Some types of spiders build webs that consist of threads made of dry silk coated with a solution of a variety of compounds. This coating leaves the threads, which are used to capture prey, hygroscopic - that is, they attract water from the atmosphere. It has been hypothesized that this aqueous coating makes the threads good electrical conductors. To test the electrical properties of coated thread, researchers placed a 5-mm length of thread between two electrical contacts. The researchers stretched the thread in 1-mm increments to more than twice its original length, and then allowed it to return to its original length, again in 1-mm increments. Some of the resistance measurements are shown.If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area A of the coating compare when the thread is 13 mm long versus the starting length of 5 mm? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread.If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area of the coating compare when the thread is 13 long versus the starting length of 5 ? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread.A13mm is about 1/10 A5mm.A13mm is about 1/4 A5mm. === correct answer... I figured it out. R = pL/A. L is 2.5 times. Therefore, A must be 1/4 times.A13mm is about 2/5 A5mm.A13mm is the same as A5mm.
Answers
Answer:
A13 mm is about 1/4 A5 mm
Explanation:
Find the attachment
Related Questions
first law of equilibrium
Answers
Answer:
For an object to be an equilibrium it must be experiencing no acceleration.
Explanation:
Hope it helps.
A uniform ladder stands on a rough floor and rests against a frictionless wall. Since the floor is rough, it exerts both a normal force N1 and a frictional force f1 on the ladder. However, since the wall is frictionless, it exerts only a normal force N2 on the ladder. The ladder has a length of L = 4.6m, a weight of WL= 69.0N , and rests against the wall a distance d = 3.75 m above the floor. If a person with a mass of m = 90 kg is standing on the ladder, determine the forces exerted on the ladder when the person is halfway up the ladder.
Required:
Solve of N1, N2 and f1
Answers
Answer:
The normal force N1 exerted by the floor is [tex]N_1 = 951 \ N[/tex]
The normal force N2 exerted by the wall is [tex]N_2= 616.43 \ N[/tex]
The frictional force exerted by the wall is [tex]f = N_2 = 616.43 \ N[/tex]
Explanation:
From the question we are told that
The length of the ladder is [tex]L = 4.6 \ m[/tex]
The weight of the ladder is
The distance of the ladder position on the wall from the floor is [tex]D = 3.75 \ m[/tex]
The mass of the person is [tex]m = 90 kg[/tex]
Applying Pythagoras theorem
The length of the position the ladder on the ground from the base of the wall is
[tex]A = \sqrt{L^ 2 - D^2}[/tex]
substituting values
[tex]A = \sqrt{(4.6^2)-(3.75^2)}[/tex]
[tex]A = 2.66 \ m[/tex]
In order the for the ladder not to shift from the ground the sum of the moment about the position of the ladder on the ground must be equal to zero this is mathematically represented as
[tex]\sum M = 0 = N_2 * D - [\frac{1}{2} * W_L ] * [(mg) *A ][/tex]
[tex]\sum M = 0 = N_2 * 3.75 - [\frac{1}{2} * 69.0 ] * [(90*9.8) * \frac{4.6}{2.66} ][/tex]
[tex]N_2 * 3.75 =2311.62[/tex]
[tex]N_2 * 3.75 =2311.62[/tex]
[tex]N_2= 616.43 \ N[/tex]
Now the force exerted by the floor on the ladder is mathematically represented as
[tex]N_1 = W_L + (m * g )[/tex]
substituting values
[tex]N_1 = 951 \ N[/tex]
Now the horizontal forces acting on the ladder are [tex]N_2 \ and \ f[/tex] and they are in opposite direction so
[tex]f = N_2 = 616.43 \ N[/tex]
uring a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball? During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball? 26.7 N 16.7 N 13.3 N 107 N 40.0 N
Answers
Answer:
107 N, option d
Explanation:
Given that
mass of the ball, m = 0.2 kg
initial velocity of the ball, u = 20 m/s
final velocity of the ball, v = -12 m/s
time taken, Δt = 60 ms
Solving this question makes us remember "Impulse Theorem"
It states that, "that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object"
Mathematically, it is represented as
FΔt = m(v - u), where
F = the average force
Δt = time taken
m = mass of the ball
v = final velocity of the ball
u = initial velocity of the ball
From the question we were given, if we substitute the values in it, we have
F = ?
Δt = 60 ms = 0.06s
m = 0.2 kg
v = -12 m/s
u = 20 m/s
F = 0.2(-12 - 20) / 0.06
F = (0.2 * -32) / 0.06
F = -6.4 / 0.06
F = -106.7 N
Thus, the magnitude is 107 N
An accident in a laboratory results in a room being contaminated by a radioisotope with a half life of 4.5 hours. If the radiation is measured to be 64 times the maximum permissible level, how much time must elapse before the room is safe to enter? The mass of Helium atom is 4.002602 u (where u = 1.66 x 10-27 kg) but the mass of 1 proton is 1.00730 u and 1 neutron is 1.00869 u. Calculate the binding energy per nucleon in MeV.
Answers
Answer:
a) t = 27.00 h
b) B = 6.84 MeV/nucleon
Explanation:
a) The time can be calculated using the following equation:
[tex] R = R_{0}e^{-\lambda*t} [/tex]
Where:
R: is the radiation measured at time t
R₀: is the initial radiation
λ: is the decay constant
t: is the time
The decay constant can be calculated as follows:
[tex] t_{1/2} = \frac{ln(2)}{\lambda} [/tex]
Where:
t(1/2): is the half life = 4.5 h
[tex] \lambda = \frac{ln(2)}{t_{1/2}} = \frac{ln(2)}{4.5 h} = 0.154 h^{-1} [/tex]
We have that the radiation measured is 64 times the maximum permissible level, thus R₀ = 64R:
[tex] \frac{R}{64R} = e^{-\lambda*t} [/tex]
[tex] t = -\frac{ln(1/64)}{\lambda} = -\frac{ln(1/64)}{0.154 h^{-1}} = 27.00 h [/tex]
b) The binding energy (B) can be calculated using the following equation:
[tex]B = \frac{(Z*m_{p} + N*m_{n} - M_{A})}{A}*931.49 MeV/u[/tex]
Where:
Z: is the number of protons = 2 (for [tex]^{4}_{2}He[/tex])
[tex]m_{p}[/tex]: is the proton mass = 1.00730 u
N: is the number of neutrons = 2 (for [tex]^{4}_{2}He[/tex])
[tex]m_{n}[/tex]: is the neutron mass = 1.00869 u
[tex]M_{A}[/tex]: is the mass of the He atom = 4.002602 u
A = N + Z = 2 + 2 = 4
The binding energy of [tex]^{4}_{2}He[/tex] is:
[tex]B = \frac{(2*1.00730 + 2*1.00869 - 4.002602)}{4}*931.49 MeV/u = 7.35\cdot 10^{-3} u*931.49 MeV/u = 6.84 MeV/nucleon[/tex]
Hence, the binding energy per nucleon is 6.84 MeV.
I hope it helps you!
assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charged the battery
Answers
Answer:
The amount of water that will power a battery with that rating = 7.35 m³
Explanation:
The power rating for the battery is missing from the question.
Complete Question
Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charged the battery with power rating, 12 V, 50 Ampere-minutes
Solution
Potential energy possessed by water at that height = mgH
m = mass of the water = ρV
ρ = density of water = 1000 kg/m³
V = volume of water = ?
g = acceleration due to gravity = 9.8 m/s²
H = height of water = 50 cm = 0.5 m
Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J
Energy of the battery = qV
q = 50 A.h = 50 × 60 = 3,000 C
V = 12 V
qV = 3,000 × 12 = 36,000 J
Energy = 36,000 J
At a 100% conversion rate, the energy of the water totally powers the battery
(4900V) = (36,000)
4900V = 36,000
V = (36,000/4900)
V = 7.35 m³
Hope this Helps!!!
A layer of ethyl alcohol (n = 1.361) is on top of water (n = 1.333). To the nearest degree, at what angle relative to the normal to the interface of the two liquids is light totally reflected?
a. 78 degree
b. 88 degree
c. 68 degree
d. 49 degree
e. the critical angle isundefined
Answers
Answer:
a. 78 degree
Explanation:
According to Snell's Law, we have:
(ni)(Sin θi) = (nr)(Sin θr)
where,
ni = Refractive index of medium on which light is incident
ni = Refractive index of ethyl alcohol = 1.361
nr = Refractive index of medium from which light is refracted
nr = Refractive index of ethyl alcohol = 1.333
θi = Angle of Incidence
θr = Angle of refraction
So, the Angle of Incidence is know as the Critical Angle (θc), when the refracted angle becomes 90°. This is the case of total internal reflection. That is:
θi = θc
when, θr = 90°
Therefore, Snell's Law becomes:
(1.361)(Sin θc) = (1.333)(Sin 90°)
Sin θc = 1.333/1.361
θc = Sin⁻¹ (0.9794)
θc = 78.35° = 78° (Approximately)
Therefore, correct answer will be:
a. 78 degree
The angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.
From the information given;
the refractive index of the ethyl alcohol [tex]\mathbf{n_1= 1.361}[/tex]the refractive index of the water [tex]\mathbf{n_2 = 1.333}[/tex] the angle of incidence is the critical angle [tex]\theta_i = \theta_c[/tex] the angle of refraction [tex]\theta _r = 90^0[/tex]According to Snell's Law of refraction;
[tex]\mathbf{n_1 sin \theta _c = n_2 sin \theta_r}[/tex]
[tex]\mathbf{1.361 \times sin \theta _c = 1.333 \times sin 90}[/tex]
[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times sin 90}{1.361}}[/tex]
[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times 1}{1.361}}[/tex]
[tex]\mathbf{ \theta _c = sin^{-1} (0.9794)}[/tex]
[tex]\mathbf{ \theta _c =78.35^0}[/tex]
[tex]\mathbf{ \theta _c \simeq78^0}[/tex]
Therefore, we can conclude that the angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.
Learn more about Snell Law of refraction here:
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A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 28.0 m/s. Then the vehicle moves for 41.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.
(a) How long is the ride-sharing car in motion (in s)?
(b) What is the average velocity of the ride-sharing car for the motion described? (Enter the magnitude in m/s.)
Answers
Answer:
Explanation:
Time taken to accelerate to 28 m /s
= 28 / 2 = 14 s
a ) Total length of time in motion
= 14 + 41 + 5
= 60 s .
b )
Distance covered while accelerating
s = ut + 1/2 at²
= 0 + .5 x 2 x 14²
= 196 m .
Distance covered while moving in uniform motion
= 28 x 41
= 1148 m
distance covered while decelerating
v = u - at
0 = 28 - a x 5
a = 5.6 m / s²
v² = u² - 2 a s
0 = 28² - 2 x 5.6 x s
s = 28² / 2 x 5.6
= 70 m .
Total distance covered
= 196 + 1148 + 70
= 1414 m
total time taken = 60 s
average velocity
= 1414 / 60
= 23.56 m /s .
Plaskett's binary system consists of two stars that revolve In a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal . Assume the orbital speed of each star is |v | = 240 km/s and the orbital period of each is 12.5 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 times 1030 kg Your answer cannot be understood or graded.
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The mass is [tex]M =1.43 *10^{32} \ kg[/tex]
Explanation:
From the question we are told that
The mass of the stars are [tex]m_1 = m_2 =M[/tex]
The orbital speed of each star is [tex]v_s = 240 \ km/s =240000 \ m/s[/tex]
The orbital period is [tex]T = 12.5 \ days = 12.5 * 2 4 * 60 *60 = 1080000\ s[/tex]
The centripetal force acting on these stars is mathematically represented as
[tex]F_c = \frac{Mv^2}{r}[/tex]
The gravitational force acting on these stars is mathematically represented as
[tex]F_g = \frac{GM^2 }{d^2}[/tex]
So [tex]F_c = F_g[/tex]
=> [tex]\frac{mv^2}{r} = \frac{Gm_1 * m_2 }{d^2}[/tex]
=> [tex]\frac{v^2}{r} = \frac{GM}{(2r)^2}[/tex]
=> [tex]\frac{v^2}{r} = \frac{GM}{4r^2}[/tex]
=> [tex]M = \frac{v^2*4r}{G}[/tex]
The distance traveled by each sun in one cycle is mathematically represented as
[tex]D = v * T[/tex]
[tex]D = 240000 * 1080000[/tex]
[tex]D = 2.592*10^{11} \ m[/tex]
Now this can also be represented as
[tex]D = 2 \pi r[/tex]
Therefore
[tex]2 \pi r= 2.592*10^{11} \ m[/tex]
=> [tex]r= \frac{2.592*10^{11}}{2 \pi }[/tex]
=> [tex]r= 4.124 *10^{10} \ m[/tex]
So
[tex]M = \frac{v^2*4r}{G}[/tex]
=> [tex]M = \frac{(240000)^2*4*(4.124*10^{10})}{6.67*10^{-11}}[/tex]
=> [tex]M =1.43 *10^{32} \ kg[/tex]
A man pushes a 25kg box up an incline 2.0m by applying a steady force of 95N parallel to the incline. The box moves up the incline at a steady speed. The incline makes an angle 15 degrees to the horizontal
a) What is the force of friction between the incline and the box
b)If the box is released at the top of the incline, what will its speed be at the bottom
Answers
Answer:
a) Ff = 19.29 N
b) v = 3.00 m/s
Explanation:
a) To calculate the friction force you use the second Newton Law in the incline plane, with an acceleration equal to zero, because the motion of the box has a constant velocity:
[tex]F-F_f-Wsin(\theta)=0\\\\[/tex] (1)
F: force applied by the man = 95N
Ff: friction force
W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N
θ: degree of the inclined plane = 15°
You solve the equation (1) for Ff and you replace the values of all variables in the equation (1):
[tex]F_f=-Wsin(\theta)+F\\\\F_f=-(245N)sin18\°+95N=19.29N[/tex]
b) To fins the velocity of the box at the bottom you use the following formula:
[tex]W_N=\Delta K[/tex] (2)
That is, the net work over the box is equal to the change in the kinetic energy of the box.
The net work is:
[tex]W_N=Mgsin(18\°)d-Ffd[/tex]
d: distance traveled by the box = 2.0m
[tex]W_N=245sin18\°(2.0m)N-19.29(2.0m)N=112.83J[/tex]
You use this value of the net work to find the final velocity of the box, by using the equation (2):
[tex]112.8J=\frac{1}{2}m[v^2-v_o^2]\\\\v_o=0m/s\\\\v=\sqrt{\frac{2(112.8J)}{m}}=\sqrt{\frac{225.67J}{25kg}}=3.00\frac{m}{s}[/tex]
The speed of the box, at the bottom of the incline plane is 3.00 m/s
A ball is projected upward at time t = 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to
Answers
Answer:
The time when the ball strikes the ground is closest to [tex]t_t = 9.4 \ s[/tex]
Explanation:
From the question we are told that
The time of projection is t = 0.0 s
The distance of the point from the ground is [tex]d = 90 \ m[/tex]
The initial velocity of the ball is [tex]v _i = 36 .2 \ m/s[/tex]
generally the time required to reach maximum height is
[tex]t_r = \frac{g}{v}[/tex]
Where is the acceleration due to gravity with value [tex]g = 9.8 \ m/s^2[/tex]
Substituting values
[tex]t_r = \frac{36.2}{9.8}[/tex]
[tex]t_r = 3.69 s[/tex]
when returning the time and velocity at the roof level is t = 3.69 s and u = 36.2 m/s this due to the fact that air resistance is negligible
The final velocity at which it hit the ground is
[tex]v_f^2 = u^2 + 2ag[/tex]
So
[tex]v_f = \sqrt{ u^2 + 2gs}[/tex]
substituting values
[tex]v_f = \sqrt{ 3.69^2 + 2* 9.8 * 90}[/tex]
[tex]v_f = 55.45 \ m/s[/tex]
The time taken for the ball to move from the roof level to the ground is
[tex]t_g = \frac{v-u}{a}[/tex]
substituting values
[tex]t_g = \frac{55.45 -36.2}{9.8}[/tex]
[tex]t_g = 1.96 \ s[/tex]
The total time for this travel is
[tex]t_t = t_g + 2 t_r[/tex]
[tex]t_t = 1.96 + 2(3.69)[/tex]
[tex]t_t = 9.4 \ s[/tex]
Find the equivalent resistance from the indicated terminal pair of the networks in the attached doc
Answers
Answer:
a) R = 2.5 Ω, b) R = 1 Ω, c) R = 2R / 3 Ω
Explanation:
The resistance configuration can be in series or in parallel, for each one the equivalent resistance can be calculated
series, the equivalent resistance is the sum of the resistances
parallel, the inverse of the equivalent resistance is the inverse of the sum of the resistances
let's apply these principles to each case
case a)
equivalent series resistance
R₁ = 1 +4 = 5 ohm
R₂ = 2 +3 = 5 ohn
these two are in parallel
1 / R = 1/5 +1/5
1 / R = 2/5
R = 2.5 Ω
case B
we solve the parallel
1 / R₁ = ½ + ½ = 1
R₁ = 1 Ω
we solve the resistors in series
R₂ = 1 + 1
R₂ = 2 Ω
finally we solve the last parallel
1 / R = ½ +1/2 = 1
R = 1 Ω
case C
we solve house resistance pair in series
R₁ = R + 2R = 3R
we go to the next mesh
R₂ = R + 2R = 3R
R₃ = R + 2R = 3R
last mesh
R₄ = R + R = 2R
now we solve the parallel of this equivalent resistance
1 / R = 1 / R₁ + 1 / R₂ + 1 / R₃ + 1 / R₄
1 / R = 1 / 3R + 1 / 3R + 1 / 3R + 1 / 2R
1 / R = 3 / 3R + 1 / 2R = 1 / R + 1 / 2R
1 / R = 3 / 2R
R = 2R / 3 Ω
state Ohm`s law as applied in electricity
Answers
Answer:
Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists. When spelled out, it means voltage = current x resistance, or volts = amps x ohms, or V = A x Ω.
When you take your 1900-kg car out for a spin, you go around a corner of radius 56 m with a speed of 14 m/s. The coefficient of static friction between the car and the road is 0.88. Part A Assuming your car doesn't skid, what is the force exerted on it by static friction
Answers
Answer:
6,650 newtons
Explanation:
The computation of the force exerted on it by static friction is shown below:
Data provided in the question
Mass of car = m = 1,900 kg
speed = v = 14 m/s
radius = r = 56 m
Let us assume friction force be f
And, the Coefficient of friction = [tex]\mu[/tex]= 0.88
As we know that
[tex]f = \frac{mv^2}{r}[/tex]
[tex]= \frac{1,900 \times 14^2}{56}[/tex]
= 6,650 newtons
We simply applied the above formula so that the force exerted could come
A wire of length L is made up of two sections of two different materials connected in series. The first section of length L1 = 17.7 m is made of steel and the second section of length L2 = 28.5 m is made of iron. Both wires have the same radius of 5.30 ✕ 10−4 m. If the compound wire is subjected to a tension of 148 N, determine the time taken for a transverse pulse to move from one end of the wire to the other. The density of steel is 7.75 ✕ 103 kg/m3 and the density of iron is 7.86 ✕ 103 kg/m3.
Answers
Answer:
Explanation:
velocity of wave in a tense wire is given by the expression
[tex]v= \sqrt{\frac{T}{m} }[/tex]
v is velocity . T is tension and m is mass per unit length .
for steel wire
m = π r² ρ where r is radius and ρ is density
= 3.14 x (5.3 x 10⁻⁴)²x7.75 x 10³
= 683.57 x 10⁻⁵ kg/m
v = [tex]\sqrt{\frac{148}{683.57\times 10^{-5}} }[/tex]
= 1.47 x 10² m /s
= 147 m /s
for iron wire
m = π r² ρ where r is radius and ρ is density
= 3.14 x (5.3 x 10⁻⁴)²x7.86 x 10³
= 693.27 x 10⁻⁵ kg/m
[tex]v = \sqrt{\frac{148}{693.27\times 10^{-5}} }[/tex]
= 146 m /s
Time taken to move from one end to another
= 17.7 / 147 + 28.5 / 146
= .12 + .195
= .315 s .
Leah is moving in a spaceship at a constant velocity away from a group of stars. Which one of the following statements indicates a method by which she can determine her absolute velocity through space?
A) She can measure her increases in mass.
B) She can measure the contraction of her ship.
C) She can measure the vibration frequency of a quartz crystal.
D) She can measure the changes in total energy of her ship.
E) She can perform no measurement to determine this quantity.
Answers
Answer:
E) She can perform no measurement to determine this quantity.
Explanation:
A spacecraft is a machine used to fly in outer space.
According to Isaac Newton's third law of motion, every action produces an equal and opposite reaction. When fuel is shoot out of one end of the rocket, the rocket moves forward for which no air is required.
As Leah is moving in a spaceship at a constant velocity away from a group of stars, she cannot measure to determine this quantity.
A rocket rises vertically, from rest, with an acceleration of 3.99 m/s2 until it runs out of fuel at an altitude of 775 m. After this point, its acceleration is due to gravity downwards. What is the speed of the rocket, in m/s, when it runs out of fuel?
Answers
Answer:
Vf = 78.64 m/s
Explanation:
The rocket is travelling upward at a constant acceleration of 3.99 m/s² until it runs out of fuel. So, in order to calculate its velocity at the point, where it runs out of fuel, we can simply use 3rd equation of motion:
2as = Vf² - Vi²
where,
a = acceleration = 3.99 m/s²
s = distance or height covered by rocket till fuel runs out = 775 m
Vf = Final Velocity = ?
Vi = Initial velocity = 0 m/s (Since, rocket starts from rest)
Therefore,
2(3.99 m/s²)(775 m) = Vf² - (0 m/s)²
Vf = √(6184.5 m²/s²)
Vf = 78.64 m/s
Which factor caused higher oil prices to directly lead to inflation?
It increased demand for cars, leading to higher automobile prices.
Companies passed on production and transportation costs to consumers.
The government began to print more money.
Gas prices declined too quickly, leading to oversupply
Answers
Answer: B, Companies passed on production and transportation costs to consumers
Explanation:
A higher oil price occurred when companies passed on production and transportation costs to consumers.
Cause of high price of oilThe oil producing companies spend so much money in producing crude oil from the reservoirs to the surface. They also spend money in processing and transporting the crude oil to the end users or consumers.
The final price of the oil depends on the total amount spent by these companies in producing the hydrocarbons.
Thus, a higher oil price occurred when companies passed on production and transportation costs to consumers.
Learn more about inflations here: https://brainly.com/question/1082634
A spherical balloon has a radius of 7.40 m and is filled with helium. Part A How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 990 kg ? Neglect the buoyant force on the cargo volume itself. Assume gases are at 0∘C and 1 atm pressure (rhoair = 1.29 kg/m3, rhohelium = 0.179 kg/m3).
Answers
Answer:
The mass of the cargo is [tex]M = 188.43 \ kg[/tex]
Explanation:
From the question we are told that
The radius of the spherical balloon is [tex]r = 7.40 \ m[/tex]
The mass of the balloon is [tex]m = 990\ kg[/tex]
The volume of the spherical balloon is mathematically represented as
[tex]V = \frac{4}{3} * \pi r^3[/tex]
substituting values
[tex]V = \frac{4}{3} * 3.142 *(7.40)^3[/tex]
[tex]V = 1697.6 \ m^3[/tex]
The total mass the balloon can lift is mathematically represented as
[tex]m = V (\rho_h - \rho_a)[/tex]
where [tex]\rho_h[/tex] is the density of helium with a value of
[tex]\rho_h = 0.179 \ kg /m^3[/tex]
and [tex]\rho_a[/tex] is the density of air with a value of
[tex]\rho_ a = 1.29 \ kg / m^3[/tex]
substituting values
[tex]m = 1697.6 ( 1.29 - 0.179)[/tex]
[tex]m = 1886.0 \ kg[/tex]
Now the mass of the cargo is mathematically evaluated as
[tex]M = 1886.0 - 1697.6[/tex]
[tex]M = 188.43 \ kg[/tex]
A person is swimming in a river with a current that has speed vR with respect to the shore. The swimmer first swims downstream (i.e. in the direction of the current) at a constant speed, vS, with respect to the water. The swimmer travels a distance D in a time tOut. The swimmer then changes direction to swim upstream (i.e. against the direction of the current) at a constant speed, vS, with respect to the water and returns to her original starting point (located a distance D from her turn-around point) in a time tIn. What is tOut in terms of vR, vS, and D, as needed?
Answers
Answer:
The time taken is [tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]
Explanation:
From the question we are told that
The speed of the current is [tex]v__{R}}[/tex]
The speed of the swimmer in direction of current is [tex]v__{S}}[/tex]
The distance traveled by the swimmer is [tex]D[/tex]
The time taken to travel this distance is [tex]t_{out}[/tex]
The speed of the swimmer against direction of current is [tex]v__{s}}[/tex]
The resultant speed for downstream current is
[tex]V_{r} = v__{S}} +v__{R}}[/tex]
The time taken can be mathematically represented as
[tex]t_{out} = \frac{D}{V_{r}}[/tex]
[tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]
A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atmatm pressure is 1.00028.
Required:
How many bright-dark-bright fringe shifts are observed as the cell fills with air?
Answers
Answer:
55.3
Explanation:
The computation of the number of bright-dark-bright fringe shifts observed is shown below:
[tex]\triangle m = \frac{2d}{\lambda} (n - 1)[/tex]
where
d = [tex]3.95 \times 10^{-2}m[/tex]
[tex]\lambda = 400 \times 10^{-9}m[/tex]
n = 1.00028
Now placing these values to the above formula
So, the number of bright-dark-bright fringe shifts observed is
[tex]= \frac{2 \times3.95 \times 10^{-2}m}{400 \times 10^{-9}m} (1.00028 - 1)[/tex]
= 55.3
We simply applied the above formula so that the number of bright dark bright fringe shifts could come
Superman is jogging alongside the railroad tracks on the outskirts of Metropolis at 100 km/h. He overtakes the caboose of a 500-m-long freight train traveling at 50 km/h. At that moment he begins to accelerate at 10 m/s2. How far will the train have traveled before Superman passes the locomotive?
Answers
Answer:
d = 41.91 m
Explanation:
In order to calculate the distance traveled by the train while superman passes it, you write the equations of motion for both superman and train:
For train, you have a motion with constant speed. You write the equation of motion of the position of the front of the train:
[tex]x=x_o+v_1t[/tex] (1)
xo: initial position of the front of the train = 500m
v1: speed of the train = 50km/h
For superman, you take into account that the motion is an accelerated motion (you assume superman is at the origin of coordinates):
[tex]x'=v_2t+\frac{1}{2}at^2[/tex] (1)
v2: initial speed of superman = 100km/h
a: acceleration = 10m/s^2
When superman passes the train, both positions x and x' will be equal. Hence, you equal the equations (1) and (2) and you calculate the time t. But before you convert the units of the velocities v1 and v2 to m/s:
[tex]v_1=50\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=13.88\frac{m}{s}\\\\v_2=100\frac{km}{h}=\frac{1000m}{1km}*\frac{1h}{3600s}=27.77\frac{m}{s}[/tex]
Thus, you equal x=x'
[tex]x=x'\\\\x_o+v_1t=v_2t+\frac{1}{2}at^2\\\\500m+(13.88m/s)t=(27.77m/s)t+\frac{1}{2}(10m/s^2)t^2\\\\(50\frac{m}{s^2})t^2+(13.89\frac{m}{s})t-500m=0[/tex]
You solve the last equation for t by using the quadratic formula:
[tex]t_{1,2}=\frac{-13.89\pm \sqrt{(13.89)^2-4(50)(-500)}}{2(50)}\\\\t_{1,2}=\frac{-13.89\pm 316.53}{100}\\\\t_1=3.02s\\\\t_2=-3.30s[/tex]
You only use t1 = 3.02s because negative times do not have physical meaning.
Next, you replace this value of t in the equation (1) to calculate the position of the train (for when superman just passed it):
[tex]x=500m+(13.88m/s)(3.02s)=541.91m[/tex]
x is the position of the front of the train, then, the dstance traveled by the train is:
d = 541.91m - 500m = 41.91 m
The equation for distance is d= st. if a car has a speed of 20 m/s how long will it take to go 155m
Answers
Answer:
It will take 7.75 seconds for the car to go 155m
Explanation:
From the question, we can understand that the distance covered by the moving car is got by a product of its speed and the time it travels.
i.e distance = speed X time.
However, in this case, we have the distance travelled and the speed of the car, and we are looking for the time of travel
TO solve this, we will simply make the travel time the subject of the formula in the equation above.
i.e time = distance / speed
time = 155/20= 7.75 seconds.
Hence, it will take 7.75 seconds for the car to go 155m
A jeep starts from rest with a constant acceleration of 4m/s2.At the same time a car travels with a constant speed of 36km/h overtake and passes the jeep how far beyond the starting point will the jeep overtakes the car?
Answers
Answer:
25m
Explanation:
Let's assume the Jeep attains a velocity of 36km/h ; a constant speed same with that of the car.
While the Jeep is accelerating to that speed, the car with that speed passes it.
Now we can calculate the time taken for the Jeep to attain the velocity of 36km/h on her constant acceleration.
This time is t = v/a; from Newton's Law of Motion:
a = V-U / t ; a-acceleration
V is final velocity = 36km/h
U is initial velocity 0 since the body starts from rest.
Hence t = 36000/3600 ÷ 4 = 2.5s
Note conversting from km/h to m/s we multiply by 1000/3600.
But the distance covered by the car while the Jeep just accelerates is
S = U × t = 10× 2.5 = 25m.
Note From Newton's law of Motion, distance for constant speed is defined as: U × t
Hence the Car would be 25m off the starting point just as the Jeep accelerates. It would overtake the Jeep when it just covers 25m from the Jeep starting point.
A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above the ground and slides along a frictionless track. The car encounters a loop of radius R. Assume that the initial height h is great enough so that the car never losses contact with the track.
Required:
a. Find an expression for the kinetic energy of the car at the top of the loop. Express the kinetic energy in terms of m, g, h, and R.
b. Find the minimum initial height h at which the car can be released that still allows the car to stay in contact with the track at the top of the loop.
Answers
Answer:
Explanation:
At height h , potential energy of coaster car having mass m = mgh .
The car will lose potential energy and gain kinetic energy.
height lost by car when it is at the top of loop of radius R
= h - 2R
potential energy lost = mg ( h - 2R )
kinetic energy gained = mg ( h - 2R )
kinetic energy = 0 + mg ( h - 2R )
= mg ( h - 2R )
b )
For the car to remain in contact with the track , if v be the minimum velocity
centripetal force at top = mg
m v² / R = mg
v² = gR
kinetic energy = 1/2 mv²
= 1/2 m x gR
= mgR /
If h be the minimum height that can give this kinetic energy
mg ( h - 2R ) = mgR / 2
h - 2R = R / 2
h = 2.5 R .
World religions: Shinto
Most Shinto rituals are tied to
A) worshiping the kami.
B) the life-cycle of humans and the seasonal cycles of nature.
C) forgiveness of sins.
D) preparing for the afterlife.
Answers
A 20 g "bouncy ball" is dropped from a height of 1.8 m. It rebounds from the ground with 80% of the speed it had just before it hit the ground. Assume that during the bounce the ground causes a constant force on the ball for 75 ms. What is the force applied to the ball by the ground in N?
The following are not correct: 0.513 N, 0.317 N, 0.121 N. Please show your work so I can understand!
Answers
Answer:
F = 0.314 N
Explanation:
In order to calculate the applied force to the ball by the ground, you first calculate the speed of the ball just before it hits the ground. You use the following formula:
[tex]v^2=v_o^2+2gy[/tex] (1)
y: height from the ball starts its motion = 1.8 m
vo: initial velocity = 0 m/s
g: gravitational acceleration = 9.8 m/s^2
v: final velocity of the ball = ?
You replace the values of the parameters in the equation (1):
[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(1.8m)}=5.93\frac{m}{s}[/tex]
Next, you take into account that the force exerted by the ground on the ball is given by the change, on time, of the linear momentum of the ball, that is:
[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}=m\frac{v_2-v_1}{\Delta t}[/tex] (2)
m: mass of the ball = 20g = 20*10^-3 kg
v1: velocity of the ball just before it hits the ground = 5.93m/s
v2: velocity of the ball after it impacts the ground (80% of v1):
0.8(5.93m/s) = 4.75 m/s
Δt: time interval o which the ground applies the force on the ball = 75*10^-3 s
You replace the values of the parameters in the equation (2):
[tex]F=(20*10^{-3}kg)\frac{4.75m/s-5.93m/s}{75*10^{-3}s}=-0.314N[/tex]
The minus sign means that the force is applied against the initial direction of the motion of the ball.
The applied force by the ground on the bouncy ball is 0.314 N
A book of 500 leaves has a mass of 1kg if its thickness is 5cm what are the mass and thickness of each leaf
Answers
Answer:
0.002kg and 0.01cm
Explanation:
500 leaves has a thickness is 5cm
Means I leaf has a thickness of 5/500= 0.01cm
Similarly the mass of one leaf would be 1/500 =0.002kg
Which statement BEST explains the relationship between voltage, current, and power?
A. If voltage increases and everything else remains constant, then power will increase.
B. If voltage increases and everything else remains constant, then power will decrease.
C. If current decreases and everything else remains constant, then power will increase.
D. Voltage and power are inversely related.
Answers
Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Simultaneously, each puck is given a quick push, and they begin to slide directly toward each other. Puck A moves with a speed of 3.90 m/s , and puck B moves with a speed of 4.30 m/s . What is the distance covered by puck A by the time the two pucks collide
Answers
Answer:
The distance covered by puck A before collision is [tex]z = 8.56 \ m[/tex]
Explanation:
From the question we are told that
The label on the two hockey pucks is A and B
The distance between the two hockey pucks is D 18.0 m
The speed of puck A is [tex]v_A = 3.90 \ m/s[/tex]
The speed of puck B is [tex]v_B = 4.30 \ m/s[/tex]
The distance covered by puck A is mathematically represented as
[tex]z = v_A * t[/tex]
=> [tex]t = \frac{z}{v_A}[/tex]
The distance covered by puck B is mathematically represented as
[tex]18 - z = v_B * t[/tex]
=> [tex]t = \frac{18 - z}{v_B}[/tex]
Since the time take before collision is the same
[tex]\frac{18 - z}{V_B} = \frac{z}{v_A}[/tex]
substituting values
[tex]\frac{18 -z }{4.3} = \frac{z}{3.90}[/tex]
=> [tex]70.2 - 3.90 z = 4.3 z[/tex]
=> [tex]z = 8.56 \ m[/tex]
what statement is true according to newton’s first law of motion?
a. in the absence of unbalanced force an object at rest will stay at rest and an object in motion will come to a stop.
b. in the absence of an unbalanced force, an object will start moving and an object in motion will come to a stop.
c. in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.
d. in the absence of an unbalanced force, an object will start moving and an object in motion will stay in motion.
Answers
Answer:
c. in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.
Explanation:
First law: things keep doing what they are doing, unless force is applied.