[/tex]
[tex]f_o=\dfrac{1}{2\pi \sqrt{9.4\times 10^{-3}\times 1.9\times 10^{-6}}}[/tex]
[tex]f_o=1190.91\ Hz[/tex]
So, the resonant frequency of this circuit is 1190.91 Hz. Hence, this is the required solution.
An observer sits in a boat watching wave fronts move past the boat. The distance between successive wave crests is 0.80 m, and they are moving at 2.2 m / s.
What is the wavelength of these waves?
a. 1.6 m
b. 2.2 m
c. 0.80 m
What is the frequency of these waves?
a. 0.36 Hz
b. 2.8 Hz
c. 0.80 Hz
What is the period of these waves?
a. 0.80 s
b. 0.36 s
c. 2.8 s
To solve this problem we will use the three requested concepts: Wavelength, frequency and period.
The wavelength is the distance between each crest, therefore it is already given and is 0.8m
The correct answer is C.
The frequency can be described as a relationship between wave speed and wavelength therefore
[tex]f = \frac{v}{\lambda}[/tex]
[tex]f = \frac{2.2}{0.8}[/tex]
[tex]f = 2.75Hz \approx 2.8Hz[/tex]
The correct answer is B.
The period is the inverse of the frequency therefore
[tex]T = \frac{1}{f}[/tex]
[tex]T = \frac{1}{2.8}[/tex]
[tex]T = 0.35s[/tex]
The correct answer is B.
(a) The wavelength of the wave is 0.80m and the right option is c.
(b) The frequency of the wave is 2.8 Hz and the right option is b.
(c) The period of the wave is 0.36 s and the right option is b
(a) The distance between successive wave crests = wavelength of the wave
From the question,
(a) Wavelength = 0.80 m
Hence the wavelength = 0.80 m
(b) Using,
V = λf.............. Equation 1
Where V = Velocity of the wave, λ = wavelength of the wave, f = frequency of the wave.
f = V/λ.................... Equation 2
Given: V = 2.2 m/s, λ = 0.80 m
Substitute these values into equation 2
f = 2.2/0.8
f = 2.75 Hz.
f ≈ 2.8 Hz
Hence the frequency of the wave is 2.8 Hz
(c) f = 1/T.............. Equation 3
Where T = period.
Therefore,
T = 1/f .................. 4
Given: f = 2.8 Hz,
T = 1/2.8
T = 0.357
T ≈ 0.36 s
Hence the period of the wave = 0.36 s
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A 100 g aluminum calorimeter contains 250 g of water. The two substances are in thermal equilibrium at 10°C. Two metallic blocks are placed in the water. One is a 50 g piece of copper at 75°C. The other sample has a mass of 66 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20°C. Determine the specific heat of the unknown second sample. (Pick the answer closest to the true value.)A. 1950 joules Co/kgB. 975 joules Co/kgC. 3950 joules Co/kgD. 250 joules Co/kgE. 8500 joules Co/kg
Answer:
A. 1,950 J/kgºC
Explanation:
Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the copper and the unknown metal.
The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:
Q = c * m* Δt
where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.
In our case, we can write the following equality:
(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (ccu*mcu*Δtcu) + (cₓ*mₓ*Δtₓ)
Replacing by the givens , and taking ccu = 0.385 J/gºC and cAl = 0.9 J/gºC, we have:
Qg= 0.9 J/gºC*100g*10ºC + 4.186 J/gºC*250g*10ºC = 11,365 J(1)
Ql = 0.385 J/gºC*50g*55ºC + cₓ*66g*80ºC = 1,058.75 J + cx*66g*80ºC (2)
Based on all the previous assumptions, we have:
Qg = Ql
So, we can solve for cx, as follows:
cx = (11,365 J - 1,058.75 J) / 66g*80ºC = 1.95 J/gºC (3)
Expressing (3) in J/kgºC:
1.95 J/gºC * (1,000g/1 kg) = 1,950 J/kgºC
The specific heat of the unknown metal can be determined from the equilibrium of heat transfer in the system. The heat lost by the hot substances is equal to the heat gained by the cooler substances. Solving for the specific heat of the unknown substance involves calculating the heat gained and lost and equating their values.
Explanation:The specific heat of a substance is a measure of the amount of heat energy required to raise the temperature of a certain mass of the substance by a certain amount. In this case, we're solving for the specific heat (c) of an unknown substance. As the system is in thermal equilibrium, the heat lost by hot substances (copper and unknown metal) is equal to the heat gained by the cooler substances (water and the calorimeter).
The specific heat (c) of the unknown substance can therefore be determined by setting the heat gained (Q_gained = m*c*ΔT) by the cooler substances equal to the heat lost (Q_lost = m*c*ΔT) by the hot substances and solving for the specific heat (c) of the unknown substance. Given that ΔT is the change in temperature, m is the mass, and c is the specific heat, and using the specific heat values for water, aluminum, and copper.
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A navy seal of mass 80 kg parachuted into an enemy harbor. At one point while he was falling, the resistive force of air exerted on him was 520 N. What can you determine about the motion?
Answer:
The motion of the parachute = 3.3 m/s²
Explanation:
Weight of the parachute - Resistive force of air = ma
W - Fₐ = ma.................... Equation 1
making a the subject of formula in equation 1
a = (W- Fₐ)/m.................. Equation 2
Where W = weight of the parachute, Fₐ = resistive force of air, m = mass of the parachute, a = acceleration of the parachute
Constant: g = 9.8 m/s²
Given: Fₐ = 520 N, m = 80 kg
W = mg = 80 × 9.8 = 784 N,
Substituting these values into equation 2
a = (784-520)/80
a = 264/80
a = 3.3 m/s²
Therefore the motion of the parachute = 3.3 m/s²
A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e-0.2t T. What is the magnitude of the current induced in the coil at the time t = 2 s? A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e-0.2t T. What is the magnitude of the current induced in the coil at the time t = 2 s? 1.3 mA 7.5 mA 2.6 mA 4.2 mA 9.2 mA
Answer:
the question is incomplete, the complete question is
"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"
2.6mA
Explanation:
we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.
using the formula be low,
[tex]E=-\frac{d}{dt}(BACOS\alpha )\\[/tex]
where B is the magnitude of the field and A is the area of the circular coil.
First, let determine the area using [tex]\pi r^{2} \\[/tex] where r is the radius of 5cm or 0.05m
[tex]A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\[/tex]
since we no that the angle is at [tex]0^{0}[/tex]
we determine the magnitude of the magnetic filed
[tex]B=0.5e^{-t} \\t=2s[/tex]
[tex]E=-(0.5e^{-2} * 0.00785)[/tex]
[tex] E=-0.000532v\\[/tex]
the Magnitude of the voltage is 0.000532V
Next we determine the current using ohm's law
[tex]V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A[/tex]
[tex]I=2.6mA[/tex]
The magnitude of the induced current in the coil at t = 2s in the given scenario is 2.4 mA. This is calculated using Faraday's law of electromagnetic induction and Ohm's law.
Explanation:To find the magnitude of the current induced in the coil, we need to consider Faraday's law of electromagnetic induction. This law states that the induced electromotive force (emf) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.
In this situation, we have: B = 0.5 e-0.2t T, and the time derivative of the magnetic field is dB/dt = -0.1 e-0.2t T/s. The area A of the coil is πr²= π(0.05)² m². The induced emf (ε) equals -A dB/dt. Thus, we have ε = -π(0.05)² × -0.1 e-0.2t = 0.0007875 e-0.2t V.
Now, according to Ohm's law, I = ε/R, where R is the resistance of the coil. Substituting the given values, we have I = 0.0007875 e-0.2t / 0.2 = 0.0039375 e-0.2t A. At t=2s, we can substitute into the equation to get I = 0.0039375 e-0.4 = 0.0024 A or 2.4 mA. Therefore, the magnitude of the induced current at t = 2s is 2.4 mA.
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A proton initially traveling at 50,000 m/s is shot through a small hole in the negative plate of a parallal-plate capacitor. The electric field strength inside the capacitor is 1,500 V/m. How far does the proton travel above the negative plate before temporarily coming to rest and reversing course? Assume the proton reverses course before striking the positive plate.
Answer:
x = 8.699 10⁻³ m
Explanation:
The proton feels an electric charge that is the opposite direction of speed, let's look for acceleration using Newton's second law
F = m a
F = q E
a = q E / m
a = 1.6 10⁻¹⁹ 1500 / 1.67 10⁻²⁷
a = 1,437 10¹¹ m / s²
Now we can use kinematic relationships
v² = v₀² - 2 a x
When at rest the speed is zero (v = 0)
x = v₀² / 2 a
Let's calculate
x = 50,000² / (2 1,437 10¹¹)
x = 8.699 10⁻³ m
I am standing next to the edge of a cliff. I throw a ball upwards and notice that 4 seconds later it is traveling downwards at 10 m/s. Where is the ball located at this time? (Pick the answer closest to the true value.)A. 120 meters above me B. 30 meters below meC. 30 meters above meD. 120 meters below meE. At the same height that it started
Answer:
Explanation:
Given
Velocity after t=4 sec is v=10 m/s downward
assuming u is the initial upward velocity
[tex]v=u+at[/tex]
[tex]-10=u-gt[/tex]
[tex]u=9.8\times 4-10=29.2 m/s[/tex]
[tex]v^2-u^2=2 as[/tex]
[tex](-10)^2-(29.2)^2=2\times (-9.8)\cdot s[/tex]
[tex]s=\frac{29.2^2-10^2}{2\times 9.8}[/tex]
[tex]s=38.4 m[/tex]
i.e. 38.4 m above the initial thrown Position
Suppose you are standing a few feet away from a bonfire on a cold fall evening. Your face begins to feel hot. What is the mechanism that transfers heat from the fire to your face? (Hint: Is the air between you and the fire hotter or cooler than your face?)
•A. convection
•B. radiation
•C. conduction
•D. none of the above
A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu
To determine the steady state temperature of the wire, one can use the power dissipation formula and the convection heat transfer equation. The time for the wire to reach within 1-degree Celsius of steady state involves transient heat transfer calculations using the given material properties.
The student has asked about the steady state temperature of a 1-meter-long wire with a 1mm diameter submerged in an oil bath at 25 degrees Celsius when a current of 100A flows through it. We also need to calculate how long it takes for the wire to reach within 1-degree Celsius of the steady state temperature. To find the steady state temperature, we use the formula P = I2R, where P is the power, I is the current, and R is the resistance. Given that R = 0.01
Ω/m and I = 100A, we find P = (100A)2 x 0.01
Ω/m = 100W/m. Then, using the convection heat transfer equation Q = hA(Ts - T
bath), where Q is the heat transfer rate, h is the convection coefficient, A is the surface area, Ts is the wire surface temperature, and Tbath is the oil bath temperature, we equate Q to P since the wire is in steady state, and solve for Ts. The time to reach within 1-degree Celsius of steady state temperature requires calculating the transient heat transfer, which involves solving the heat transfer equation with the given material properties such as density, heat capacity, and thermal conductivity.
The steady-state temperature of the wire is approximately [tex]\(343.471 {°C}\)[/tex], and it takes approximately [tex]\(1.539[/tex], for the wire to reach within 1°C of the steady-state value.
Steady-State Temperature Calculation:
- Calculate the radius [tex](\(r\))[/tex] of the wire:
[tex]\[ r = \frac{d}{2} = \frac{0.001 \, \text{m}}{2} = 0.0005 \, \text{m} \][/tex]
- Calculate the surface area [tex](\(A\))[/tex] of the wire:
[tex]\[ A = 2\pi r l = 2\pi \times 0.0005 \times 1 = 0.00314 \, \text{m}^2 \][/tex]
- Calculate the heat transfer rate [tex](\(q\))[/tex]:
[tex]\[ q = I^2 R = (100)^2 \times 0.01 = 1000 \, \text{W} \][/tex]
- Calculate the steady-state temperature [tex](\(T_{\text{wire}}\))[/tex]:
[tex]\[ T_{\text{wire}} = \frac{q}{hA} + T_{\text{fluid}} \][/tex]
[tex]\[ T_{\text{wire}} \approx \frac{1000}{500 \times 0.00314} + 298.15 \][/tex]
[tex]\[ T_{\text{wire}} \approx 343.471 \, \text{°C} \][/tex]
Time to Reach Within 1°C of Steady-State:
- Calculate the volume [tex](\(V\))[/tex] of the wire:
[tex]\[ V = \pi r^2 l = \pi \times (0.0005)^2 \times 1 = 7.854 \times 10^{-7} \, \text{m}^3 \][/tex]
- Calculate the thermal time constant [tex](\(\tau\))[/tex]:
[tex]\[ \tau = \frac{\rho V c}{hA} \][/tex]
[tex]\[ \tau \approx \frac{8000 \times 7.854 \times 10^{-7} \times 500}{500 \times 0.00314} \][/tex]
[tex]\[ \tau \approx 0.7854 \, \text{s} \][/tex]
- Calculate the time [tex](\(t\))[/tex] it takes for the wire to reach within 1°C of the steady-state value:
[tex]\[ t = \tau \ln\left(\frac{T_{\text{steady}} - T_{\text{initial}}}{T_{\text{steady}} - T_{\text{fluid}}}\right) \][/tex]
[tex]\[ t \approx 0.7854 \times \ln\left(\frac{343.471 - 25}{343.471 - 298.15}\right) \][/tex]
[tex]\[ t \approx 0.7854 \times \ln\left(\frac{318.471}{45.321}\right) \][/tex]
[tex]\[ t \approx 0.7854 \times \ln(7.032) \][/tex]
[tex]\[ t \approx 1.539 \, \text{s} \][/tex]
A swimming pool heater has to be able to raise the temperature of the 40 000 gallons of water in the pool by 10.0 C°.
How many kilowatt-hours of energy are required?
(One gallon of water has a mass of approximately 3.8 kg and the specific heat of water is 4 186 J/kg⋅°C.)
a. 1 960 kWh
b. 1 770 kWh
c. 330 kWh
d. 216 kWh
Answer:
b. 1 770 kWh
Explanation:
The heat needed to change the temperature of a certain amount of a substance is given by:
[tex]Q=mC\Delta T[/tex]
Here m is the mass of the susbtance, C is the specific heat of the substance and [tex]\Delta T[/tex] is the temperature change
[tex]Q=(40000*3.8kg)(4186\frac{J}{kg\cdot ^\circ C})(10^\circ C)\\Q=6.36*10^9J[/tex]
Recall that one watt hour is equivalent to 1 watt (1 W) of power sustained for 1 hour. One watt is equal to 1 J/s. So, one watt hour is equal to 3600 J and one kilowatt hour is equal to [tex]3600*10^3 J[/tex]
[tex]Q=6.36*10^9J*\frac{1kW\cdot h}{3600*10^3J}\\Q=1766.66kW\cdot h[/tex]
Final answer:
To heat 40,000 gallons of water by 10.0 C° in a swimming pool, 1,767 kilowatt-hours of energy are required, rounding to the nearest so, option gives (b) 1,770 kWh as the answer.
Explanation:
The question asks: How many kilowatt-hours of energy are required to raise the temperature of 40,000 gallons of water in a pool by 10.0 C°? To solve this, we need to calculate the energy needed using the formula for heat energy: Q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Firstly, convert the volume of water from gallons to kilograms. 40,000 gallons is approximately 40,000 x 3.8 kg = 152,000 kg. Next, use the specific heat of water (4,186 J/kg°C) and the temperature change (10.0 C°) to find the energy in joules: Q = 152,000 kg x 4,186 J/kg°C x 10.0 C° = 6,362,720,000 J.
To convert joules to kilowatt-hours, divide the total joules by 3,600,000 (the number of joules in one kilowatt-hour): 6,362,720,000 J / 3,600,000 J/kWh = 1,767 kWh. Therefore, the energy required is 1,767 kWh, making option (b) 1,770 kWh the nearest correct answer.
Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2.0 N, and you do 12 J of work on each. Match each box to the vertical distance through which it is lifted.
Answer:
The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m
Explanation:
Worked : work can be defined as the product of force and distance.
The S.I unit of work is Joules (J).
Mathematically it can be represented as,
W = F×d.................. Equation 1
d = W/F.............................. Equation 2
where W = work, F = force, d = distance.
Given: W = 12 J
(i) for the 3.0 N weight,
using equation 2
d = 12/3
d= 4 m.
(ii) for the 4.0 N weight,
d = 12/4
d = 3 m.
(iii) for the 6.0 N weight,
d = 12/6
d = 2 m.
(iv) for the 2.0 N weight,
d = 12/2
d = 6 m
Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m
A 30 gram bullet is shot upward at a wooden block. The bullet is launched at the speed vi. It travels up 0.40 m to strike the wooden block. The wooden block is 20 cm wide and 10 cm high and its thickness gives it a mass of 500 g. The center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height. a. What is the launch speed of the bullet? b. How much mechanical energy does the bullet and the block system have before all of the processes? Use the surface the block rests on as the reference for where gravitational potential energy is zero. c. How much mechanical energy does the bullet and the block system have after all of the processes? d. How much mechanical energy was lost from beginning to end?
Answer:
Explanation:
Mass of bullet m = .03 kg
Mass of wooden block M = 0.5 kg
Since the center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height
Velocity of wooden block + bullet just after impact = √2gH
=√(2 x 9.8 x 0.6)
= 3.43 m / s
Let the launch velocity of bullet be v₁
If v₂ be the velocity with which bullet hits the block
Applying law of conservation of momentum
.03 x v₂ = .530 x 3.43
v₂ = 60.6 m /s
if v₁ be initial velocity
v₂² = v₁² - 2 gh
v₁² = v₂² + 2 gh
= 60.6 ² + 2 x 9.8 x 0.4
v₁ = 60.65 m /s this is launch speed.
b )
Initial kinetic energy of bullet
= 1/2 m v²
= .5 x .03 x 3680
= 55 J
Potential energy of bullet + block = 0
Total energy = 5 J
c)
Kinetic energy of bullet block system
1/2 m v²
= .5 x .53 x 3.43
= 3.11 J
d )
Loss of energy in the impact = Total mechanical energy lost from beginning to end?
3.11 J - 5
= 1.89 J