Answer:
s = 2.16 x 10¹¹ m
Explanation:
Since, the waves travelling from Earth to the Mars rover are electromagnetic. Therefore, there speed must be equal to the speed of light. So, from the equation given below:
s = vt
where,
s = the distance between Earth and Mars = ?
v = speed of the wave = speed of light = 3 x 10⁸ m/s
t = time taken by the radio signals to reach the rover from Earth
t = (12 min)(60 s/1 min) = 720 s
Therefore,
s = (3 x 10⁸ m/s)(720 s)
s = 2.16 x 10¹¹ m
In which example is kinetic friction most involved? a sled stuck on a snowy hill a bottle of water wedged in a vending machine an explorer unsuccessfully pushing on a massive stone that is blocking the entrance to a cave a volleyball player sliding across the court while diving for the ball
Answer:
I believe the answer is A volleyball player sliding across the court while diving for the ball.
Explanation:
Kinetic friction is a body moving on the surface experiences a force in the opposite direction of its movement.
Hope this helps! （づ￣3￣）づ╭❤～
Calculate the work performed by an ideal Carnot engine as a cold brick warms from 150 K to the temperature of the environment, which is 300 K. (Use 300 K as the temperature of the hot reservoir of the engine). The heat capacity of the brick is C
Answer
Work done is 57.9KJ
Explanation
First solve the problem according to work done due to variation in temperature
So W= intergral Cu( 1-Tu/T). at Tu and T
So Given that
C = Heat capacity of the Brick
TEPc= Cold Temperature
TEPh = Hot Temperature
W = C ( TEPh-TEP) - TEPhCln ( TEPh/TEPc)
So
W= (1)-(300-150)-300 (1) ln 2
W= -57.9KJ
ametal of mass 0.6kg is heated by an electric heater connected to 15v batter when the ammeter reading is 3A its tempeeature rises feom 20c to 85c in 10 minutes calculate the s.h.c of metal cylinder
Answer:
692 J/kg/°C
Explanation:
Electric energy added = amount of heat
Power × time = mass × SHC × increase in temperature
Pt = mCΔT
(15 V × 3 A) (10 min × 60 s/min) = (0.6 kg) C (85°C − 20°C)
C = 692 J/kg/°C
One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious space camper and you launch a serious rocket: it reaches an altitude of 211 km . What gain Δ???? in gravitational potential energy does the launch accomplish? The mass and radius of the Moon are 7.36×1022 kg and 1740 km, respectively.
Answer:
ΔP.E = 6.48 x 10⁸ J
Explanation:
First we need to calculate the acceleration due to gravity on the surface of moon:
g = GM/R²
where,
g = acceleration due to gravity on the surface of moon = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of moon = 7.36 x 10²² kg
R = Radius of Moon = 1740 km = 1.74 x 10⁶ m
Therefore,
g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²
g = 2.82 m/s²
now the change in gravitational potential energy of rocket is calculated by:
ΔP.E = mgΔh
where,
ΔP.E = Change in Gravitational Potential Energy = ?
m = mass of rocket = 1090 kg
Δh = altitude = 211 km = 2.11 x 10⁵ m
Therefore,
ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)
ΔP.E = 6.48 x 10⁸ J
A homeowner purchases insulation for her attic rated at R-15. She wants the attic insulated to R-30. If the insulation she purchased is 10 cm thick, what thickness does she need to use
Answer:
she need to use 20 cm thick
Explanation:
given data
wants the attic insulated = R-30
purchased = 10 cm thick
solution
as per given we can say that
10 cm is for the R 15
but she want for R 30
so
R 30 thickness = [tex]\frac{30}{15} \times 10[/tex]
R 30 thickness = 20 cm
so she need to use 20 cm thick
An electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Which of the following are also the same for the two particles?
(A) speed
(B) kinetic energy
(C) frequency
(D) momentum
Explanation:
The De-Broglie wavelength is given by :
[tex]\lambda=\dfrac{h}{p}[/tex]
h is Planck's constant
p is momentum
In this case, an electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Mass of electron and proton is different. It means their velocity and energy are different.
Only momentum is the factor that remains same for both particles i.e. momentum.
What happens to the deflection of the galvanometer needle (due to moving the magnet) when you increase the number of loops
Answer:
If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil which will increase total induced electromotive force
Explanation:
galvanometer is an instrument that can detect and measure small current in an electrical circuit.
If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil. If it is move in a way into the coil,the needle deflect in that way and if it move in another way, it will deflect in the other way.
The total induced emf is equal to the emf induced in each loop by the changing magnetic flux, then multiplied by the number of loops and an increase in the number of loops will cause increase in the total induced emf.
Consider 1 mol an ideal gas at 28∘ C and 1.06 atm pressure. To get some idea how close these molecules are to each other, on the average, imagine them to be uniformly spaced, with each molecule at the center of a small cube.
A) What is the length of an edge of each cube if adjacent cubes touch but do not overlap?
B) How does this distance compare with the diameter of a typical molecule? The diameter of a typical molecule is about 10-10 m. (in l/dmolecule)
C) How does their separation compare with the spacing of atoms in solids, which typically are about 0.3 nm apart? (in l/lsolid)
Answer:
A) Length of an edge = 3.38 × 10^(-9) m
B) 34 times the diameter of a molecule.
C) 11 times the atomic spacing in solids.
Explanation:
A) We will use Avogadro's hypothesis to solve this. It states that 1 mole of gas occupies 22.4 L at STP.
We want to find the volume occupied by 1 mole of gas at 1.06 atm pressure and temperature of 28 °C (= 301 K).
Thus, by the ideal gas equation, we have;
V_mole = (1 × 22.4/273) × (301/1.06) = 23.3 L = 0.0233 m³
Now, since from avogadros number, 1 mole of gas contains 6.02 x 10^(23) molecules, then volume occupied by a molecule is given by;
V_molecule = 0.0233/(6.02 × 10^(23)) m³ = 3.87 x 10^(-26) m³
Thus, length of an edge of the cube = ∛(3.87 × 10^(-26)) = 3.38 × 10^(-9) m
B) We are told that The diameter of a typical molecule is about 10^(-10) m.
Thus, the distance is about;
(3.38 × 10^(-9))/(10^(-10)) ≈ 34 times the diameter of a molecule.
C) We are told that the spacing of atoms is typically are about 0.3 nm apart
Thus;
The separation will be about;
(3.38 × 10^(-9))/(0.3 × 10^(-9)) ≈ 11 times the atomic spacing in solids.
g suppose he used an alpha particle with an energy of 8.3 MeV, what would be the speed of this alpha particle
Answer:
speed of the alpha particle is 2 x 10^7 m/s.
Explanation:
energy of alpha particle = 8.3 Mev
1 Mev = 1.602 x 10^-13 J
8.3 Mev = [tex]x[/tex]
solving, [tex]x[/tex] = 8.3 x 1.602 x 10^-13 = 1.329 x 10^-12 J
mass of a alpha particle = 6.645 x 10^−27 kg
The energy of the alpha particle is the kinetic energy KE of the alpha particle
KE = [tex]\frac{1}{2}mv^{2}[/tex]
where m is the mass of the alpha particle
v is the velocity of the alpha particle
substituting values, we have
1.329 x 10^-12 = [tex]\frac{1}{2}*6.645*10^{-27}*v^{2}[/tex]
[tex]v^{2}[/tex] = 4 x 10^14
[tex]v = \sqrt{4*10^{14} }[/tex] = 2 x 10^7 m/s
An electron has an initial velocity to the south but is observed to curve upward as the result of a magnetic field. This magnetic field must have a component:___________
a) north
b) upwards
c) downwards
d) east
e) west
Answer:
e) west
Explanation:
According to Lorentz left hand rule, the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.
In this case, if we point the thumb to the south (towards your body), with the palm facing up, then the fingers will point west.
7. A sound wave begins traveling through a thin metal rod at one end with a speed that is 15 times the speed of sound in air. If an observer at the other end of the rod hears the sound twice, one from the sound traveling through the rod and one from the sound traveling through the air, with a time delay of 0.12 s, how long is the rod? The speed of sound in air is 343 m/s.
Answer:
L = 44,096 m
Explanation:
The speed of the sound wave is constant therefore we can use the relations of uniform kinematics
v = x / t
the speed of the wave in the bar is
v = 15 v or
v = 15 343
v = 5145 m / s
The sound at the bar goes the distance
L = v t
Sound in the air travels the same distance
L = v_air (t + 0.12)
as the two recognize the same dissonance,
v t = v_air (t +0.12)
t (v- v_air) = 0.12 v_air
t = 0.12 v_air / (v -v_air)
l
et's calculate
t = 0.12 343 / (5145 - 343)
t = 8.57 10-3 s
The length of the bar is
L = 5145 8.57 10-3
L = 44,096 m
The temperature of the hot spots caused by the impact of transferred matter onto the surface of a pulsar can be 108 K. What is the peak wavelength in the blackbody spectrum of such a spot, and in what range of the electromagnetic spectrum does it occur
Given that,
Temperature = 10⁸ K
We need to calculate the peak wavelength in the blackbody spectrum
Using formula of peak wavelength
[tex]peak\ wavelength = \dfrac{2.898\times10^{-3}}{T}[/tex]
Where, T= temperature
Put the value into the formula
[tex]peak\ wavelength = \dfrac{2.898\times10^{-3}}{10^{8}}[/tex]
[tex]peak\ wavelength = 2.90\times10^{-11}\ m[/tex]
[tex]peak\ wavelength = 290\ nm[/tex]
This range of wavelength is ultraviolet.
Hence, The peak wavelength in the blackbody spectrum is 290 nm and the range of wavelength is ultraviolet electromagnetic spectrum .
When light travels from one medium to another with a different index of refraction, how is the light's frequency and wavelength affected
Answer:
The frequency does not change, but the wavelength does
Explanation:
Here are the options
A. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, the frequency changes and the wavelength does not.
B. The frequency does change, but the wavelength remains unchanged.
C. Both the frequency and wavelength change.
D. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, neither the wavelength nor the frequency changes.
E. The frequency does not change, but its wavelength does.
When light goes through one medium to the next, the frequency doesn't really change seeing as frequency is dependent on wavelength and light wave velocity. And when the wavelength shifts from one medium to the next.
[tex]n= \frac{C}{V} \ and\ \frac{\lambda_o}{\lambda_m}[/tex]
where [tex]\lambda_o[/tex] indicates wavelength in vacuum
[tex]\lambda_m[/tex] indicates wavelength in medium
n indicates refractive index
v indicates velocity of light wave
c indicates velocity of light
And wavelength is medium-dependent. Frequency Here = v[tex]\lambda[/tex] and shift in wavelength and velocity, not shifts in overall frequency.
Therefore the correct option is E
The element sodium can emit light at two wavelengths, λ1 = 588.9950 nm and λ2 = 589.5924 nm. Light in sodium is being used in a Michelson interferometer. Through what distance must mirror M 2 be moved if the shift in the fringe pattern for one wavelength is to be 1.00 fringe more than the shift in the fringe pattern for the other wavelength?
Answer:
The distance is [tex]d = 0.00029065 \ m[/tex]
Explanation:
From the question we are told that
The first wavelength is [tex]\lambda _1 = 588.9950 nm = 588.9950 *10^{-9} \ m[/tex]
The second wavelength is [tex]\lambda _2 = 589.5924 nm = 589.5924 *10^{-9} \ m[/tex]
The difference in the fringe pattern is n = 1.0
Generally the equation defining the effect of the movement of the mirror M 2 in a Michelson interferometer is mathematically represented as
[tex]2 * d = [\frac{\lambda _1 * \lambda_2 }{\lambda_2 - \lambda _1 } ] * n[/tex]
Here d is the mirror M 2 must be moved
substituting values
[tex]2 * d = [\frac{(588.9950*10^{-9} ) * (589.5924 *10^{-9}) }{(589.5924 *10^{-9}) - (588.9950*10^{-9} ) } ] * 1.0[/tex]
[tex]d = 0.00029065 \ m[/tex]
A rod on a compressed spring exerts 12 N of force on a 0.05-kg steel ball. The
rod pushes the ball 0.03 m. How much work does the spring do on the ball?
A) 36
B) 36 N
C) 60 N
D)1.00
Answer:
Work = 0.36N
Explanation:
Given
Force = 12N
Distance = 0.03m
Weight = 0.05kg
Required
Determine the work done
Workdone is calculated as thus;
Work = Force * Distance
Substitute 12N for Force and 0.03m for Distance
Work = 12N * 0.03m
Work = 0.36Nm
Using proper S.I units
Work = 0.36N
Hence, work done by the spring on the ball is 0.36N
A plastic rod that has been charged to − 15 nC touches a metal sphere. Afterward, the rod's charge is − 5.0 nC.
1) What kind of charged particle was transferred between the rod and the sphere, and in which direction?
A) electrons transferred from rod to sphere.
B) electrons transferred from sphere to rod.
C) protons transferred from rod to sphere.
D) protons transferred from sphere to rod.
2) How many charged particles were transferred?
Answer:
B) electrons transferred from sphere to rod.
(2) 1.248 x 10¹¹ electrons were transferred
Explanation:
Given;
initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let the charge acquired by the plastic rod = q
q + 15nC = -5nC
q = -5nC - 15nC
q = -20 nC
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Hence, electrons transferred from sphere to rod
B) electrons transferred from sphere to rod.
2) How many charged particles were transferred?
1.602 x 10⁻¹⁹ C = 1 electron
20 x 10⁻⁹ C = ?
= 1.248 x 10¹¹ electrons
Thus,1.248 x 10¹¹ electrons were transferred
1. Electrons transferred from sphere to rod.
Option B is correct.
2. There are [tex]6.24*10^{10}[/tex] electrons transferred from sphere to rod.
Given that initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let us consider that the charge absorbed by the plastic rod is q
[tex]q - 15nC = -5nC\\q = -5nC +15nC\\q = -10 nC[/tex]
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Therefore, electrons transferred from sphere to rod
The charge on one electron is, 1.602 x 10⁻¹⁹ C .
Number of electrons, [tex]n=\frac{10*10^{-9} }{1.602*10^{-19} }= 6.24*10^{10}[/tex]
Thus,[tex]6.24*10^{10}[/tex] electrons were transferred.
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An electron in the first energy level of the electron cloud has an electron in the third energy level
Answer:
a lower energy than
Explanation:
sorry im a month late but is lower energy than
Two long, parallel wires are separated by a distance of 2.60 cm. The force per unit length that each wire exerts on the other is 4.30×10^−5 N/m, and the wires repel each other. The current in one wire is 0.520 A.Required:a. What is the current in the second wire? b. Are the two currents in the same direction or in opposite directions?
Answer:
10.75 A
The current is in opposite direction since it causes a repulsion force between the wires
Explanation:
Force per unit length on the wires = 4.30×10^−5 N/m
distance between wires = 2.6 cm = 0.026 m
current through one wire = 0.52 A
current on the other wire = ?
Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as
[tex]F/l[/tex] = [tex]\frac{u_{0}I_{1} I_{2} }{2\pi r }[/tex]
where [tex]F/l[/tex] is the force per unit length on the wires
[tex]u_{0}[/tex] = permeability of vacuum = 4π × 10^−7 T-m/A
[tex]I_{1}[/tex] = current on the first wire = 0.520 A
[tex]I_{2}[/tex] = current on the other wire = ?
r = the distance between the two wire = 0.026 m
substituting the value into the equation, we have
4.30×10^−5 = [tex]\frac{4\pi *10^{-7}*0.520*I_{2} }{2\pi *0.026}[/tex] = [tex]\frac{ 2*10^{-7}*0.520*I_{2} }{0.026}[/tex]
4.30×10^−5 = 4 x 10^-6 [tex]I_{2}[/tex]
[tex]I_{2}[/tex] = (4.30×10^-5)/(4 x 10^-6) = 10.75 A
The current is in opposite direction since it causes a repulsion force between the wires.
The primary difference between a barometer and a manometer is
A. a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.
B. a barometer uses mercury, while a manometer can use any liquid. a barometer is used to measure atmospheric pressure, and a manometer is used to measure absolute pressure.
C a barometer reads in mm, while a manometer reads in Pa.
D a barometer can measure either positivee or negative pressure, while a manometer only
E positive pressure. measures
Answer:
a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.
Explanation:
A barometer measures air pressure at any locality with sea level as the reference.
However, a manometer is used to measure all pressures especially gauge pressures. Thus, if the aim is to measure the pressure at any point below a fluid surface, a barometer is used to determine the air pressure. The manometer may now be used to determine the gauge pressure
The algebraic sum of these two values gives the absolute pressure.
You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
most likely final temperature of the mixture?
O A. 80°C
OB. 10-C
OC. 20°C
O D. 60°C
Answer:
Option (c) : 20°C
Explanation:
[tex]t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2} [/tex]
T(final) = 500* 10 + 100*70/600 = 20°C
A car moving at 36 m/s passes a stationary police car whose siren has a frequency of 500 Hz. What is the change in the frequency (in Hz) heard by an observer in the moving car as he passes the police car? (The speed of sound in air is 343 m/s.)
Answer:
Change in the frequency (in Hz) = 104.96 Hz
Explanation:
Given:
Speed of sound in air (v) = 343 m/s
Speed of car (v1) 36 m/s
Frequency(f) = 500 Hz
Find:
Change in the frequency (in Hz)
Computation:
Frequency hear by the observer(before)(f1) = [f(v+v1)] / v
Frequency hear by the observer(f1) = [500(343+36)] / 343
Frequency hear by the observer(f1) = 552.48 Hz
Frequency hear by the observer(after)(f2) = [f(v-v1)] / v
Frequency hear by the observer(f2) = [500(343-36)] / 343
Frequency hear by the observer(f2) = 447.52 Hz
Change in the frequency (in Hz) = f1 - f2
Change in the frequency (in Hz) = 552.48 Hz - 447.52 Hz
Change in the frequency (in Hz) = 104.96 Hz
You are holding on to one end of a long string that is fastened to a rigid steel light pole. After producing a wave pulse that was 5 mm high and 4 em wide, you want to produce a pulse that is 4 cm wide but 7 mm high. You must move your hand up and down once,
a. a smaller distance up, but take a shorter time.
b. the same distance up as before, but take a shorter time.
c. a greater distance up, but take a longer time.
d. the same distance up as before, but take a longer time.
e. a greater distance up, but take the same time.
Answer:
It will take. the same distance up as before, but take a longer time
A nozzle with a radius of 0.22 cm is attached to a garden hose with a radius of 0.89 cm that is pointed straight up. The flow rate through hose and nozzle is 0.55 L/s.
Randomized Variables
rn = 0.22 cm
rh = 0.94 cm
Q = 0.55
1. Calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle in m.
2. Calculate the maximum height (in cm) to which water could be squirted with the hose if it emerges with the nozzle removed assuming the same flow rate.
Answer:
1. 0.2m
1. 66m
Explanation:
See attached file
The expressions of fluid mechanics allows to find the result for the maximum height that the water leaves through the two points are;
1) The maximum height when the water leaves the hose is: Δy = 0.20 m
2) The maximum height of the water leaves the nozzle is: Δy = 68.6m
Given parameters
The flow rate Q = 0.55 L/s = 0.55 10⁻³ m³ / s Nozzle radius r₁ = 0.22 cm = 0.22 10⁻² m Hose radius r₂ = 0.94 cm = 0.94 10⁻² mTo find
1. Maximum height of water in hose
2. Maximum height of water at the nozzle
Fluid mechanics studies the movement of fluids, liquids and gases in different systems, for this it uses two expressions:
The continuity equation. It is an expression of the conservation of mass in fluids.
A₁v₁ = A₂.v₂
Bernoulli's equation. Establishes the relationship between work and the energy conservation in fluids.P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y₂
Where the subscripts 1 and 2 represent two points of interest, P is the pressure, ρ the density, v the velocity, g the acceleration of gravity and y the height.
1, Let's find the exit velocity of the water in the hose.
Let's use subscript 1 for the nozzle and subscript 2 for the hose.
The continuity equation of the flow value that must be constant throughout the system.
Q = A₁ v₁
v₁ = [tex]\frac{Q}{A_1 }[/tex]
The area of a circle is:
A = π r²
Let's calculate the velocity in the hose.
A₁ = π (0.94 10⁻²) ²
A₁ = 2.78 10⁻⁴ m²
v₁ = [tex]\frac{0.55 \ 10^{-3}}{2.78 \ 10^{-4}}[/tex]
v₁ = 1.98 m / s
Let's use Bernoulli's equation.
When the water leaves the hose the pressure is atmospheric and when it reaches the highest point it has not changed P1 = P2
½ ρ v₁² + ρ g y₁ = ½ ρ v₂² + ρ g v₂
y₂-y₁ = ½ [tex]\frac{v_i^2 - v_2^2}{g}[/tex]
At the highest point of the trajectory the velocity must be zero.
y₂- y₁ = [tex]\frac{v_1^2}{2g}[/tex]
Let's calculate
y₂-y₁ = [tex]\frac{1.98^2}{2 \ 9.8}[/tex]
Δy = 0.2 m
2. Let's find the exit velocity of the water at the nozzle
A₁ = π r²
A₁ = π (0.22 10⁻²) ²
A₁ = 0.152 10⁻⁴ m / s
With the continuity and flow equation.
Q = A v
v₁ = [tex]\frac{Q}{A}[/tex]
v₁ = [tex]\frac{0.55 \ 10{-3} }{0.152 \ 10^{-4} }[/tex]
v₁ = 36.67 m / s
Using Bernoulli's equation, where the speed of the water at the highest point is zero.
y₂- y₁ = [tex]\frac{v^1^2}{g}[/tex]
Let's calculate.
Δy = [tex]\frac{36.67^2 }{2 \ 9.8 }[/tex]
Δy = 68.6m
In conclusion using the expressions of fluid mechanics we can find the results the maximum height that the water leaves through the two cases are:
1) The maximum height when the water leaves the hose is:
Δy = 0.20 m
2) The maximum height of the water when it leaves the nozzle is:
Δy = 68.6 m
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A man using a 70kg garden roller on a level surface, exerts a force of 200N at 45 degrees to the ground. find the vertical force of the roller on the ground if,
i.he pulls
ii.he pushes the roller
Answer:
i) 545.2 N upwards
ii) 828.2 N downwards
Explanation:
mass of the roller = 70 kg
force exerted = 200 N
angle the force makes with the ground ∅ = 45°
weight of the roller W = mg
where
m is the mass of the roller
g is the acceleration due to gravity = 9.81 m/s^2
weight of the roller = 70 x 9.81 = 686.7 N
The effective vertical force exerted by the man = F sin ∅ = 200 x sin 45°
==> F = 200 x 0.707 = 141.5 N
i) if the man pulls, then the exerted force will be in opposite direction to the weight of the roller vertically upwards
Resultant vertical force = 686.7 N - 141.5 N = 545.2 N upwards
ii) if he pushes, then the exerted force will be in the direction of the weight vertically downwards
Resultant vertical force = 686.7 N + 141.5 N = 828.2 N downwards
QUESTION 27
The titanium shell of an SR-71 airplane would expand when flying at a speed exceeding 3 times the speed of sound. If the skin of the
plane is 400 degrees C and the linear coefficient of expansion for titanium is 5x10-6/C when flying at 3 times the speed of sound, how
much would a 10-meter long (originally at oC) portion of the airplane expand? Write your final answer in centimeters and show all of your
work.
Answer:
2 cm.
Explanation:
Data obtained from the question include the following:
Original Length (L₁ ) = 10 m
Initial temperature (T₁) = 0°C
Final temperature (T₂) = 400°C
Linear expansivity (α) = 5×10¯⁶ /°C
Increase in length (ΔL) =..?
Next, we shall determine the temperature rise (ΔT).
This can be obtained as follow:
Initial temperature (T₁) = 0°C
Final temperature (T₂) = 400°C
Temperature rise (ΔT) =..?
Temperature rise (ΔT) = T₂ – T₁
Temperature rise (ΔT) = 400 – 0
Temperature rise (ΔT) = 400°C
Thus, we can obtain the increase in length of the airplane by using the following formula as illustrated below:
Linear expansivity (α) = increase in length (ΔL) /Original Length (L₁ ) × Temperature rise (ΔT)
α = ΔL/(L₁ × ΔT)
Original Length (L₁ ) = 10 m
Linear expansivity (α) = 5×10¯⁶ /°C
Temperature rise (ΔT) = 400°C
Increase in length (ΔL) =..?
α = ΔL/(L₁ × ΔT)
5×10¯⁶ = ΔL/(10 × 400)
5×10¯⁶ = ΔL/4000
Cross multiply
ΔL = 5×10¯⁶ × 4000
ΔL = 0.02 m
Converting 0.02 m to cm, we have:
1 m = 100 cm
Therefore, 0.02 m = 0.02 × 100 = 2 cm.
Therefore, the length of the plane will increase by 2 cm.
"A satellite requires 88.5 min to orbit Earth once. Assume a circular orbit. 1) What is the circumference of the satellites orbit
Answer:
circumference of the satellite orbit = 4.13 × 10⁷ m
Explanation:
Given that:
the time period T = 88.5 min = 88.5 × 60 = 5310 sec
The mass of the earth [tex]M_e[/tex] = 5.98 × 10²⁴ kg
if the radius of orbit is r,
Then,
[tex]\dfrac{V^2}{r} = \dfrac{GM_e}{r^2}[/tex]
[tex]{V^2} = \dfrac{GM_e r}{r^2}[/tex]
[tex]{V^2} = \dfrac{GM_e }{r}[/tex]
[tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]
Similarly :
[tex]T = \sqrt{\dfrac{ 2 \pi r} {V} }[/tex]
where; [tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]
Then:
[tex]T = {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {GM_e }} }[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {6.674\times 10^{-11} \times 5.98 \times 10^{24} }} }[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ 3.991052 \times 10^{14} }}[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {19977617.48}[/tex]
[tex]5310 \times 19977617.48= 2 \pi r^{3/2}}[/tex]
[tex]1.06081149 \times 10^{11}= 2 \pi r^{3/2}}[/tex]
[tex]\dfrac{1.06081149 \times 10^{11}}{2 \pi}= r^{3/2}}[/tex]
[tex]r^{3/2}} = \dfrac{1.06081149 \times 10^{11}}{2 \pi}[/tex]
[tex]r^{3/2}} = 1.68833392 \times 10^{10}[/tex]
[tex]r= (1.68833392 \times 10^{10})^{2/3}}[/tex]
[tex]r= 2565.38^2[/tex]
r = 6579225 m
The circumference of the satellites orbit can now be determined by using the formula:
circumference = 2π r
circumference = 2π × 6579225 m
circumference = 41338489.85 m
circumference of the satellite orbit = 4.13 × 10⁷ m
21.-Una esquiadora olímpica que baja a 25m/s por una pendiente a 20o encuentra una región de nieve húmeda de coeficiente de fricción μr =0.55. ¿Cuánto desciende antes de detenerse?
Answer:
y = 12.82 m
Explanation:
We can solve this exercise using the energy work theorem
W = ΔEm
friction force work is
W = fr . s = fr s cos θ
the friction force opposes the movement, therefore the angle is 180º
W = - fr s
we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular
N -Wy = 0
N = mg cos θ
the friction force remains
fr = μ N
fr = μ mg cos θ
work gives
W = - μ mg s cos θ
initial energy
Em₀ = ½ m v²
the final energy is zero, because it stops
we substitute
- μ m g s cos θ = 0 - ½ m v²
s = ½ v² / (μ g cos θ)
let's calculate
s = ½ 20² / (0.55 9.8 cos 20)
s = 39.49 m
this is the distance it travels along the plane, to find the vertical distance let's use trigonometry
sin 20 = y / s
y = s sin 20
y = 37.49 sin 20
y = 12.82 m
Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first dark fringes on either side of the central maximum is 4.70 mm . Part A What is the width of the slit
Answer:
The width of the slit is 0.4 mm (0.00040 m).
Explanation:
From the Young's interference expression, we have;
(λ ÷ d) = (Δy ÷ D)
where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.
Thus,
d = (Dλ) ÷ Δy
D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×[tex]10^{-9}[/tex] m)
d = (3.30 × 563 ×[tex]10^{-9}[/tex] ) ÷ (0.0047)
= 1.8579 × [tex]10^{-6}[/tex] ÷ 0.0047
= 0.0003951 m
d = 0.00040 m
The width of the slit is 0.4 mm (0.00040 m).
The orbital motion of Earth around the Sun leads to an observable parallax effect on the nearest stars. For each star listed, calculate the distance in parsecs before converting that distance to astronomical units. A. Sirius (0.38") B. Alpha Centauri A (0.75") C. Procyon (0.28") D. Wolf 359 (0.42") E. Epsilon Eridani (0.31") D(pc) = 1/parallax(arcsecs), D(a.u.) = D(pc) * 206265 (arcsecs per radian)
Answer:
Following are the answer to this question:
Explanation:
Formula:
[tex]D(PC) =\frac{1}{parallax}\\\\D(av)=D(PC) \times 20.626\ J[/tex]
Calculating point A:
when the value is [tex]0.38[/tex]
[tex]\to 0.38 \toD(PC)= \frac{1}{0.38}\\\\[/tex]
[tex]=2.632[/tex]
[tex]\to D(a.v) = \frac{1}{0.38} \times 206265\\[/tex]
[tex]=542,802.6[/tex]
Calculating point B:
when the value is [tex]0.75[/tex]
[tex]\to D(PC)=\frac{1}{0.75}[/tex]
[tex]=1.33[/tex]
[tex]\to D(a.v) = \frac{1}{0.75} \times 206265\\[/tex]
[tex]=275,020[/tex]
Calculating point C:
when the value is [tex]0.28[/tex]
[tex]\to D(PC)=\frac{1}{0.28}[/tex]
[tex]=3.571[/tex]
[tex]\to D(a.v) = \frac{1}{0.28} \times 206265\\[/tex]
[tex]=736660.7[/tex]
Calculating point D:
when the value is [tex]0.42[/tex]
[tex]\to D(PC)=\frac{1}{0.42}[/tex]
[tex]=2.38[/tex]
[tex]\to D(a.v) = \frac{1}{0.42} \times 206265\\[/tex]
[tex]=490910.7[/tex]
Calculating point E:
when the value is [tex]0.31[/tex]
[tex]\to D(PC)=\frac{1}{0.31}[/tex]
[tex]=3.226[/tex]
[tex]\to D(a.v) = \frac{1}{0.31} \times 206265\\[/tex]
[tex]=665370.97[/tex]
A plastic balloon that has been rubbed with wool will stick to a wall.
a. Can you conclude that the wall is charged? If not, why not? If so, where does the charge come from?
b. Draw a series of charge diagrams showing how the balloon is held to the wall.
Answer:
Explanation:
When plastic balloon is rubbed with wool , charges are created on both balloon and silk in equal amount . Rubber balloon will acquire negative charge and silk will acquire positive charge .
Now when balloon is brought near a wall , there is induction of charge on the wall due to charge on the balloon . On the near surface of wall positive charge is produced and on the surface deep inside the wall negative charge is produced . The charge deep inside goes inside the earth but the positive charge near the surface of wall can not escape . It remains trapped by negative charge on the balloon .
hence there is mutual attraction between balloon and surface of wall is just like attraction between opposite charges . But once the ballon due to mutual attraction comes in contact with the wall , the charge on balloon and on wall neutralises each other and hence after some time the balloon falls off from the wall on the ground . It does not remain attracted to wall for ever . It happens due to neutralisation of charges on balloon and wall .
