Answer
3.340J
Explanation;
Using the relation. Energy stored in capacitor = U = 7.72 J
U =(1/2)CV^2
C =(eo)A/d
C*d=(eo)A=constant
C2d2=C1d1
C2=C1d1/d2
The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.
Initial separation between the plates =d1= 3.30mm .
Final separation = d2 = 1.45 mm
(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same
Energy=U =(1/2)q^2/C
U2C2 = U1C1
U2 =U1C1 /C2
U2 =U1d2/d1
Final energy = Uf = initial energy *d2/d1
Final energy = Uf =7.72*1.45/3.30
(A) Final energy = Uf = 3.340J
A fish appears to be 2.00 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish
Answer:
2,66
Explanation:
The refractive index= real depth/ apparent depth
real depth = refractive index * apparent depth
Let's assume index for water is 1.33
real depth = 2*1,33 = 2,66
When using a crowbar to remove a nail, the person should hold it at which of the following spots to increase the IMA of the lever? nearest the end prying out the nail furthest from the end prying out the nail right in the middle the location where the person holds it will not affect the IMA
Answer: the furthest from the end prying out the nail
Answer:
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Explanation:
g In the atmosphere, the shortest wavelength electromagnetic waves are called A. infrared waves. B. ultraviolet waves. C. X-rays. D. gamma rays. E.
Answer:gamma ray
Explanation:
An atom in the ground state has a collision with an electron, then emits a photon with a wavelength of 1240 nm. What conclusion can you draw about the initial kinetic energy of the electron
Answer:
attached below is the free body diagram of the missing illustration
Initial kinetic energy of the electron = 3 eV
Explanation:
The conclusion that can be drawn about the kinetic energy of the electron is
[tex]E_{e} = E_{3} - E_{1}[/tex]
E[tex]_{e}[/tex] = initial kinetic energy of the electron
E[tex]_{1}[/tex] = -4 eV
E[tex]_{3}[/tex] = -1 eV
insert the values into the equation above
[tex]E_{e}[/tex] = -1 -(-4) eV
= -1 + 4 = 3 eV
Four friends push on the same block in different directions. Allie pushes on the block to the north with a force of 18 N. Bill pushes on the block to the east with a force of 14 N. Chris pushes on the block to south with a force of 23 N. Debra pushes on the block to the west with a force of 20 N. Assuming it does not move vertically, in which directions will the block move? north and west south and east south and west north and east
Answer:
South and West
Explanation:
Those people are pushing the hardest. It will move south faster than it moves west.
You have three resistors: R1 = 1.00 Ω, R2 = 2.00 Ω, and R3 = 4.00 Ω in parallel. Find the equivalent resistance for the combination
Answer:
4 / 7
Explanation:
1/total resistance = 1/1 + 1/2 + 1/4
= 1¾
total resistance = 1 ÷ 1¾
= 4/7
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.11 mm and place your screen 8.63 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.71 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength lambda expressed in nanometers?
Answer:
λ = 605.80 nm
Explanation:
These double-slit experiments the equation for constructive interference is
d sin θ = m λ
where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.
In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm
Let's use trigonometry to find the angle
tan θ = y / L
as the angles are very small
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
we substitute in the first equation
d y / L = m λ
λ = d y / m L
let's calculate
λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)
λ = 6.05805 10⁻⁷ m
let's reduce to nm
λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)
λ = 605.80 nm
We've seen that for thermal radiation, the energy is of the form AVT4, where A is a universal constant, V is volume, and T is temperature. 1) The heat capacity CV also is proportional to a power of T, Tx. What is x
Answer:
this raise the temperature is x = 3
Explanation:
Heat capacity is the relationship between heat and temperature change
C = Q / ΔT
if the heat in the system is given by the change in energy and we carry this differential formulas
[tex]c_{v}[/tex] = dE / dT
In this problem we are told that the energy of thermal radiation is
E = A V T⁴
Let's look for the specific heat
c_{v} = AV 4 T³
the power to which this raise the temperature is x = 3
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2, separated by a distance of 1.70 mm. A 25.0-V potential difference is applied to these plates. Calculate: a. the electric field between the plates b. the surface charge density c. the capacitance d. the charge on each plate.
Answer:
(a) 1.47 x 10⁴ V/m
(b) 1.28 x 10⁻⁷C/m²
(c) 3.9 x 10⁻¹²F
(d) 9.75 x 10⁻¹¹C
Explanation:
(a) For a parallel plate capacitor, the electric field E between the plates is given by;
E = V / d -----------(i)
Where;
V = potential difference applied to the plates
d = distance between these plates
From the question;
V = 25.0V
d = 1.70mm = 0.0017m
Substitute these values into equation (i) as follows;
E = 25.0 / 0.0017
E = 1.47 x 10⁴ V/m
(c) The capacitance of the capacitor is given by
C = Aε₀ / d
Where
C = capacitance
A = Area of the plates = 7.60cm² = 0.00076m²
ε₀ = permittivity of free space = 8.85 x 10⁻¹²F/m
d = 1.70mm = 0.0017m
C = 0.00076 x 8.85 x 10⁻¹² / 0.0017
C = 3.9 x 10⁻¹²F
(d) The charge, Q, on each plate can be found as follows;
Q = C V
Q = 3.9 x 10⁻¹² x 25.0
Q = 9.75 x 10⁻¹¹C
Now since we have found other quantities, it is way easier to find the surface charge density.
(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e
σ = Q / A
σ = 9.75 x 10⁻¹¹ / 0.00076
σ = 1.28 x 10⁻⁷C/m²
The temperature difference between the inside and the outside of a house on a cold winter day is 33°F. (a) Express this difference on the Celsius scale. 0.55 Incorrect: Your answer is incorrect. °C (b) Express this difference on the Kelvin scale. 273.7 Incorrect: Your answer is incorrect. K
Answer:
a) 0.56°C
b) 273.56 K
Explanation:
If we want to convert from Fahrenheit scale to Celcius scale we use the formula;
T(°C) = (T(°F) - 32) × 5/9
Where T(°F) = 33°F
Hence;
T(°C) = (33°F - 32) × 5/9
T°C = 0.56°C
b)
T(K) = T°C + 273
T(K) = 0.56 + 273
T(K) = 273.56 K
Sammy is 5 feet and 5.3 inches tall.tall.what is sammy's height in metres?
Answer:
65.3
Explanation:
1 foot = 12 inches
Sammy is 5 feet tall.
5 feet = ? inches
Multiply the feet value by 12 to find in inches.
5 × 12
= 60
Add 5.3 inches to 60 inches.
60 + 5.3
= 65.3
Answer:
It will be 》》》》1.664716m
Give an example of hypothesis for an experiment and then identify its dependent and independent variables. Write all the steps of the scientific method. Explain why it is good to limit an experiment to test only one variable at a time whenever possible ?
Please somebody !!!!
Which notation is better to use? (Choose between 4,000,000,000,000,000 m and 4.0 × 1015 m)
Answer:
4 x 10¹⁵
Explanation:
The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the pool from above. How deep (in cm) will it appear to be
Answer:
d' = 75.1 cm
Explanation:
It is given that,
The actual depth of a shallow pool is, d = 1 m
We need to find the apparent depth of the water in the pool. Let it is equal to d'.
We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,
[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m[/tex]
or
d' = 75.1 cm
So, the apparent depth is 75.1 cm.
At a rock concert, a dB meter registered 131 dB when placed 2.6 m in front of a loudspeaker on the stage. The intensity of the reference level required to determine the sound level is 1.0×10−12W/m2.
a) What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?
b) How far away would the sound level be 86 dB?
Answer:
Explanation:
A) 131 dB = 10*log(I / 1e-12W/m²)
where I is the intensity at 2.6 m away.
13.1 = log(I / 1e-12W/m²
1.25e13= I / 1e-12W/m²
I = 1.25 x10^1W/m²
power = intensity * area
P = I * A = 12.5W/m² * 4π(2.6m)² =1061 W ◄
B) 86 dB = 10*log(I / 1e-12W/m²)
8.6 = log(I / 1e-12W/m²)
3.98e8 = I / 1e-12W/m²
I = 3.98e-4 W/m²
area A = P / I = 1061W / 3.98e-4W/m² = 2.66e6 m²
A = 4πr²
2.66e6 m² = 4πr²
r = 14.5m ◄
A 6.7 cm diameter circular loop of wire is in a 1.27 T magnetic field. The loop is removed from the field in 0.16 ss . Assume that the loop is perpendicular to the magnetic field.
Required:
What is the average induced emf?
Answer:
The induced emf is [tex]\epsilon = 0.0280 \ V[/tex]
Explanation:
From the question we are told
The diameter of the loop is [tex]d = 6.7 cm = 0.067 \ m[/tex]
The magnetic field is [tex]B = 1.27 \ T[/tex]
The time taken is [tex]dt = 0.16 \ s[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{\Delta \phi}{dt}[/tex]
Where N = 1 given that it is only a circular loop
[tex]\Delta \phi = \Delta B * A[/tex]
Where [tex]\Delta B = B_f - B_i[/tex]
where [tex]B_i[/tex] is 1.27 T given that the loop of wire was initially in the magnetic field
and [tex]B_f[/tex] is 0 T given that the loop was removed from the magnetic field
Now the area of the of the loop is evaluated as
[tex]A = \pi r^2[/tex]
Where r is the radius which is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.067}{2}[/tex]
[tex]r = 0.0335 \ m[/tex]
So
[tex]A = 3.142 * (0.0335)^2[/tex]
[tex]A = 0.00353 \ m^2[/tex]
So
[tex]\Delta \phi = (0- 127)* (0.00353)[/tex]
[tex]\Delta \phi = -0.00448 Weber[/tex]
=> [tex]\epsilon = - 1 * \frac{-0.00448}{0.16}[/tex]
=> [tex]\epsilon = 0.0280 \ V[/tex]
A pair of narrow, parallel slits separated by 0.230 mm is illuminated by green light (λ = 546.1 nm). The interference pattern is observed on a screen 1.50 m away from the plane of the parallel slits.
A) Calculate the distance from the central maximum to the first bright region on either side of the central maximum.
B) Calculate the distance between the first and second dark bands in the interference pattern.
Answer:
A) y = 3.56 mm
B) y = 3.56 mm
Explanation:
A) The distance from the central maximum to the first bright region can be found using Young's double-slit equation:
[tex] y = \frac{m\lambda L}{d} [/tex]
Where:
λ: is the wavelength = 546.1 nm
m: is first bright region = 1
L: is the distance between the screen and the plane of the parallel slits = 1.50 m
d: is the separation between the slits = 0.230 mm
[tex] y = \frac{m\lambda L}{d} = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
B) The distance between the first and second dark bands is:
[tex] \Delta y = \frac{\Delta m*\lambda L}{d} [/tex]
Where:
[tex] \Delta m = m_{2} - m_{1} = 2 - 1 = 1 [/tex]
[tex] \Delta y = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
I hope it helps you!
Two parallel slits are illuminated with monochromatic light of wavelength 567 nm. An interference pattern is formed on a screen some distance from the slits, and the fourth dark band is located 1.83 cm from the central bright band on the screen. (a) What is the path length difference corresponding to the fourth dark band? (b) What is the distance on the screen between the central bright band and the first bright band on either side of the central band? (Hint: The angle to the fourth dark band and the angle to the first bright band are small enough that tan θ ≈ sin θ.)
Answer:
a)1984.5nm
b)523mm
Explanation:
A)A destructive interference can be explained as when the phase shifting between the waves is analysed by the path lenght difference
θ=(m+0.5)λ where m= 1,2.3....
Where given from the question the 4th dark Fringe which will take place at m= 3
θ=7/2y
Where y= 567nm
= 7/2(567)=1984.5nm
But
B)tan θ ≈ y/d
And sinθ = mλ/d
y=mλd when m= 1 which is the first bright we have
Then y=(1× 567.D)/d
But the distance from Central to the 4th dark Fringe is 1.83cm then
y= 7λD/2d= 1.83cm
D/d=(2)×(1.83×10^-2)/(7×567×10^-9)
=92221.5
y= (567×10^-9)× (92221.5)
=0.00523m
Therefore, the distance between the first and center is y1-y0= 523mm
the atomic number of a nucleus increases during which nuclear reactions
Answer:
Answer A : Fusion followed by beta decay (electron emission)
Explanation:
Notice that you want the Atomic number to increase, that is the number of protons in a nucleus. So if all four cases given experience the same fusion of nuclei, the only one that net increases the number of protons in the last stage, is the reaction that undergoes a beta decay (with emission of an electron) thus leaving a positive imbalance of positive charge (proton generated in the beta decay of a neutron).
Therefore, answer A is the correct one.
Answer:
A : Fusion followed by beta decay (electron emission)
Explanation:
Ap3x
Coherent light that contains two wavelengths, 660 nm (red) and 470 nm (blue), passes through two narrow slits that are separated by 0.310 mm. Their interference pattern is observed on a screen 4.40 m from the slits. What is the disatnce on the screen between the first order bright fringe for each wavelength?
Answer:
0.002699 m or 2.699 mm
Explanation:
y = Fringe distance
d= Distance between slits = 0.310mm
L = Screen distance = 4.40m
λ= Wavelength
Given from question
λ₁= 660 nm = 6.6 x 10^-9 m
λ₂= 470 nm = 4.7 x 10^-9 m
d = 0.340 mm = 3.4 x 10^-3 m
L = 4.40 m
In the case of constructive interference, we use below formula
y/L = mλ/d
For first order wavelength
(y₁/4.40) =(1×660x10⁻⁹)/(0.310*10⁻³)
y₁= (0.310*10⁻³)×(4.40)/(0.310*10⁻³)
y₁=0.00937m
(y2/4.40) =(1×470x10⁻⁹)/(0.310*10⁻³)
y2= =(1×470x10⁻⁹)×(4.40)/(0.310*10⁻³)
y2=0.00667m
distance between the fringes is given by (y₁ -y2)
=0.00937-0.00667=0.002699m
Therefore, distance on the screen between the first-order bright fringes for the two wavelengths is 0.002699 m or 2.699 mm
A wire along the z axis carries a current of 4.9 A in the z direction Find the magnitude and direction of the force exerted on a 3.3 cm long length of this wire by a uniform magnetic field pointing in the x direction having a magnitude 0.43T
Answer:
0.069 N, in the X directionExplanation:
According to Flemming's left hand rule, it sates that if the first three fingers of the left hand are held mutually at right angles to one another, the fore finger will point in the direction of magnetic field, the middle finger will point in direction of current, while the thumb will point to the direction of force.
Mathematically the law is stated as
F= BIL
given data
Magnetic field B= 0.43T
Current I= 4.9 A
length of conductor L= 3.3cm to meter , 3.3/100= 0.033 m
Applying the formula the force is calculated as
F= 0.43*4.9* 0.033= 0.069 N
According to Flemming's rule the direction of all parameters are mutually perpendicular to one another, then the Force is in the X direction
An inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA. Part A What is the resistance RR of the inductor
i
CHECK COMPLETE QUESTION BELOW
inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.
Part A)What is the resistance RR of the inductor
PART B) what is inductance L of the conductor
Answer:
A)R=1818.18 ohms
B)L=1.0446H
Explanation:
We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.
There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,
.
a)A)What is the resistance RR of the inductor?
The current flowing into RL circuit can be calculated using below expresion
i=ε/R[1-e⁻(R/L)t]
at t=∞ there is maximum current
i(max)= ε/R
Where ε emf of the battery
R is the resistance
R=ε/i(max)
= 12V/(6.60*10⁻³A)
R=1818.18 ohms
Therefore, the resistance R=1818.18 ohms
b)what is inductance L of the conductor?
i(t=0.80ms and 4.96mA
RT/L = ⁻ln[1- 1/t(max)]
Making L subject of formula we have
L=-RT/ln[1-i/i(max)]
If we substitute the values into the above expresion we have
L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]
L=1.0446H
Therefore, the inductor L=1.0446H
A simple pendulum is 3.00 m long. (a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2? s (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2? s (c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2? s
Answer:
a,)3.042s
b)4.173s
c)3.281s
Explanation:
For a some pendulum the period in seconds T can be calculated using below formula
T=2π√(L/G)
Where L = length of pendulum in meters
G = gravitational acceleration = 9.8 m/s²
Then we are told to calculate
(a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2?
Since oscillations for this pendulum is located in the elevator that is accelerating upward at 3.00 then
use G = 9.8 + 3.0 = 12.8 m/s²
Period T=2π√(L/G)
T= 2π√(3/12.8)
T=3.042s
b) (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2?
G = 9.8 – 3.0 = 6.8 m/s²
T= 2π√(3/6.8)
T=4.173s
C)(c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2?
Net acceleration is
g'= √(g² + a²)
=√(9² + 3²)
Then period is
T=2π√(3/11)
T=3.281s
A long solenoid of radius 3 cm has 1100 turns per meter. If the solenoid carries a current of 1.5 A, then calculate the magnetic field at the center of the solenoid.a. 2.1E^-3T b. 1.0E^-3 T c. 1.7E^-4T d. 7.0E^-2 T
Answer:
The magnetic field at the center of the solenoid is 2.1 × 10⁻³ T
Explanation:
The magnetic field B at the center of the solenoid is given by
B = μ₀ni where μ₀ = permeability of free space = 4π × 10⁻⁷H/m, n = number of turns per unit length of the solenoid = 1100 turns per meter and i = current in the solenoid = 1.5 A.
So B = μ₀ni
= 4π × 10⁻⁷H/m × 1100 × 1.5 A
= 4π × 10⁻⁷H/m × 1650 A-turns/m
= 20734.5 × 10⁻⁷T
= 2.07345 × 10⁻³ T
≅ 2.1 × 10⁻³ T
So the magnetic field at the center of the solenoid is 2.1 × 10⁻³ T
You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 38.5∘ from the ground with an initial speed of 27.5 m/s. At what height above its initial position does the softball hit the wall? Ignore any effects of air resistance.
The ball's horizontal position in the air is
[tex]x=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t[/tex]
It hits the wall when [tex]x=17.5\,\mathrm m[/tex], which happens at
[tex]17.5\,\mathrm m=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t\implies t\approx0.813\,\mathrm s[/tex]
Meanwhile, the ball's vertical position is
[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ t-\dfrac g2t^2[/tex]
where [tex]g[/tex] is the acceleration due to gravity, 9.80 m/s^2.
At the time the ball hits the wall, its vertical position (relative to its initial position) is
[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ(0.813\,\mathrm s)-\dfrac g2(0.813\,\mathrm s)^2\approx\boxed{10.7\,\mathrm m}[/tex]
Which unbalanced force accounts for the direction of the net force of the rocket?
a. Air resistance
b. Friction
c. Gravity
d. Thrust of rocket engine
It depends on what stage of the mission you're talking about.
==> While it's sitting on the pad before launch, the forces on the rocket are balanced, so there's no net force on it.
==> When the engines ignite, their thrust (d) is greater than the force of gravity. So the net force on the rocket is upward, and the spacecraft accelerates upward.
==> After the engines shut down, the net force acting on the rocket is due to Gravity (c).
. . . If the rocket has enough vertical speed, it escapes the Earth completely, and just keeps going.
. . . If it has enough horizontal speed, it enters Earth orbit.
. . . If it doesn't have enough vertical or horizontal speed, it falls back to Earth.
A rocket will preserve to speed up so long as there's a resultant pressure upwards resulting from the thrust of the rocket engine.
What unbalanced force bills for the course of the internet pressure of the rocket?A rocket launches whilst the pressure of thrust pushing it upwards is greater than the burden force because of gravity downwards. This unbalanced pressure reasons a rocket to accelerate upwards. A rocket will maintain to hurry up so long as there's a resultant force upwards resulting from the thrust of the rocket engine.
What's the net pressure of unbalanced?
If the forces on an item are balanced, the net pressure is zero. If the forces are unbalanced forces, the results do not cancel each difference. Any time the forces acting on an object are unbalanced, the net pressure is not 0, and the movement of the item modifications.
Learn more about the thrust of the rocket engine. here: https://brainly.com/question/10716695
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A fireperson is 50 m from a burning building and directs a stream of water from a fire hose at an angle of 300 above the horizontal. If the initial speed of the stream is 40 m/s the height that the stream of water will strike the building is
Answer:
We can think the water stream as a solid object that is fired.
The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)
The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).
The initial speed of the stream is 40m/s.
First, using the fact that:
x = R*cos(θ)
y = R*sin(θ)
in this case R = 40m/s and θ = 30°
We can use the above relation to find the components of the velocity:
Vx = 40m/s*cos(30°) = 34.64m/s
Vy = 20m/s.
First step:
We want to find the time needed to the stream to hit the buildin.
The horizontal speed is 34.64m/s and the distance to the wall is 50m
So we want that:
34.64m/s*t = 50m
t = 50m/(34.64m/s) = 1.44 seconds.
Now we need to calculate the height of the stream at t = 1.44s
Second step:
The only force acting on the water is the gravitational one, so the acceleration of the stream is:
a(t) = -g.
g = -9.8m/s^2
For the speed, we integrate over time and we get:
v(t) = -g*t + v0
where v0 is the initial speed: v0 = 20m/s.
The velocity equation is:
v(t) = -g*t + 20m/s.
For the position, we integrate again over time:
p(t) = -(1/2)*g*t^2 + 20m/s*t + p0
p0 is the initial height of the stream, this data is not known.
Now, the height at the time t = 1.44s is
p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po
= 16.57m + p0
So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.
Two identical rooms in a house are connected by an open doorway. The temperatures in the two rooms are maintained at different values. Which room contains more air
Answer:
The room with the lower temperature
Explanation:
Using
PV=nRT
Since both the rooms same volume and are connected, so they will have same pressure
PV=nRT=constant
nT=Constant/R=constant
If T is more n has to be less
Thus, lower the temperature, more the number molecules.
In a front-end collision, a 1500 kg car with shock-absorbing bumpers can withstand a maximumforce of 80 000 N before damage occurs. If the maximum speed for a non-damaging collision is4.0 km/h, by how much must the bumper be able to move relative to the car
Answer:
The bumper will be able to move by 0.01155m.
Explanation:
The magnitude of deceleration of the car in the front end collision.
[tex]a = \frac{F_m}{m} \\[/tex]
[tex]a = \frac{80000}{1500} \\[/tex]
[tex]a = 53.33[/tex]
This is the deceleration of the car that is generated to stop due to a front end collision.
4 km/h = 1.11 m/s
Now, the initial speed of the bumper in the relation of car, Vi = 0
Now, the initial speed of the bumper in the relation of car, Vf = 1.11 m/s
Use the below equation:
[tex]s = \frac{(Intitial \ speed)^2 – (Final \ speed)^2}{2a} \\[/tex]
[tex]s = \frac{(1.11)^2 – (0)}{2 \times 53.33} \\[/tex]
[tex]s = 0.01155 \\[/tex]
Thus, the bumper can move relative to the car is 0.01155 m .
A laser emits photons having an energy of 3.74 × 10–19 J. What color would be expected for the light emitted by this laser? (c = 3.00 × 108 m/s, h = 6.63 × 10–34 J ⋅ s)
Answer:
The wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.
Explanation:
Given;
energy of the emitted photons, E = 3.74 x 10⁻¹⁹ J
speed of light, c = 3 x 10⁸ m/s
Planck's constant, h = 6.63 x 10⁻³⁴ J.s
The wavelength of the emitted light will be calculated by applying energy of photons;
[tex]E = hf[/tex]
where;
E is the energy emitted light
h is Planck's constant
f is frequency of the emitted photon
But f = c / λ
where;
λ is the wavelength of the emitted photons
[tex]E = \frac{hc}{\lambda} \\\\\lambda = \frac{hc}{E} \\\\\lambda = \frac{6.63*10^{-34} *3*10^{8}}{3.74*10^{-19}} \\\\\lambda = 5.318 *10^{-7} \ m\\\\\lambda = 531.8 *10^{-9} \ m\\\\\lambda = 531.8 \ nm[/tex]
λ ≅ 532 nm
the wavelength of the emitted photons is 532 nm.
Therefore, the wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.
