The voltage across the element is = 240 V
I hope you understand....
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You are a venture capitalist that is asked to invest in a startup company that claims it will be able to launch tiny "micro space probes" into space at close to the speed of light using a massive electromagnetic rail gun system2. You are cynical about their cost estimates and decide to analyze the problem in more detail before you invest in their company. Neglect air resistance for this worksheet.
1. A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?
2. Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?
43. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.
4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the railgun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).
a. Will the recoil momentum of the ship be relativistic? Justify your argument.
b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?
Answer:
1. 5.825 × 10¹⁷ J
2. i. $ 29.125 billion ii. I would not invest in the company
3. A nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.
4. a i. The momentum will not be relativistic
ii. This is because objects with large masses do not move at relativistic speeds
b i. 155 m/s
ii. This speed wouldn't be a problem for the ship.
Explanation:
1. A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?
The kinetic energy of the payload is K = (γ - 1)mc² where m = mass of payload = 1 kg, c = speed of light = 3 × 10⁸ m/s and γ = 1/√(1 - β²) where β = 0.9 (since the payload moves at 90 % speed of light)
So, K = (γ - 1)mc²
= (1/√(1 - β²) - 1)mc²
= (1/√(1 - (0.9)²) - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/√(1 - 0.81) - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/√0.19 - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/0.436 - 1) × 1 kg × (3 × 10⁸ m/s)²
= (2.294 - 1) × 1 kg × (3 × 10⁸ m/s)²
= 1.294 × 1 kg × 9 × 10¹⁶ m²/s²
= 11.65 × 10¹⁶ kgm²/s²
= 1.165 × 10¹⁷ J
Let E be the total electrical energy of the rail gun. Since 20 % of this energy is converted to kinetic energy of the payload, we have
20 % of E = K
0.2E = K
E = K/0.2
= 1.165 × 10¹⁷ J/0.2
= 5.825 × 10¹⁷ J
2 Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?
i. how much would this launch cost?
Since the total energy required is E = 5.825 × 10¹⁷ J = 5.825 × 10¹¹ MJ and it costs 5 cent/MJ. So the total cost of energy will be total energy rate = 5.825 × 10¹¹ MJ × 5 cent/MJ = 29.125 × 10¹¹ = 2.9125 × 10¹² cents. Converting this to dollars, we have 2.9125 × 10¹² cents/100 cents/dollar = 2.9125 × 10¹⁰ dollars = 29.125 × 10⁹ dollars = 29.125 billion dollars = $ 29.125 billion
ii. Would you invest in this company?
I would not invest in the company
3. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.
Since the kinetic energy of the payload is 1.165 × 10¹⁷ J and a nuclear explosion generates about 10¹⁵ J of energy, then a nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.
4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the rail gun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).
a. Will the recoil momentum of the ship be relativistic? Justify your argument.
i. Will the recoil momentum of the ship be relativistic?
The momentum will not be relativistic.
ii. Justify your argument.
This is because objects with large masses do not move at relativistic speeds. Since the speed cannot be relativistic, its momentum which is the product of mass and speed is non-relativistic
b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?
i. At what speed will the ship recoil after it launches a probe?
Since the total energy of the payload E' = K + mc² = 1.165 × 10¹⁷ J + 1 kg × (3 × 10⁸ m/s)² = 1.165 × 10¹⁷ J + 1 kg × 9 × 10¹⁶ m²/s² = 11.65 × 10¹⁶ J + 9 × 10¹⁶ J = 20.65 × 10¹⁶ J
Also, E'² = (pc)² + (mc²)² where p = momentum of payload
So, making p subject of the formula, we have
(pc)² = E'² - (mc²)²
pc = √[E'² - (mc²)²]
p = √[E'² - (mc²)²]/c
substituting the values of the variables into the equation, we have
p = √[E'² - (mc²)²]/c
p = √[(20.65 × 10¹⁶ J)² - 1kg × (3 × 10⁸ m/s²)²]/3 × 10⁸ m/s
p = √[(20.65 × 10¹⁶ J)² - (1kg × 9 × 10⁸ m²/s²)²]/3 × 10⁸ m/s
p = √[426.4225 × 10³² J² - 81 × 10³² J²]/3 × 10⁸ m/s
p = √[345.4225 × 10³² J²]/3 × 10⁸ m/s
p = 18.59 × 10¹⁶/3 × 10⁸ m/s
p = 6.20 × 10⁸ kgm/s
From the law of conservation, this momentum of the payload equals the momentum of recoil of the ship.
So, p = m'v where m' = mass of navy ship = 4,000 metric tons = 4,000 × 1000 kg = 4 × 10⁶ kg and v = speed of navy ship
So, v = p/m'
= 6.20 × 10⁸ kgm/s ÷ 4 × 10⁶ kg
= 1.55 × 10² m/s
= 155 m/s
ii. Do you think that this is a problem for the ship?
Since the ship's speed is 155 m/s, which is small for an object with such a large mass, this speed wouldn't be a problem for the ship.
What is the resistance of a bulb of 4ow
connected in a line of 220v?
2
Answer:
1210 ohm
Explanation:
Given :
P=40 W
V=220 V
Now,
[tex]P=\frac{V^{2} }{R} \\40=\frac{(220)^{2} }{R} \\40R=48400\\R=\frac{48400}{40} \\R=1210 ohm[/tex]
Therefore, resistance of bulb will be 1210 ohm
A rock is thrown from the edge of the top of a 51 m tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 74 m from the base of the building 8 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.
Answer:
The speed of projection is 34 m/s.
Explanation:
Height of building, h = 51 m
horizontal distance, d = 74 m
time, t = 8 s
Let the angle is A and the speed is u.
d = u cos A x t
74 = u cos A x 8
u cos A = 9.25 .... (1)
Use second equation of motion
[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]
Squaring and adding both the equations
[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]
Which describes a characteristic of metallic bonds?
Answer:
arge number of electrons free to move between the charged ions in the lattice.
Explanation:
The metallic bond occurs when an atom with few electrons is united in its last level, therefore the best way to decrease the total energy of the system is to lose all its electrons to remain with the configuration of a noble gas. The electrons that it loses cannot be acquired by other atoms since they all have few electrons, thus leaving a large number of electrons free to move between the charged ions in the lattice.
Some important characteristics emerge from this description of the metallic bond:
* It has many free electrons therefore its electrical conductivity is high
* As the charged ions are fixed, the material can be malleable, bent without breaking since the free electrons create the bond that keeps the system stable.
* As the electrons are free when heating a part of the material, these electrons acquire energy and rapidly propagate it to the other side, giving a high thermal conductivity
* As the temperature increases, the electrons acquire more kinetic energy, which is why there are more collisions between them and consequently the resistivity of the material increases.
When rebuilding her car's engine, a physics major must exert 405 N of force to insert a dry steel piston into a steel cylinder. What is the magnitude of the normal force in newtons between the piston and cylinder
Answer:
[tex]N=675N[/tex]
Explanation:
From the question we are told that:
Force [tex]F=405N[/tex]
Generally the equation for Normal force in this case is is mathematically given by
[tex]F=\mu_s N[/tex]
Where
Static Friction=[tex]\mu_s[/tex]
[tex]\mu_s=0.6[/tex]
Therefore
[tex]N=\frac{F}{\mu_s}[/tex]
[tex]N=\frac{405}{0.6}[/tex]
[tex]N=675N[/tex]
A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that is perpendicular to the platform at its center. The angular speed of the merry-go-round is constant, and a child at a distance of 1.4 m from the axis has a tangential speed of 2.2 m/s. What is the tangential speed of another child, who is located at a distance of 2.1 m from the axis?
(a) 1.5 m/s
(b) 3.3 m/s
(c) 2.2 m/s
(d) 5.0 m/s
(e) 0.98 m/s
Answer:
[tex]V_2=3.3m/s[/tex]
Explanation:
From the question we are told that:
Distance [tex]d_1=1.4m[/tex]
Tangential speed [tex]V=2.2m/s[/tex]
Distance 2 [tex]d_2=2.1m[/tex]
Generally the equation for Angular velocity is mathematically given by
[tex]w=\frac{v}{r}[/tex]
Therefore
[tex]\frac{v_1}{r_1}=\frac{v_2}{r_2}[/tex]
[tex]V_2=\frac{2.2*2.1}{1.4}[/tex]
[tex]V_2=3.3m/s[/tex]
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.
Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?
Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
(4.56 x 10^-13)-(1.17 x 10^-13)
Consider a point on a bicycle wheel as the wheel turns about a fixed axis, neither speeding up nor slowing down. Compare the linear and angular accelerations of the point.
a. Both are zero.
b. Only the angular acceleration is zero.
c. Only the linear acceleration is zero.
d. Neither is zero
Explain your choice
Answer:
c. Only the linear acceleration is zero.
Explanation:
The linear acceleration is defined as the rate of change of linear velocity. Since the bicycle is moving in the same direction, with the same speed, without speeding up or slowing down. Therefore, there will be no change in linear velocity and as a result, linear acceleration will be zero.
The angular acceleration is the rate of change of angular velocity. Since the angular velocity is changing its direction constantly. Therefore, it has a certain component of acceleration at all times called centripetal acceleration.
Therefore, the correct option is:
c. Only the linear acceleration is zero.
HELP ME PLS
Chlorine (chemical symbol Cl) is located in Group 17, Period 3. Which is
chlorine most likely to be?
A. A metalloid with properties of both metals and nonmetals
B. A gaseous, highly reactive nonmetal
C. A soft, shiny, highly reactive nonmetal
D. A soft, shiny, highly reactive metal
Answer:
Option B
Explanation:
A gaseous, highly reactive non-metal
Answer:B
Explanation:I just took the test
An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the point of origin of the flight. The plane flies with an airspeed of 120 m/s. If a constant wind blows at 10 m/s toward the west during the flight, what direction must the plane fly relative to the air to arrive at the destination
Answer:
The right solution is "4.8° east of north".
Explanation:
Given:
Distance,
= 500 km
Speed,
[tex]\vec{v}=120 \ m/s[/tex]
Wind (towards west),
[tex]v_0=10 \ m/s[/tex]
According to the question, we get
The angle will be:
⇒ [tex]\Theta=Cos^{-1}(\frac{v_0}{v_1} )[/tex]
[tex]=Cos^{-1}(\frac{10}{120} )[/tex]
[tex]=85.21[/tex] (north of east)
hence,
The direction must be:
⇒ [tex]\Theta'=90-85.21[/tex]
[tex]=4.79^{\circ}[/tex]
or,
[tex]=4.8^{\circ}[/tex] (east of north)
Assignment: 06.05 Infections and Health
Estimate the force a person must exert on a massless string attached to a 0.15 kg ball to make the ball revolve in horizontal circle of radius 0.6 m. The ball makes 2 revolutions per second.
Answer:
[tex]F = 14.2 N[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=0.15kg[/tex]
Radius [tex]r=0.6[/tex]
Angular Velocity [tex]\omega=2rev/s[/tex]
[tex]\omega= =2x2 \pi rad/s=>4 \pi rad/s[/tex]
Generally the equation for Force applied is mathematically given by
[tex]F =mrw2[/tex]
[tex]F=0.15*0.6* (4*x3.14^)2[/tex]
[tex]F = 14.2 N[/tex]
How does exhailing remove waste from the body? Explain the systems that make this happen‚ using complete sentences
When we exhale, 90% waste material is Carbon Dioxide ( CO2 ) , so, it gets exhaled out in the form of CO2 rich air and it gets removed from the body, therefore our internal body becomes more pure and helps in making our internal temperature constant at a suitable level.
How does the density of water change when: (a) it is heated from 0o
C to
4o
C; (b) it is heated from 4o
C to 10o
C ?
Answer:
[b] it id heated from 4o
Explanation:
A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?
Answer:
First of all the formula is F= uR,( force= static friction× reaction)
mass= 5+25=30
F= 50
R= mg(30×10)=300
u= ?
F=UR
u= F/R
u= 50/300=0.17N
A bowling ball with a mass of 8 kg is moving at a speed of 5 m/s. What is its
kinetic energy?
Answer:
100J
Explanation:
Kinetic energy=1/2mv^2
Kinetic energy=(1/2 x 8)x5^2
Kinetic energy=4x25
Kinetic energy=100
100J
The moon Phobos orbits Mars
(mass = 6.42 x 1023 kg) at a distance
of 9.38 x 106 m. What is its period of
orbit?
Answer:
The moon Phobos orbits Mars (m = 6.42 x 1023 kg) at a distance of 9.38 x 106 m.
what was the significance of jumping a.keep the snake b.keep feet cleans c.avoid the hot water d.avoid the Bumbo stick
Answer:
D I think I’m not for sure
Explanation:
Falls often cause injuries, so one of the significant aspects of falls is the displacement and motion of a body. It is defined as how far a body moves vertically during the fall. Accordingly, there are three important factors that should be known to compute the kinetic energy (KE) of a falling body. These factors are:
Answer:
We need, mass, gravity and height.
Explanation:
When a body falls freely from a height, its initial velocity is zero, but due to the height it has some potential energy at the top and the kinetic energy is zero.
As it falls, the potential energy is gradually converted in to the kinetic energy so that the total energy of the falling body is conserved.
At the time as the body strikes the ground, the entire potential energy is converted into the kinetic energy.
Potential energy is given by
U = m g h
where, m is the mass, g is the gravity and h is the height,
So, to get the kinetic energy we require mass, gravity and height of the body.
The total mass of the wheelbarrow and the road is 80 kg calculate the weight of the wheelbarrow and the road
Force required to lift the wheelbarrow
Answer:
The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.
Explanation:
Given that,
The total mass of the wheelbarrow and the road is 80 kg.
The weight of an object is given by :
W = mg
where
g is acceleration due to gravity
So,
W = 80 × 9.8
= 784 N
So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.
At a playground, Maryam a 3-year old girl and Zahirah a 6-year old girl are playing with the swings. Maryam is sitting while Zahirah is standing on the swing. Both of them were given the same push by their mother. Choose the CORRECT statements:
A. Maryam is swinging faster than Zahirah.
B. Zahirah is swinging faster than Maryam.
C. Both swings at the same pace.
D. Maryam is swinging faster since she is younger.
E. Zahirah is swing faster since she is older.
Answer:
both swings at the same place
Explanation:
because there mother is giving same amount of force to both.
What is the Y-component of a vector A, which is of magnitude
16-12 and at a 45° angle to the horizontal?
Explanation:
the answer is in the image above
The Y-component of a vector A, which is of magnitude 16√2 and at a 45° angle to the horizontal would be 16
What is a vector quantity?The quantities that contain the magnitude of the quantities along with the direction are known as the vector quantities.
Examples of vector quantities are displacement, velocity acceleration, force, etc.
As given in the problem we have to find out the Y-component of a vector A, which is of magnitude 16√12 and at a 45° angle to the horizontal,
Y component of the vector A = 16√2 sin45°
=16√2 ×1/√2
=16
Thus, the Y component of vector A would be 16.
To learn more about the vector quantity here, refer to the link;
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Mark pushes his broken car 140 m down the block to his friend's house. He has to exert a 110 N horizontal force to push the car at a constant speed. How much thermal energy is created in the tires and road during this short trip?
Answer:
Explanation:
This is a work problem...energy is created and used in the form of work.
W = FΔx where W is work, F is the force needed to move the object Δx in meters.
W = 110(140) ∴
W = 15000 J
How much time will it take for a person to walk the length of a football field (100 yards)
at a constant speed of 5 ft/s ?
The speed is in feet per seconds so change the length of the field from yards to feet.
1 yard = 3 feet
100 yards x 3 = 300 feet
The field is 300 feet long
Time = distance / speed
Time = 300 feet / 5 feet per second = 60 seconds = 1 minute
It will take 1 minute
Answer:
A person will take 60 Seconds to walk the distance of 100 yards.
Explanation:
Data Given ;
Distance ( d ) = 100 yards = 300 Ft
Speed ( v ) = 5 Ft/s
Time ( t ) = ?
What is speed ?The distance travelled in unit time is called speed.
formula ; [tex]v = \frac{d}{t} \\[/tex]
On putting values,
[tex]5 = \frac{300}{t}[/tex]
[tex]t =\frac{300}{5}[/tex]
[tex]t = 60 sec[/tex]
Hence the time taken by the person is 60 sec.
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What happens to the water when you throw rock into a pond
Answer:
The water usually rushes back too enthusiastically, causing a splash – and the bigger the rock, the bigger the splash. The splash then creates even more ripples that tend to move away from where the rock went into the water.
It is said that a gas fills all the space available to it. Why then doesn't the atmosphere go off into space?
A string of length 3m and total mass of 12g is under a tension of 160N. A transverse harmonic wave with wavelength 25cm and amplitude 2cm travels to the right along the string. It is observed that the displacement at x=0 at t=0 is 0.87cm. a) What is the wave? b) Wrote the wave function, y(x,t) c) Find the particle velocity at position x=0 at time t=10s. What is the maximum particle velocity?
Answer:
What is the answer bro idont now
What type of Earth scientist would be interested in understanding volcanic eruptions on
the seafloor?
Answer:
Marine geologists learn about the rocks and geologic processes of the ocean basins.
A 3.0 kg pendulum swings from point A of height ya = 0.04 m to point B of height yb = 0.12 m, as seen in the diagram below.
Answer:
3.0−0.12=2.88 or 2.88÷0.04=72
0.04×3.0=0.12 and 0.04+3.0=3.04
Answer: 2.4 J
Explanation: Khan Academy