Answer:
Velocity of moon = 586.23 km/h
Explanation:
We are given;
Distance of moon from the Earth = 384403 km
Time taken to orbit earth;t = 27.32166 days
24 hours make 1 day, thus 27.32166 days = 27.32166 × 24 = 655.72 hours
Formula for velocity is distance/time
Thus,
Velocity of moon = distance from moon to earth/time taken to orbit the earth
Velocity of moon = 384403/655.72 = 586.23 km/h
Please answer the following questions about uniform circular motion.?
Part (a) A planet orbits a star in a circular orbit at a constant orbital speed, which of the following statements is true?
All of these are correct.
The magnitude of the orbital velocity of the planet is unchanged, thus there is no acceleration and therefore no force action on the planet.
None of these are correct.
The planet experiences a centripetal force pulling towards its star.
The planet experiences no centripetal force.
The planet experiences a centripetal force pushing it away from its star.
Part (b) When a planet is orbiting a star, which force plays the role of the centripetal force?
The force resulting from the planets’ velocity around the star.
The force resulting from the centripetal acceleration.
The gravitational force
Part (c) Which of the following are true statements about uniform circular motion?
An object in uniform circular motion experiences a tangential force.
An object in uniform circular motion experiences a centripetal force, an equal and opposite centrifugal force, and a tangential force.
An object in uniform circular motion experiences a centripetal force and a tangential force.
An object in uniform circular motion experiences a centripetal force.
None of these choices are true.
An object in uniform circular motion experiences no forces.
Answer:
a) The planet experiences a centripetal force towards its star
b) The universal attractive force (Gravitational force)
c)None of these choices are true.
Explanation:
This problem raise several claims, let's review some aspects of circular motion
F = m a
the centripetal acceleration is
a = v² / r
where v is the speed (modulus of velocity) that is constant and r is the radius
The direction of the acceleration is perpendicular to the motion.
Let's review the different claims
Part a) the orbital velocity is constant
The correct statement is: The planet experiences a centripetal force towards its star
Part b) what is the centripetal force
The correct statement: The universal attractive force (Gravitational force)
Part c) which statement is true
1) False. There can be no tangential force
2) False. There is a centripetal force that creates the movement, but there is no centrifugal force because the system is accelerated and there is no tangential force because the movement is circular.
3) False. There is no tangential force
4) True none is true
5) False. There is a force because movement has acceleration
A liquid of density 1250 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.93 m/s and the pipe diameter d1 is 11.1 cm. At Location 2, the pipe diameter d2 is 16.7 cm. At Location 1, the pipe is Δy=8.89 m higher than it is at Location 2. Ignoring viscosity, calculate the difference ΔP between the fluid pressure at Location 2 and the fluid pressure at Location 1.
Answer:
The pressure difference is [tex]\Delta P = 1.46 *10^{5}\ Pa[/tex]
Explanation:
From the question we are told that
The density is [tex]\rho = 1250 \ kg/m^3[/tex]
The speed at location 1 is [tex]v_1 = 9.93 \ m/s[/tex]
The diameter at location 1 is [tex]d_1 = 11.1\ cm = 0.111 \ m[/tex]
The diameter at location 2 is [tex]d_1 = 16.7\ cm = 0.167 \ m[/tex]
The height at location 1 is [tex]h_1 = 8.89 \ m[/tex]
The height at location 2 is [tex]h_2 = 1 \ m[/tex]
Generally the cross- sectional area at location 1 is mathematically represented as
[tex]A_1 = \pi * \frac{d^2}{4}[/tex]
=> [tex]A_1 = 3.142 * \frac{ 0.111^2}{4}[/tex]
=> [tex]A_1 = 0.0097 \ m^2[/tex]
Generally the cross- sectional area at location 2 is mathematically represented as
[tex]A_2 = \pi * \frac{d_1^2}{4}[/tex]
=> [tex]A_2= 3.142 * \frac{ 0.167^2}{4}[/tex]
=> [tex]A_2 =0.0219 \ m^2[/tex]
From continuity formula
[tex]v_1 * A_1 = v_2 * A_2[/tex]
=> [tex]v_2 = \frac{A_1 * v_1}{A_2 }[/tex]
=> [tex]v_2 = \frac{0.0097 * 9.93}{0.0219 }[/tex]
=> [tex]v_2 = 4.398 \ m/s[/tex]
Generally according to Bernoulli's theorem
[tex]P_1 + \rho * g * h_1 + \frac{1}{2} \rho * v_1^2 = P_2 + \rho * g * h_2 + \frac{1}{2} \rho * v_2^2[/tex]
=> [tex]P_2 - P_1 = \frac{1}{2} \rho (v_1 ^2 - v_2^2 ) + \rho* g (h_1 - h_2)[/tex]
=> [tex]\Delta P = \frac{1}{2}* 1250* (9.93 ^2 - 4.398^2 ) + 1250* 9.8 (8.89- 1)[/tex]
=> [tex]\Delta P = 1.46 *10^{5}\ Pa[/tex]
An airplane is traveling at 400 mi/h. It touches down at an airport 2000 miles away. How long was the airplane airborne?
Answer:
5 hours
Explanation:
Given that
Speed of the airplane, v = 400 mile/hr
Distance of the airport, s = 2000 miles
This is quite a straightforward question that deals with one of the basic formulas in physics.
Speed.
speed is said to the the ratio of distance covered with respect to the time taken. This can be mathematically expressed as
Speed, v = distance covered, d / time taken, t
v = d / t
In the question above, we're looking for the time taken. So, so make t, subject of formula.
t = d / v, now we proceed to substituting the earlier given values into this equation.
t = 2000 / 400
t = 5 hrs,
therefore we can conclude that the airplane was airborne for 5 hours
If a thermometer measured the temperature in an oven as 400oF five days in a row when the temperature was actually 397oF, this measuring instrument would be considered quite:
Answer:
It can be said to be reliable although it is not valid
Explanation:
This is because Reliability means an indicator of consistency, A measure should produce similar or the same results consistently if it measures the same quantity. So does the thermometer measures over 5days but it is not valid because it deviates from the real value
Two strings are respectively 1.00 m and 2.00 m long. Which of the following wavelengths, in meters, could represent harmonics present on both strings?
1) 0.800, 0.670, 0.500
2) 1.33, 1.00, 0.500
3) 2.00, 1.00, 0.500
4) 2.00, 1.33, 1.00
5) 4.00, 2.00, 1.0
Answer:
5) 4.00, 2.00, 1.0
Explanation:
wave equation is given as;
F₀ = V / λ
Where;
F₀ is the fundamental frequency = first harmonic
Length of the string for first harmonic is given as;
L₀ = (¹/₂) λ
λ = 2 L₀
when L₀ = 1
λ = 2 x 1 = 2m
when L₀ = 2m
λ = 2 x 2 = 4m
For First harmonic, the wavelength is 2m, 4m
For second harmonic;
L₁ = (²/₂)λ
L₁ = λ
When L₁ = 1
λ = 1 m
when L₁ = 2
λ = 2 m
For second harmonic, the wavelength is 1m, 2m
Thus, the wavelength that could represent harmonics present on both strings is 4m, 2m, 1 m
Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 5.00 m from two double slits 0.500 mm apart illuminated by 500-nm light.
Answer:
y = 1.75 cm
Explanation:
In the double-slit experiment the equation for destructive interference is
d sin tea = (m + ½)
λ
let's use trigonometry to find the angle
tan θ = y / L
as all the experiment does not occur at small angles
tan θ = sin θ / cos θ = sin θ = y / L
we substitute
y = (m + 1/2 ) λ L / d
we calculate
y = (3 + ½) 500 10⁻⁹ 5.00 / 0.5 10⁻³
y = 1.75 10⁻² m
y = 1.75 cm
Two gratings A and B have slit separations dA and dB, respectively. They are used with the same light and the same observation screen. When grating A is replaced with grating B, it is observed that the first-order maximum of A is exactly replaced by the second-order maximum of B.
a.) I already found that the ratio of db/da is 2
b.) Find the next two principal maxima of grating A and the principal maxima of B that exactly replace them when the gratings are switched. Identify these maxima by their order numbers
Answer:
mA=2,mB=4
mA=2,mB=4 and
mA=3,mB=6
Explanation:
First of all we need to write the equation of the networks
sin θ = mA λ / dA
sin θ = mB λ / dB
Equating we have
mA λ/ dA = mB λ / dB
We are given the ratio as
dB / dA = 2
So
mA 2 = mB
Finally overlapping orders
We have
mA=2,mB=4
mA=2,mB=4
and mA=3,mB=6
Joe is hiking through the woods when he decides to stop and take in the view. He is particularly interested in three objects: a squirrel sitting on a rock next to him, a tree a few meters away, and a distant mountain. As Joe is taking in the view, he thinks back to what he learned in his physics class about how the human eye works.
Light enters the eye at the curved front surface of the cornea, passes through the lens, and then strikes the retina and fovea on the back of the eye. The cornea and lens together form a compond lens system. The large difference between the index of refraction of air and that of the aqueous humor behind the cornea is responsible for most of the bending of the light rays that enter the eye, but it is the lens that allows our eyes to focus. The ciliary muscles surrounding the lens can be expanded and contracted to change the curvature of the lens, which in turn changes the effective focal length of the cornea-lens system. This in turn changes the location of the image of any object in one's field of view. Images formed on the fovea appear in focus. Images formed between the lens and the fovea appear blurry, as do images formed behind the fovea. Therefore, to focus on some object, you adjust your ciliary muscles until the image of that object is located on the fovea.
A) Joe first focuses his attention (and his eyes) on the tree. The focal length of the cornea-lens system in his eye must be __________ the distance between the front and back of his eye.
a. greater than
b. less than
c. equal to
B) Joe's eyes are focused on the tree, so the squirrel and the mountain appear out of focus. This is because the image of the squirrel is formed ______ and the image of the mountain is formed _____.
a. between the lens and fovea / between the lens and fovea
b. between the lens and fovea / behind the fovea
c. behind the fovea / between the lens and fovea
d. behind the fovea / behind the fovea
C) Joe now shifts his focus from the tree to the squirrel. To do this, the ciliary muscles in his eyes must have _____ the curvature of the lens, resulting in a(n) _______ focal length for the cornea-lens system. Note that curvature is different from radius of curvature.
a. increased / increased
b. increased / decreased
c. decreased / increased
d. decreased / decreased
Answer:
A) correct answer is C, B) correct answer is b and C) The correct answer is b
Explanation:
In the exercises of geometric optics, the equation of the constructor tells us the location of the image.
1 / f = 1 / p + 1 / q
where f is the focal length of the cornea-crystalline system, p and q are the distances to the object and the image.
In this case, the distance to the image on the retina is constant, about 3 cm. Therefore depending on the distance to the object) p = the focal length must change
1 / q = 1 / f 1 / p
let's apply this expression to our case
A) indicates that the tree is at a medium distance
so that the image is formed on the retina THE SAME AS
correct answer is C
B) The squirrel is at a smaller distance (p ') than the tree (p), therefore if we substitute in the equation above we find that q must decrease. Consequently the image is in front of the retina
The mountain is very far, suppose in infinity, so the image is BEHIND THE RETINA
therefore the correct answer is b
C) The squirrel is very close so the curvature of the lens INCREASES, resulting in a DECREASE in the focal length
The correct answer is b
If the plane travels at 600 km/h for 2 hours, how far has it traveled?
Explanation:
Distance = speed × time
d = (600 km/hr) (2 hr)
d = 1200 km
50 g of pieces of brass are heated to 200 ° C and then placed in the aluminum container of a 50 g calorimeter containing 160 g of water. What is the equilibrium temperature, if the temperature of the container and the water is initially 20 ° C?
Answer:
24.7°C
Explanation:
Heat lost by brass = heat gained by aluminum and water
-q = q
-mCΔT = mCΔT + mCΔT
-(50 g) (0.380 J/g/°C) (T − 200°C) = (50 g) (0.900 J/g/°C) (T − 20°C) + (160 g) (4.186 J/g/°C) (T − 20°C)
-(19 J/°C) (T − 200°C) = (45 J/°C) (T − 20°C) + (670 J/°C) (T − 20°C)
-(19 J/°C) (T − 200°C) = (715 J/°C) (T − 20°C)
-19 (T − 200°C) = 715 (T − 20°C)
-19T + 3800°C = 715T − 14300°C
18100°C = 734T
T = 24.7°C
A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Part A What will be their mutual speed immediately after the collision? Express your answer to two significant figures and include the appropriate units.
Given that,
Mass of trackler, m₁ = 100 kg
Speed of trackler, u₁ = 2.6 m/s
Mass of halfback, m₂ = 92 kg
Speed of halfback, u₂ = -5 m/s (direction is opposite)
To find,
Mutual speed immediately after the collision.
Solution,
The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s[/tex]
So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.
What kind of wave is formed (transverse or longitudimal wave, pick one) is formed by ripples on a calm pond? With explanation! Please help, most detailed answer will get brainliest and many points.
Answer:
Transverse
Explanation:
It's tranverse because the water molecules are moving repeatedly up and down vertically when the waves move horizontally across the waters surface.
pls help me solve dis
Answer:
9 Ω
Explanation:
The following data were obtained from the question:
Resistor 1 (R1) = 3 Ω
Resistor 2 (R2) = 3 Ω
Resistor 3 (R3) = 3 Ω
Resistor 4 (R4) = 3 Ω
Resistor 5 (R5) = 3 Ω
Resistor 6 (R6) = 3 Ω
Resistor 7 (R7) = 3 Ω
Resistor 8 (R8) = 3 Ω
Resistor 9 (R9) = 3 Ω
Resistor 10 (R10) = 3 Ω
Resistor 11 (R11) = 3 Ω
Resistor 12 (R12) = 3 Ω
Equivalent Resistance (R) =.?
From the above diagram,
Resistor 1, 2, 3, 4, 5 and 6 are in series connection and in parralle connections with Resistor 7, 8, 9, 10, 11 and 12 which are also in series connection.
Thus we shall determine the equivalent resistance of Resistor 1, 2, 3, 4, 5 and 6
This is illustrated below:
Resistance Ra = R1 + R2 + R3 + R4 + R5 + R6
Ra = 3 + 3 + 3 + 3 + 3 + 3
Ra = 18 Ω
Next, we shall determine the equivalent resistance of Resistor 7, 8, 9, 10, 11 and 12.
This is illustrated below:
Resistance Rb = R7 + R8 + R9 + R10 + R11 + R12
Rb = 3 + 3 + 3 + 3 + 3 + 3
Rb = 18 Ω
Thus, Ra and Rb are in parallel connections. The equivalent resistance between A and B can be obtained as shown below :
Ra = 18 Ω
Rb = 18 Ω
Equivalent resistance R =?
1/R = 1/Ra + 1/Rb
1/R = 1/18 + 1/18
1/R = 2/18
1/R = 1/9
Invert
R = 9 Ω
Therefore, the equivalent resistance between A and B is 9 Ω.
Why is a flask is wider on the bottom than on the top? allows for a more precise measurement allows for better thermal equilibrium prevents spillage to hold a thermometer
Answer:
allows for better thermal equilibrium
Explanation:
Due to the cone shape, most of the liquid will be closer to the bottom than the top. The large surface area of the bottom allows for faster heating.
In the growth stage of a product's life cycle, demand has its own momentum through word of mouth and interactive marketing.
a. True
b. False
Answer:
Advertising, events and experiences, and personal selling all become more important in the maturity stage.
Explanation:
In the growth stage, demand has its own momentum through word of mouth and interactive marketing. hope this helps you :)
A speeding motorist traveling down a straight highway at 100 km/h passes a parked police car. It takes the police constable 1.0 s to take a radar reading and to start up his car. The police vehicle accelerates from rest at 2 m/s2 and finally catches up with the speeder. a) How much time has elapsed when the two cars meet?
Answer:
t = 7.5 s
Explanation:
The distance traveled by the car at the time of meeting of the two cars must be the same. First, we calculate the distance traveled by the police car. For that we use 2nd equation of motion. Here, we take the time when police car starts to be reference. So,
s₁ = Vi t + (0.5)gt²
where,
s₁ = distance traveled by police car
Vi = Initial Velocity = 0 m/s
t = time taken
Therefore,
s₁ = (0 m/s)(t) + (0.5)(9.8 m/s²)t²
s₁ = 4.9 t²
Now, we calculate the distance traveled by the car. For constant speed and time to be 1 second more than the police car time, due to car starting time, we get:
s₂ = Vt = V(t + 1)
where,
s₂ = distance traveled by car
V = Velocity of car = (100 km/h)(1000 m/1 km)(1 h/ 3600 s) = 27.78 m/s
Therefore,
s₂ = 27.78 t + 27.78
Now, we know that at the time of meeting:
s₁ = s₂
4.9 t² = 27.78 t + 27.78
4.9 t² - 270.78 t - 27.78 = 0
solving the equation and choosing the positive root:
t = 6.5 s
since, we want to know the time from the moment car crossed police car. Therefore, we add 1 second of starting time in this.
t = 6.5 s + 1 s
t = 7.5 s
what happens to the speed of the
Skateboard/refrigerator when there is no longer a force being applied ?
Answer:
The speed stays constant after the force stops pushing.
Explanation:
Speed always stays constant when the force stops pushing it.
A truck pulls a block 8 meters across a level surface at a force of 216 N over the course of 12 seconds. How much power did the truck use
Answer:
work = 1728
Power = 134
Explaination:
by using the formula,
Work(W)= Force(F)×Distance(D)
andPower(P)= Work(W)/Time taken(T)
Parallel rays of monochromatic light with wavelength 582 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm. If the intensity at the center of the central maximum is 4.40×10^−4 W/m^2.
Required:
What is the intensity at a point on the screen that is 0.710 mm from the center of the central maximum?
Answer:
What is the intensity is 1.3349 × 10⁻⁷ w/m²
Explanation:
Given that;
λ = 582 nm = 582 × 10⁻⁹
R = 75.0 cm = 0.75 m
d = 0.640 mm = 0.000640 m
a = 0.434 mm = 0.000434 m
I₀ = 4.40×10⁻⁴ W/m²
y = 0.710 mm = 0.00071 m
Now to get our tanФ we say
tanФ = y/R = 0.00071 / 0.75 = 0.0009466
Ф is so small
∴ tanФ ≈ sinФ
So
∅ = 2πdsinФ / λ
we substitute
∅ = ( 2π × 0.000640 × 0.0009466 ) / 582 × 10⁻⁹
= 6.54 rad
Now
β = 2πasinФ / λ
we substitute
β = ( 2π × 0.000434 × 0.0009466 ) / 582 × 10⁻⁹
β = 4.435 rad
I = I₀ cos²(∅/2) [(sin(β/2))/(β/2)]²
we substitute
I = 4.40×10⁻⁴ cos(3.27)² [ (sin(2.2175)) / (2.2175) ]²
= 4.40×10⁻⁴ × 0.9967 × 0.0003044
= 1.3349 × 10⁻⁷ w/m²
What effect does the velocity of a rotating object have on the centripetal force exerted on it?
Answer:
The velocity of a rotating object and the centripetal force exerted on it are proportional to each other. Centripetal force is equal to mv /r. If the velocity increases, the centripetal force increases as well.
Explanation:
The velocity has the direct effect on the centripetal force. Hence, increasing the velocity, centripetal force also increases or vice - versa.
The given problem is based on the basic concept of centripetal force and fundamentals involved in the centripetal force. The centripetal force is a center seeking force which comes to play, when any object undergoes the rotational motion around the circular path.
The mathematical expression for the centripetal force is given as,
[tex]F= \dfrac{mv^{2}}{r}[/tex]
Here,
m is the mass of an object.
v is the velocity of an object undergoing the motion around the circular path.
r is the radius of curvature.
From the concept of centripetal force, the velocity of a rotating object and the centripetal force exerted on it are proportional to each other. Centripetal force is equal to mv /r. If the velocity increases, the centripetal force increases as well.
Thus, we can conclude that the velocity has the direct effect on the centripetal force. Hence, increasing the velocity, centripetal force also increases or vice - versa.
Learn more about the centripetal force here:
https://brainly.com/question/14249440
The coefficient of static friction between a 3.00 kg crate and the 35.0o incline is 0.300. What minimum force F must be applied perpendicularly to the incline to prevent the crate from sliding down
Answer:
32.13 N
Explanation:
Given that
mass of the crate, m = 3 kg
angle of inclination, = 35°
coefficient of static friction, = 0.3
To solve this, we can assume that the minimum force is F Newton, then use the formula
mgsinA = coefficient of static friction * [F + mgcosA]
=>3 * 9.8 * sin35 = 0.3 * [F + 3 * 9.8 * cos35]
=> 29.4 * 0.5736 = 0.3 * [F + 29.4 * 0.8192]
=> 16.86 = 0.3 [F + 24.08]
=> 16.86 = 0.3F + 7.22
=> 16.86 - 7.22 = 0.3F
=> 0.3F = 9.64
=> F = 9.64/0.3
=> F = 32.13 N
Therefore, the Force that must be applied is 32.13 N
The net force that must be applied is 9.8 N.
The minimum force required is Fnet.
Fnet = -Ff + mgsinθ
But Ff = μN = μmgcosθ
Fnet = - μmgcosθ + mgsinθ
Where;
m = 3.00 kg
μ = 0.300
θ = 35.0o
Substituting values;
Fnet = mgsinθ - μmgcosθ
Fnet = (3 × 10 × sin 35.0o) - (3 × 0.300 × 10 × cos 35.0o)
= 17.2 - 7.4
Fnet = 9.8 N
Learn more: https://brainly.com/question/2510654
If you wanted to find the area of the hot filament in a light bulb, you would have to know the temperature (determinable from the color of the light), the power input, the Stefan-Boltzmann constant, and what property of the filament
Answer:
To find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and Emissivity of the Filament.
Explanation:
The emissive power of a light bulb can be given by the following formula:
E = σεAT⁴
where,
E = Power Input or Emissive Power
σ = Stefan-Boltzmann constant
ε = Emissivity
A = Area
T = Absolute Temperature
Therefore,
A = E/σεT⁴
So, to find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and Emissivity of the Filament.
An electron accelerates through a 12.5 V potential difference, starting from rest, and then collides with a hydrogen atom, exciting the atom to the highest energy level allowed. List all the possible quantum-jump transitions by which the excited atom could emit a photon. 4 rightarrow 3 4 rightarrow 2 4 rightarrow 1 3 rightarrow 2 3 rightarrow 1 2 rightarrow 1
Answer:
Initial state Final state
3 ⇒ 2
3 ⇒ 1
2 ⇒ 1
Explanation:
For this exercise we must use Bohr's atomic model
E = - 13.606 / n²
where is the value of 13.606 eV is the energy of the ground state and n is the integer.
The energy acquired by the electron in units of electron volt (eV)
E = e V
E = 12.5 eV
all this energy is used to transfer an electron from the ground state to an excited state
ΔE = 13.6060 (1 / n₀² - 1 / n²)
the ground state has n₀ = 1
ΔE = 13.606 (1 - 1/n²)
1 /n² = 1 - ΔE/13,606
1 / n² = 1 - 12.5 / 13.606
1 / n² = 0.08129
n = √(1 / 0.08129)
n = 3.5
since n is an integer, maximun is
n = 3
because it cannot give more energy than the electron has
From this level there can be transition to reach the base state.
Initial state Final state
3 ⇒ 2
3 ⇒ 1
2 ⇒ 1
The possible quantum-jump transitions by which the excited atom emits a photon are :
Initial state Final state
3 ---> 2
3 ----> 1
2 ----> 1
Given data :
Potential difference through which an electron accelerates = 12.5 V
Energy acquired by the the electrons = 12.5 eV ( e * 12.5 )
The Model we will use to determine the possible quantum jump transition is Bohr's atomic model
E = - 13.606 / n²
where ; n = integer
energy at ground state = 13.606 eV
The energy acquired by the electrons ( 12.5 eV ) is used to move the electron from its ground state to an excited state.
Therefore
ΔE = 13.606 * (1 / n₀² - 1 / n²) ---- ( 1 )
where n₀ = 1
Back to equation ( 1 )
ΔE = 13.606 (1 - 1/n²) -- ( 2 )
Resolving equation ( 2 )
1 / n² = 0.08129
n = 3.5 . Therefore the maximum integer = 3
Hence The collision between the electron and the hydrogen atom will undergo three ( 3 ) transition to reach the base state.
In Conclusion The possible quantum-jump transitions by which the excited atom emits a photon are :
Initial state Final state
3 ---> 2
3 ----> 1
2 ----> 1
Learn more : https://brainly.com/question/7289718
A Carnot heat engine and an irreversible heat engine both operate between the same high temperature and low temperature reservoirs. They absorb the same energy from the high temperature reservoir as heat. The irreversible engine:__________. A. does more work. B. has the same efficiency as the reversible engine. C. has the greater efficiency. D. transfers more energy to the low temperature reservoir as heat. E. cannot absorb the same energy from the high temperature reservoir as heat without violating the second law of thermodynamics.
Answer:
the answers d
A Carnot heat engine and an irreversible heat engine both operate between the same high-temperature and low-temperature reservoirs. They absorb the same energy from the high-temperature reservoir as heat. The irreversible engine cannot absorb the same energy from the high-temperature reservoir as heat without violating the second law of thermodynamics, therefore the correct option is E.
What is a Carnot engine?It is a theoretical engine assumed to have the maximum efficiency among all engines out there in the universe.
No other engine can have greater efficiency than the Carnot engine.
The Carnot engine work on the principle of the Carnot cycle which assume that the heat and work tranfer processes are reversible in nature. there is no loss due to the friction of the piston.
As we know from the second law of thermodynamics that no other engine can have greater efficiency than the Carnot engine.
Thus, the irreversible engine cannot absorb the same energy from the high-temperature reservoir as heat without violating the second law of thermodynamics, therefore the correct option is E.
Learn more about the Carnot engine from here
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An object–spring system undergoes simple harmonic motion. If the amplitude increases but the mass of the object is not changed, the total energy of the system:______.a. undergoes a sinusoidal change. b. decreases exponentially. c. decreases. d. doesn't change. e. increases.
Answer:
The correct answer is:
doesn't change (d)
Explanation:
The total energy in a system is the sum of Kinetic and Potential energies in a system, assuming that energy is not lost to an external procedure. Now, let us define what potential and kinetic energies are:
Potential Energy: this is energy at rest or stored energy
Kinetic Energy: this is energy in motion
In a simple harmonic motion of a mass-spring system, there is no dissipative force, hence the total energy is equal to the potential and kinetic energies. The total energy is not changed rather, it varies between potential and kinetic energies depending on the point at which the mass is. The kinetic energy is greatest at the point of lowest amplitude (highest velocity) and lowest at the point of greatest amplitude (lowest velocity), while potential energy is greatest at the point of highest amplitude (lowest velocity) and lowest at the point of smallest amplitude ( highest velocity). However, at every point, the sum of kinetic and potential energies equals total energy.
A car traveling initially at 8.0 m/s accelerates to a velocity of 14 m/s in 2.0s. What is the average acceleration of the car?
Answer:
3.0 m/s²
Explanation:
Given:
v₀ = 8.0 m/s
v = 14 m/s
t = 2.0 s
Find: a
v = at + v₀
14 m/s = a (2.0 s) + 8.0 m/s
a = 3.0 m/s²
A tire swing hanging from a branch reaches nearly to the ground. How could you estimate the height of the branch using only a stopwatch?
Answer:
Explanation:
With the help of expression of time period of pendulum we can calculate the height of the branch . The swinging tire can be considered equivalent to swinging bob of a pendulum . Here length of pendulum will be equal to height of branch .
Let it be h . Let the time period of swing of tire be T then from the formula of time period of pendulum
[tex]T = 2\pi\sqrt{\frac{l}{g} }[/tex] where l is length of pendulum .
here l = h so
[tex]T = 2\pi\sqrt{\frac{h}{g} }[/tex]
[tex]h = \frac{T^2g}{4\pi^2}[/tex]
If we calculate the time period of swing of tire , we can calculate the height of branch .
The time period of swing of tire can be estimated with the help of a stop watch . Time period is time that the tire will take in going from one extreme point to the other end and then coming back . We can easily estimate it with the help of stop watch .
A small candle is 34 cm from a concave mirror having a radius of curvature of 22 cm. Where will the image of the candle be located? Follow the sign conventions.
Answer:
16.26 cm in front of the mirror
Explanation:
Using,
1/f = 1/u+1/v....................... Equation 1
Where f = focal focal length of the concave mirror, u = object distance, v = image distance.
make v the subject of the equation
v = fu/(u-f)................... Equation 2
Note: The focal length of a concave mirror is positive
Using the real- is- positive convention
Given: f = 22/2 = 11 cm, u = 34 cm.
Substitute into equation 2
v = (34×11)/(34-11)
v = 374/23
v = 16.26 cm.
The image will be formed 16.26 cm in front of the mirror.
What is the tension in the cord after the system is released from rest? Both masses (A and B) are 10-kg.
Answer:
98 N.
Explanation:
Given data: mass= 10 kg, gravity= 9.8 m/s2
required: tension in the cord= ?
solution:
formula of tension= mass x gravity
by putting values of mass and gravity, we get
tension= 10 x 9.8
tension= 98 N. Ans
If the mass of the object which is attached with both ends of cord is 10 kg, so the tension which is a opposite force of weight is 98 N.
During an earthquake, the top of a building oscillates with an amplitude of 30cm at 1.5Hz.
Part A What is the magnitude of the maximum displacement from the equilibrium of the top of the building? (in cm)
Part B What is the maximum velocity of the top of the building?(in cm/s)
Part C What is the maximum acceleration of the top of the building?(in cm/s^2
A.30cm
B.282.7cm/s
C.26655cm/s²
Explanation:
A. Maximum displacement= Amplitude= 0.3m
B.mum velocity=
2πAf= 2π x 1.5x30 =282.7cm/s
C. Maximum acceleration
= ( angular velocity)² A= (2πf)² A
=( 2 x 3.142 x 1.5)² x 30
2665.5cm/s²