Answer:
4 C
Explanation:
The strength of an electric field can be defined as: [tex]E = \frac{f}{q}[/tex] where f=force and q=charge, and e=strength
Since we're given the strength and the force we can simply rearrange the equation so that we solve for Q:
Original Equation:
[tex]E=\frac{f}{q}[/tex]
Multiply both sides by q
[tex]E*q = f[/tex]
Divide both sides by E
[tex]q=\frac{f}{e}[/tex]
So now plug the known values into the equation
[tex]q=\frac{1000 N}{250 N/C}[/tex]
Simplify:
[tex]q = 4 c[/tex]
Light is incident normally on the short face of a 30∘−60∘−90∘ prism (Figure 1). A drop of liquid is placed on the hypotenuse of the prism.
a) If the index of the prism is 1.50, find the maximum index that the liquid may have for the light to be totally reflected.
Express your answer using three significant figures.
1.06 is the maximum refractive index that the liquid may have for the light to be totally reflected.
Only when a light source passes from a denser to a rarer medium can it completely reflect.
When the angle of incidence surpasses a specific critical value, specular reflection occurs in the more highly refractive of the two mediums at their interface, and this reflection is known as total reflection.
sin [tex]i_{c}[/tex] = μ[tex]_{r}[/tex] / μ[tex]_{d[/tex]
From the diagram
Angle of incidence = 60°
sin60° ≥ sin[tex]i_{c}[/tex] = μ[tex]_{r}[/tex]/μ[tex]_{d}[/tex]
μ[tex]_{r}[/tex] ≤ μ[tex]_{d}[/tex] sin60°
μ[tex]_{r}[/tex] ≤ √1.5 × √3/2
= 1.06
Hence, the maximum index that the liquid may have for the light to be totally reflected is 1.06
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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N. Calculate the magnitude of the tension in the string marked A.
The magnitude of the tension in the string marked A is 39.5 N.
What is the tension in A?The tension in A is determined thus:
The angle at A, θ = tan⁻¹(3/8) = 20.56
When extrapolated below negative x, the angle at B, α = tan⁻¹(5/4) = 51.34
When extrapolated below negative x, the angle at C, β = tan⁻¹(1/6) = 9.46
Taking the horizontal components of tension;
56.3cos(9.46) = A * cos(20.56) + B * cos(51.34)
0.6247B= 55.53 - 0.936A
B = (55.53 - 0.936A)/0.6247 ----(1)
Taking the vertical components of tension;
56.3 * sin(9.46) + A * sin(20.6) = B * sin(51.3)
9.25 + 0.35A = 0.78B ---- (2)
substitute the value (1) in (2)
9.25 + 0.35A = 0.78{(55.53 - 0.936A)/0.6247}
(9.25 + 0.35A) * 0.6247 = 43.31 - 0.73A
0.22A + 0.73A = 43.31 - 5.78
0.93A = 37.53
A = 39.5 N
In conclusion, the tension in A is determined by solving for the vertical and horizontal components of tension.
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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km. The First Cosmic Speed i.e. the speed of a satellite on a low lying circular orbit around this planet is 7.34 km/s.
1. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
2. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
10,378.82 m/s is the second cosmic speed.
69,801 km is the radius of the synchronous orbit of a satellite.
Given
Mass of planet = 4.74 × [tex]10^{24}[/tex] kg
Radius of planet = 5870 km = 5870000m
First Cosmic speed = 7.34 km/sec
1) Second cosmic speed i.e. the minimum speed required for a satellite to break free permanently from the planet is also known as the escape velocity of a satellite.
It can be calculated by
v = √2GM/r where,
v= Escape velocity of the satellite
G = Gravitational constant
M = Mass of planet
r = Radius of planet
v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 10,378.82 m/s
2) Speed of the satellite at the given period
v = 2πr/T where,
T= Time period of rotation = 16.6 × 3600 seconds
r = v×T/2π
r = (7,338.93 x 16.6 x 3600 s) / (2π)
r = 69,801 km
Hence
The Second Cosmic Speed i.e. the minimum speed required for a satellite to break free permanently from the planet is 10,378.82 m/s.
And the radius of the synchronous orbit of a satellite is 69,801 km.
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I need to figure out how to extend a single paragraph's worth of answers that I can make into 3 paragraphs, requesting a LOT of help here.
Question:
Discuss one feature of a spacesuit that protects astronauts outside the spacecraft.
An astronaut's spacesuit is considerably more than just a pair of garments they put on during spacewalks. An entire spacesuit is actually a single-person spacecraft. The Extravehicular Mobility Unit, or EMU, is the official name for the spacesuit used on the International Space Station and Space Shuttle. "Extravehicular" refers to something that is outside of a vehicle or spaceship. To be "mobile" implies to be able to move around while wearing an astronaut suit. The astronaut is shielded by the spacesuit from the perils of being outdoors in space.
The human body cannot survive in space's hostile atmosphere without protection. A spacesuit must shield the astronaut from the vacuum of space, the drastic temperature changes in space, and, if at all possible, it must lessen the astronaut's exposure to radiation. Therefore, the suit's primary function is to act as a pressure vessel. It must keep an air environment close to the skin constant, deliver a constant supply of clean air to the lungs, and expel stale, carbon dioxide-rich air. Every 90 minutes or so, an astronaut in low Earth orbit (LEO) experiences day and night. They can fast chill to -250 F during the night and reach 250 F while the sun is shining on them (-156 C). The body's normal temperature of 98.6 F must be maintained by the suit (37 C).
To do all this, the current NASA suit has 14 layers. The liquid ventilation and cooling garment is the first three layers. It resembles a body-hugging spandex garment that has tubes inside that carry cool water across the body to dissipate extra heat. Air below the next layer cannot escape since it is a pressure vessel made of nylon coated with urethane. The following layer resembles a tent since it is constructed of Dacron. Its goal is to exert pressure on the pressure garment so that it keeps its form and doesn't expand. Neoprene coasted nylon ripstop is the following layer. It is quite resilient to tears. Seven layers of mylar film and foil blanket are used to cover it. These restrict warmth transfer into or out of the suit by acting like a thermos. Due to its several layers, it also provides defense against very small micrometeroids, which drain energy as they pierce and break each layer. The top layer is made of orthofabric, a goretex, nomex, and kevlar mixture. It is very impervious to tearing and thermally reflecting to help regulate temperature. The suits are made by ILC Dover. Incorporating materials that minimize radiation penetration through the suit is something they are always working on.
According to the "Law of Increasing Opportunity Costs," what would be the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning?
A. Food or Drink.
B. Money or income.
C. Sleep or rest.
C. The opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.
What is law of opportunity cost?
The law of increasing opportunity cost is an economic principle that describes how opportunity costs increase as resources are applied.
As the student gives up his sleep or night rest in the place of his exam preparation, we say that the opportunity cost is the sleep or rest.
Thus, the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.
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When a scientific calculator shows the quantity below, what does it mean?
1.5E8
A 40.0-kg child running at 8.00 m/s suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. The child is traveling tangential to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 600 kg ∙ m2 and very little friction at its rotation axis. What is the angular speed of the merry-go-round just after the child has jumped onto it?
The angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.
What is principle of conservation of angular momentum?
The principle of conservation of angular momentum states that the sum of initial angular momentum is equal to final angular momentum.
Li = Lf
Li = Ii ωi + Ic ωc
Li = Iiωi + MR²(V/R)
Li = Iiωi + MRV ----- (1)
Angular speed of the merry go round after the child jumps on itLf = If ωf
ωf = Lf/If
If = Im + MR²
ωf = Lf / ( Im + MR²)
Recall, Lf = Li
ωf = (Iiωi + MRV) / ( Im + MR²)
where;
Iiωi is the initial angular momentum of the merry - go round = 0M is mass of the childR radius of rotationV is tangential speed of the childIm is the moment of inertia of the merry go roundωf = (Iiωi + MRV) / ( Im + MR²)
ωf = (0 + 40 x 1.5 x 8) / (600 + 40(1.5)²)
ωf = (480) / (690)
ωf = 0.696 rad/s
Thus, the angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.
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What does it mean to say that two systems are in thermal equilibrium
Answer:
In simple words, thermal equilibrium means that the two systems are at the same temperature.
37 POINTS PLEASE HELP MEE. The distance from A to B consists of an uphill section and a downhill section. A cyclist rides from A to B and then returns to A for a period of 4 hours and 30 minutes. When traveling as well as when returning the uphill speed is 15 km / h and the speed at the downhill is 20 km / h. Do you calculate the length of the distance from A to B?
The distance from A to B is 39.375 Km.
What is the total distance travelled by the cyclist?
The total distance travelled by the cyclist is given by the formula below:
Total distance = average speed * total time takenTotal distance = (15 + 20)/2 * 4.5
Total distance = 78.75 Km
Thus, distance from A to B = 78.75/2 = 39.375 Km
In conclusion, the total distance travelled is determined from the average speed and total time taken.
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An object with a density of 941.0 kg/m3 and a mass of 1039.0 kg is thrown into the ocean. Find the volume that sticks out of the water. (use ρseawater = 1024 kg/m3)
The volume that sticks out of the water is 83 m³.
To find the answer, we need to know about the archimedes principle.
What's archimedes principle?It says that when an object is on a water surface, the amount of force on the object is equal to the weight of water displaced by it.Mathematically, weight of the object= weight of water displacedWhat's the volume of an object remain on the water surface, if the density and mass of the object are 941.0 kg/m³, 1039.0 kg respectively?Let V = volume of the object, v= volume of water displacedV-v = volume that sticks out of the waterWeight of the object = V× density of object × gWeight of water displaced= v× density of water × gAs per archimedes principle, V× density of object × g=v× density of water × gV-v = density of water - density of object= 1024 - 941 = 83 m³
Thus, we can conclude that the volume that sticks out of the water is 83 m³.
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Four canisters contain helium gas.
If all the canisters had the same amount of particles, which canister would have the fastest moving particles?
W
X
Y
Z
D. The canister with the fastest moving particle is Z due to concentration of particle.
Canister with the fastest moving particle
The speed of the particles depend on the mean distance of the particles.
The canister with the largest concentration per particle will contain particles with the greatest speed.
If the particle concentration is increasing from W to Z, then Z will have fastest moving particle.
Thus, the canister with the fastest moving particle is Z due to concentration of particle.
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Answer:
d. Z
Explanation:
If all the canisters had the same amount of particles, which canister would have the fastest moving particles?
Z
Question 3 of 10
What is the primary means by which heat is transferred through fluids?
O A. Direct contact of particles
OB. Radiation
OC. Electromagnetic waves
OD. Convection currents
The primary means by which heat is transferred through fluids is convection currents (option D).
What is convection current?Convection is the transmission of heat in a fluid by the circulation of currents.
Heat can be transferred by different methods depending on the medium. Fluids like gases and liquids transfer heat through the process of convection.
Therefore, the primary means by which heat is transferred through fluids is convection currents.
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During long jump athlete runs before taking the jump by doing so he
A:provides himself a larger inertia
B:decreases his inertia
C:decreases his momentum
D:decreases his K.E
A. During long jump athlete runs before taking the jump by doing so he provides himself a larger inertia.
What is inertia?
Inertia is the reluctance of an object to stop moving once in motion or start moving when it is at rest.
When an athlete runs before taking the jump, he is trying to increase his inertia, that is his reluctance to stop, thereby increasing his forward motion or jump.
Thus, during long jump athlete runs before taking the jump by doing so he provides himself a larger inertia.
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A golf ball is hit so that it leaves the club face at a velocity of 45m/s at an angle of 40° to the horizontal. by ignoring the effect of air resistance and spin on the ball, use the information to answer the question (a) to (d).
a) Calculate the horizontal component of the velocity
b) Determine the vertical component of the velocity.
c) Find the time taken for the ball to reach its maximum height.
d) Calculate the horizontal distance travelled when the ball is at its maximum height.
Answer:a)45cos40 b)45sin40 c) 2.95 d)102
Explanation:
a) [tex]v_{x} = 45 cos40[/tex]
b) [tex]v_{y}=45sin40[/tex]
c) Since air resistance is negligible, the only force acting on the golf ball is gravity. Thus, its vertical acceleration is -g. We know the final velocity must be 0 m/s, because this will be when the golf ball reaches the maximum height and starts to change direction (it falls back to the ground). We also know initial velocity in the vertical direction (see part b). Thus, we can use this equation: [tex]v_{f} = v_{i} + at[/tex].
[tex]0 m/s = 45sin40 + (-9.8m/s^2)t\\t = 2.95s[/tex]
d) The horizontal distance traveled is dependent on (1) how long the ball is in the air and (2) what the horizontal velocity is. (1) was found in part c, and (2) was found in part a.
[tex]x=vt\\x=45cos40(2.95) =102m[/tex]
You apply a force on an object that is 100 times its mass. The acceleration of the object will be 100m/s^2. After a while, it perpendicularly hits another object with a mass 10 times smaller than the first object. When they collide, in theory, the momentum of the first object will all get transferred into the small object. And that means, the heavy object will completely be at rest, while the smaller object will have a force applied to it by the big object equal to ma. In real life, it seems that when two objects collide, the bigger object doesn’t completely stop and just pushes the smaller or the same massed object. In real life, it seems that when the collision happens, the first object continues moving towards the other object, and doesn’t completely get at rest right after the collision. Why is that?
The bigger object will move some distance because the initial and final momentum of the colliding particles are not zero.
What is momentum?The term momentum has to do with the product of mass and velocity. We know that during a collision, momentum is conserved. This implies that the momentum before collision is equal to the momentum after collision. Thus, the total momentum of the system is constant.
Given the fact that the initial and final momentum of the colliding particles are not zero, the the big object is not going to stop immediately but must move some distance towards the smaller object.
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A bullet of mass 50g moving with an initial speed of 500m/s penetrates a wall and comes to rest at in 0.2seconds. calculate the deceleration of the bullet over the 0.2second.
The deceleration of the bullet over 0.2 second, given the data from the question is –2500 m/s²
What is acceleration?This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
a is the acceleration
v is the final velocity
u is the initial velocity
t is the time
NOTE: Deceleration is the opposite of acceleration
With the above equation for acceleration, we can obtain the deceleration of the bullet. Details below:
How to determine the deceleration of the bulletInitial velocity (u) = 500 m/sFinal velocity (v) = 0 m/sTime (t) = 0.2 sDecelration (a) =?a = (v – u) / t
a = (0 – 500) / 0.2
a = –500 / 0.2
a = –2500 m/s²
Thus, the deceleration of the bullet is –2500 m/s²
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Three ropes A, B and C are tied together in one single knot K.
If the tension in rope A is 65.3 N, then what is the tension in rope B?
The tension in the rope B is determined as 10.9 N.
Vertical angle of cable Btanθ = (6 - 4)/(5 - 0)
tan θ = (2)/(5)
tan θ = 0.4
θ = arc tan(0.4) = 21.8 ⁰
Angle between B and Cθ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰
Apply cosine rule to determine the tension in rope B;
A² = B² + C² - 2BC(cos A)
B = C
A² = B² + B² - (2B²)(cos A)
A² = 2B² - 2B²(cos 43.6)
A² = 0.55B²
B² = A²/0.55
B² = 65.3/0.55
B² = 118.73
B = √(118.73)
B = 10.9 N
Thus, the tension in the rope B is determined as 10.9 N.
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Hello!
This is an example of a force summation in the vertical direction.
We have the tension of rope A upward (+), and the equal vertical components of the tensions of rope B and C downward (-).
These forces sum to zero, since the knot is stationary.
[tex]\Sigma F = T_A - T_{By} - T_{Cy} \\\\0 = T_A - T_{By} - T_{Cy}[/tex]
Ropes 'B' and 'C' form equivalent angles from the vertical. (If you were to draw a line from rope A down). We can use right-triangle trig to determine the angle:
[tex]tan^{-1}(\frac{O}{A}) = \theta[/tex]
The ropes are 5 m long and 2 m tall, which are the opposite and adjacent sides respectively:
[tex]tan^{-1}(\frac{5}{2}) = 68.2^o[/tex]
The vertical components are the adjacent sides from this angle, so, we would use cosine.
[tex]0 = T_A - T_Bcos\theta - T_Ccos\theta[/tex]
Rope 'B' and 'C' have the same tensions since they form the same angle with the vertical and are the same length, so we can call them 'T'.
[tex]0 = T_A - 2Tcos\theta[/tex]
Solving for 'T':
[tex]2Tcos\theta = T_A \\\\T = \frac{T_A}{2cos\theta}\\\\T = \frac{65.3}{2cos(68.2)} = \boxed{87.92 N}[/tex]
Two hockey pucks are moving towards each other. (Assume no friction.) The first one is 0.13 kg and moving at a speed of 1.11 m/s, while the second puck is 0.16 kg and moving at 1.21 m/s, and they collide. (Assume elastic collision.) After collision, the second puck ends up with a speed of 1.16m/s at an angle of 42 degrees below its original path, while the first puck ends up with an unknown speed at an angle above its original path. Find the final speed and angle of the first puck.
The final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.
What happened in an Elastic Collision ?In an elastic collision, both momentum and energy are conserved. But only momentum is conserved in inelastic collision.
Given that two hockey pucks are moving towards each other. The first one is 0.13 kg and moving at a speed of 1.11 m/s, while the second puck is 0.16 kg and moving at 1.21 m/s, and they collide. (Assume elastic collision.) After collision, the second puck ends up with a speed of 1.16m/s at an angle of 42 degrees below its original path, while the first puck ends up with an unknown speed at an angle above its original path.
The given parameters are;
M1 = 0.13 kgM2 = 0.16 kgU1 = 1.11 KgU2 = 1.21 KgV1 = ?V2 = 1.16 kgФ1 = ?Ф2 = 42°The mathematical representation of the above question will be in two components.
Horizontal component
M1U1 - M2U2 = M1V1cosФ - M2V2cosФ
Substitute all the parameters
0.13 x 1.11 - 0.16 x 1.21 = 0.13 x V1 cosФ - 0.16 x 1.16cos42
0.1443 - 0.1936 = 0.13V1cosФ - 0.1379
0.13V1cosФ = 0.0886
V1cosФ = 0.0886/0.13
V1cosФ = 0.6815 ........ (1)
Vertical component
0 = M1V1sinФ - M2V2sinФ
M1V1sinФ = M2V2sinФ
Substitute all the parameters
0.13 x V1 sinФ = 0.16 x 1.16sin42
V1 sinФ = 0.1242/0.13
V1 sinФ = 0.9553 ......... (2)
Divide equation 2 by 1
V1 sinФ / V1 cosФ = 0.9553/ 0.6815
Tan Ф = 1.40
Ф = [tex]Tan^{-1}[/tex](1.4)
Ф = 54.5°
Substitute Ф into equation 2
V1 sin54.5 = 0.9553
V1 = 0.9553 / 0.8141
V1 = 1.17 m/s
Therefore, the final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.
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8. The pulley is assumed massless and frictionless and rotates freely about its axle. The blocks have masses m = 40 g and m₂ = 20 g, and block mi is pulled to the right by a horizontal force of magnitude F = 0.03 N. Find the magnitude of the acceleration of block m2 and the tension in the cord if the surface is frictionless. (2pt) m₂ a₂ T₂ T₂ T₁ m₁
The magnitude of the acceleration of block m2 is 0.09 m/s² and the tension in the cord is 1.8 x 10⁻³ N.
Acceleration of the blocksThe acceleration of the blocks is calculated from the net force on the blocks.
∑F = ma
a = ∑F/m
a = (F) / (m₁ + m₂)
where;
F is the horizontal force appliedm₁ is mass of first block = 40 g = 0.04 kgm₂ is mass of the second block = 20 g = 0.02 kga = (0.03)/(0.04 + 0.02)
a = 0.09 m/s²
Tension due to block m₂T = m₂a
T = (0.02 x 0.09) = 1.8 x 10⁻³ N
Thus, the magnitude of the acceleration of block m2 is 0.09 m/s² and the tension in the cord is 1.8 x 10⁻³ N.
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What force pulls a falling apple down to the ground?
A. Spring force
B. Tension
C. Normal force
D. Gravity
Answer:
d
Explanation:
Gravitational force is a type of force that pulls objects to the Earth's surface.
In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The force required to pull the two hemispheres apart is 4.2×10⁴ N and 29 number of horses are needed to pull these hemispheres apart.
What's the expression of force in terms of pressure?Mathematically, force = pressure/areaTotal area of the two hemispheres = 4π×(0.43)²= 2.3 m²Total pressure on the hemispheres= 15 milibar (directed inward) + 940 milibar (atmospheric pressure) = 955 milibar=955×100 N/m²= 9.55×10⁴ N/m²
Force on the hemispheres= 9.55×10⁴/2.3 = 4.2×10⁴ NWhat's the minimum number of horses required to get 4.2×10⁴ N of force, if each horse can pull with a force of 1450N?No. of horses required to separate the hemispheres = 4.2×10⁴/1450 = 29
Thus, we can conclude that the 29 horses are needed to pull the two hemispheres with a force of 4.2×10⁴ N.
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Drag each label to the correct location on the image. Jessica is visiting a park with her mother. Jessica sits on a swing. Her mother pulls the swing to a height of 3 meters above the ground and lets it go. The image shows Jessica at three positions on the swing. Jessica‘s mass is 44 kilograms and the maximum velocity of the swing is 5 meters/second. What’s the energy she has at each position shown? Ignore friction and air resistance. Use g = 9.8 m/s2, PE = m × g × h, and KE = 0 joules KE = 550 joules PE = 862.4 joules
The energy she has at each position shown are:
Position at maximum height -- 1,294 JPosition at minimum height --0 JPosition at maximum velocity - 550 JPosition at minimum velocity - 0 JWhat does velocity implies?Velocity is known to be a term that connote the direction of any kind of a moving body or an object.
Note that the Speed is known to be a a scalar quantity and as such, Velocity is said to be a vector quantity.
Note also that from the question given, Jessica's height of the swing 3 meters above the ground, therefore:
Jessica Position at maximum height :
PE = mgh
PE = 44kg x 9.8m/s² x 3
PE = 1,294 J
Jessica Position at minimum height:
PE = 0 J
Jessica Position at maximum velocity:
KE = 1/2 x mv²
KE = 1/2 x 44kg x (5m/s²)²
KE = 550 J
Jessica Position at minimum velocity:
KE = 0 J
Therefore, The energy she has at each position shown are:
Position at maximum height -- 1,294 JPosition at minimum height --0 JPosition at maximum velocity - 550 JPosition at minimum velocity - 0 JLearn more about velocity from
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A random selection of volunteers at a research institute has been exposed to a weak flu virus. After the volunteers began to have flu symptoms, of them were given multivitamin tablets daily that contained gram of vitamin C and grams of various other vitamins and minerals. The remaining volunteers were given tablets containing grams of vitamin C only. For each individual, the length of time taken to recover from the flu was recorded. At the end of the experiment the following data were obtained. Days to recover from flu Treated with multivitamin , , , , , , , , , Treated with vitamin C , , , , , , , , , Suppose that it is known that the population standard deviation of recovery time from the flu is days when treated with multivitamins and that the population standard deviation of recovery time from the flu is days when treated with vitamin C tablets. Suppose also that both populations are approximately normally distributed. Construct a confidence interval for the difference between the mean recovery time when treated with multivitamins () and the mean recovery time when treated with vitamin C only (). Then find the lower limit and upper limit of the confidence interval.
The confidence interval is given as lower interval is -0.77 while The upper interval is 1.67
How to solve for the confidence intervalMultivitamin treatment
n1 is 10
σ1 = 1.8
x21 = 5
vitamin C treatment
n2 = 10
σ2 = 1.5
x2 = 4.55
zα/2 = 1.645
The formula for the confidence interval is given as
(x1 - x2) ± zα/2[tex]\sqrt{\frac{sd1^2}{n1}+\frac{sd2^2}{n2} }[/tex]
When we input the values we have above we would have
CI = (-0.7688550 , 1.668855 )
The the lower interval is -0.77
The upper interval is 1.67
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Complete questionA random selection of volunteers at a research institute has been exposed to a weak flu virus. After the volunteers began to have flu symptoms, 10 of them were given multivitamin tablets daily that contained 1 gram of vitamin C and 3 grams of various other vitamins and minerals. The remaining 10 volunteers were given tablets containing 4 grams of vitamin C only. For each individual, the length of time taken to recover from the flu was recorded. At the end of the experiment the following data were obtained:
Treated with multivitamin
Days to recover from flu
2.4, 6.4, 9.1, 4.1, 4.6, 6.4, 6.4, 3.2, 6.9, 0.5
Treated with Vitamin C
5.2, 3, 3.6, 5.5, 7.5, 6.7, 1.3, 1.9, 5.3, 5.5
Suppose that it is known that the population standard deviation of recovery time from the flu is 1.8 days when treated with multivitamins and that the
population standard deviation of recovery time from the flu is 1.5 days when treated with vitamin C tablets. Suppose also that both populations are
approximately normally distributed. Construct a 90% confidence interval for the difference µµ₂ between the mean recovery time when treated with
multivitamins (μ,) and the mean recovery time when treated with vitamin C only (H2). Then complete the table below.
Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)
What is the lower limit of the 90% confidence interval?
What is the upper limit of the 90% confidence interval?
A 15.0-kg child descends a slide 2.40 m high and reaches the bottom with a speed of 1.10 m/s .
How much thermal energy due to friction was generated in this process?
Express your answer to three significant figures and include the appropriate units.
The thermal energy that is generated due to friction is 344J.
What is the thermal energy?Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.
Thus, the kinetic energy is given as;
KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J
PE = mgh = 15.0-kg * 9.8 m/s^2 * 2.40 m = 352.8 J
The thermal energy is; 352.8 J - 9.1 J = 344J
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please need answers ASAP.
Answer:
overloading can be avoided if two many appliances are not connected to a single socket short circuiting is a name given to a situation in which they live and the natural voice accidentally coming contact
A traffic light hangs from a pole as shown in (Figure 1). The uniform aluminum pole AB is 7.60 m long and has a mass of 10.0 kg . The mass of the traffic light is 23.5 kg. 1.) Determine the tension in the horizontal massless cable CD. 2) Determine the vertical component of the force exerted by the pivot A on the aluminum pole. 3.) Determine the horizontal component of the force exerted by the pivot A on the aluminum pole.
There is a 446N force in the horizontal massless cable CD. The vertical component of the pivot A's force on the aluminum pole, which is 328.3N. The force that the pivot A applies to the aluminum pole has a horizontal component of 446N.
We need to be aware of the force in order to discover the solution.
How can I determine the tension in the CD's horizontal massless cable?The free body diagram of the masses must be drawn in order to determine the tension in the horizontal cable CD.To obtain the tension on CD in the free body diagram, let's balance all the vertical and horizontal forces. [tex]TH-mg\frac{l}{2}cos\alpha -Mglcos\alpha =0\\T=\frac{glcos\alpha (\frac{m}{2}+M )}{h} \\T=446N[/tex]where, H=3.8m, l=7.6m, m=10kg, M=23.5 kg and alpha= 37 degree,
How to calculate the force the pivot A exerted on aluminum's vertical and horizontal components?The overall force acting vertically is,[tex]F_V-mg-Mg=0\\F_v=328.2N\\[/tex]
The overall force acting horizontally is,[tex]F_H=T=446N[/tex]
In light of this, we may say that the tension in the horizontal massless cable CD and the horizontal component of the force applied by the pivot A to the aluminum pole are identical and each exert 466N of force, whereas the vertical component of that force is 328.3N.
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A closed curve encircles several conductors. The line integral around this curve is (image attached below)
a) What is the net current in the conductors?
b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral?
The net current in the conductors and the value of the line integral
[tex]I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A[/tex]The resultant remains same 3.2 *10^4 TmThis is further explained below.
What is the net current in the conductors?Generally,
To put it another way, the total current In flowing across a surface S (contained by C) is proportional to the line integral of the magnetic B-field (in tesla, T).
[tex]\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \iint_S \mathbf{J} \cdot \mathrm{d}\mathbf{S} = \mu_0I_\mathrm{enc}[/tex]
[tex]I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A[/tex]
B)
In conclusion, It is possible for the line integral to go around the loop in either direction (clockwise or counterclockwise), the vector area dS to point in either of the two normal directions and Ienc, which is the net current passing through the surface S, to be positive in either direction—but both directions can be chosen as positive in this example. The right-hand rule solves these ambiguities.
The resultant remains the same at 3.2 *10^4 Tm
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converting 67 m•s¹ to km•h¹
Answer:
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Two bullets are fired at the same time with the same kinetic energy.
If one bullet has twice the mass of the other, what is the ratio of the speed of the lighter bullet to the speed of the heavier?
Which can do the most work?
The ratio of the speed of the lighter bullet to the speed of the heavier will be four times the heavier bullet.
Both can do same amount of work.
What is kinetic energy?Kinetic energy of a body is the energy due to the motion of the body.
Kinetic energy = mv²/2
where m is mass and v is velocity of the object.
Since both objects have the same kinetic energy but one bullet has twice the mass of the other, the ratio of the speed of the lighter bullet to the speed of the heavier will be four times the heavier bullet.
The kinetic energy of both bullets is the same. Hence, they can do equal amount of work.
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A roller-coaster car shown in the figure below is pulled up to point 1 where it is released from rest. Take y = 39 m .
(Figure 1)
Assuming no friction, calculate the speed at point 2.
Express your answer to two significant figures and include the appropriate units.
Calculate the speed at point 3.
Express your answer to two significant figures and include the appropriate units.
Calculate the speed at point 4.
Express your answer to two significant figures and include the appropriate units.
The speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.
Speed of the roller coasterThe speed of the roller coaster at any position is calculated by applying the principle of conservation of energy.
K.E = P.E
¹/₂mv² = mgh
v = √2gh
where;
h is vertical displacementSpeed at point 2v(2) = √[(2 x 9.8)(39 - 0)]
v(2) = 28 m/s
Speed at point 3v(3) = √[(2 x 9.8)(39 - 26)]
v(3) = 16 m/s
Speed at point 4v(4) = √[(2 x 9.8)(39 - 14)]
v(4) = 22 m/s
Thus, the speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.
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