Answer:
37.91594 keV
Explanation:
[tex]E_i[/tex] = Incident energy = 400 keV
θ = 30°
h = Planck's constant = 4.135×10⁻¹⁵ eV s = 6.626×10⁻³⁴ J s
Incident photon wavelength
[tex]\lambda_i=\frac{hc}{E_i}\\\Rightarrow \lambda_i=\frac{4.135\times 10^{-15}\times 3\times 10^8}{400\times 10^3}\\\Rightarrow \lambda_i=3.101\times 10^{-12}\ m[/tex]
Difference in wavelength
[tex]\Delta \lambda=\frac{h}{m_ec}(1-cos\theta)\\\Rightarrow \Delta \lambda=\frac{6.626\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8}(1-cos30)\\\Rightarrow \Delta \lambda=3.248\times 10^{-13}\ m[/tex]
[tex]\lambda_f=\lambda_i+\Delta \lambda\\\Rightarrow \lambda_f=3.101\times 10^{-12}+3.248\times 10^{-13}\\\Rightarrow \lambda_f=3.426\times 10^{-12}[/tex]
Final photon wavelength
[tex]\lambda_f=\frac{hc}{\lambda_f}\\\Rightarrow E_f=\frac{4.135\times 10^{-15}\times 3\times 10^8}{3.426\times 10^{-12}}\\\Rightarrow E_f=362084.06\ eV = 362.08406\ keV[/tex]
Energy of the recoiling electron
[tex]\Delta E=E_i-E_f\\\Rightarrow \Delta E=400-362.08406=37.91594\ keV[/tex]
Energy of the recoiling electron is 37.91594 keV
Answer:
The energy of recoiling electron=192.44 keV
Explanation:
Energy of x-ray[tex]E_o=400 keV[/tex]
Web know that compton shift is
[tex]\Delta \lambda =\dfrac{h}{m_eC(1-cos\theta)}[/tex]
[tex]m_e[/tex] is the mass of electron and C is the velocity of sound.
Given that θ=30°
Now by putting the values
[tex]\Delta \lambda =\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8(1-cos30)}[/tex]
[tex]\Delta \lambda =2.8\times 10^{-3}nm[/tex]
[tex]\Delta \lambda_o=\dfrac{hc}{E_o}[/tex]
By putting the values
[tex]\Delta \lambda_o=\dfrac{6.63\times 10^{-34}3\times 10^8}{400\times 1.602\times 10^{-16}}[/tex]
[tex]\Delta \lambda_o=3.31times 10^{-3}nm[/tex]
[tex]\lambda =\Delta \lambda +\lambda _o[/tex]
[tex]\lambda=2.8\times 10^{-3}+3.31\times 10^{-3}nm[/tex]
[tex]\lambda=6.11`\times 10^{-3}nm[/tex]
Energy [tex]E=\dfrac{hC}{\lambda }[/tex]
So [tex]E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{6.11\times 10^{-12}}[/tex]
E=207.55 keV
The energy of recoiling electron=400-207.55 keV
The energy of recoiling electron=192.44 keV
An express subway train passes through an underground station. It enters at t = 0 with an initial velocity of 23.0 m/s and decelerates at a rate of 0.150 m/s^2 as it goes through. The station in 205 m long (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, at what time t does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?
Answer:
a) Train's nose will be present 9.19 seconds in the station.
b) The nose leaves the station at 21.62 m/s.
c) Train's end will leave after 15.33 seconds from station.
d) The end leaves the station at 20.70 m/s.
Explanation:
a) We have equation of motion s = ut + 0.5at²
Here u = 23 m/s, a = -0.15 m/s², s = 205 m
Substituting
205 = 23t + 0.5 x (-0.15) x t²
0.075t² -23 t +205 = 0
We will get t = 9.19 or t = 297.47
We have to consider the minimum time
So train's nose will be present 9.19 seconds in the station.
b) We have equation of motion v= u + at
Here u = 23 m/s, a = -0.15 m/s², t = 9.19
Substituting
v= 23 - 0.15 x 9.19 = 21.62 m/s
The nose leaves the station at 21.62 m/s.
c) We have equation of motion s = ut + 0.5at²
Here u = 23 m/s, a = -0.15 m/s², s = 205 + 130 = 335 m
Substituting
335 = 23t + 0.5 x (-0.15) x t²
0.075t² -23 t +335 = 0
We will get t = 15.33 or t = 291.33
We have to consider the minimum time
So train's end will leave after 15.33 seconds from station.
d) We have equation of motion v= u + at
Here u = 23 m/s, a = -0.15 m/s², t = 15.33
Substituting
v= 23 - 0.15 x 15.33 = 20.70 m/s
The end leaves the station at 20.70 m/s.
A circular coil of 216 turns has a radius of 2.32 cm. (a) Calculate the current that results in a magnetic dipole moment of magnitude 2.89 A·m2. (b) Find the maximum magnitude of the torque that the coil, carrying this current, can experience in a uniform 46.0 mT magnetic field.
Answer:
(a) 7.92 A
(b) 0.133 Nm
Explanation:
N = 216
r = 2.32 cm = 0.0232 m
(a) M = 2.89 Am^2
M = N i A
Where, A be the area of the coil and i be the current in the coil
2.89 = 216 x i x 3.14 x 0.0232 x 0.0232
i = 7.92 A
(b) B = 46 mT = 0.046 T
Torque, τ = M B Sin 90
τ = 2.89 x 0.046 x 1 = 0.133 Nm
The current required to produce a magnetic dipole moment of 2.89 A·m² in a coil of 216 turns and radius 2.32 cm is approximately 7.92 A. The maximum torque this coil can experience in a 46.0 mT magnetic field is about 0.133 N·m.
To solve these problems, we'll use the relationship between the magnetic dipole moment, current, and torque in a magnetic field.
Part (a): Calculate the Current
The magnetic dipole moment (μ) of a coil is given by:
μ = NIA
where N is the number of turns, I is the current, and A is the area of the coil.
Given:
N = 216 turns
μ = 2.89 A·m²
Radius (r) = 2.32 cm = 0.0232 m
The area (A) of the coil is:
A = πr² = π(0.0232 m)² ≈ 1.69 × 10⁻³ m²
To find the current (I), rearrange the formula:
I = μ / (N × A)
Substitute the values:
I = 2.89 A·m² / (216 × 1.69 × 10⁻³ m²) ≈ 7.92 A
Part (b): Calculate the Maximum Torque
The maximum torque (τ) experienced by the coil in a magnetic field (B) is given by:
τ = μB
Given the magnetic field (B) is 46.0 mT = 0.046 T, substitute the values:
τ = 2.89 A·m² × 0.046 T ≈ 0.133 N·m
Thus, the current required to achieve the given magnetic dipole moment is approximately 7.92 A, and the maximum torque experienced by the coil in a 46.0 mT magnetic field is approximately 0.133 N·m.
In the absence of friction, how much work would a child do while pulling a 12 kg wagon a distance of 3.9 m with a 22 N force? O 47J O 60J O 86 J O991
Answer:
W = 86 J
Explanation:
Work done by friction force is given by the formula
[tex]W = F_f . d[/tex]
here since the wagon is pulled by horizontal force such that speed remains constant
so here we will have
[tex]F_f = F_{ext} = 22 N[/tex]
Now we will have
[tex]W = 22\times 3.9 [/tex]
[tex]W = 86 J[/tex]
so work done against friction force will be equal to 86 J
The correct answer is option c 86.
Calculating Work Done by the Child:
In this problem, we need to determine the amount of work done by a child who is pulling a 12 kg wagon over a distance of 3.9 meters with a force of 22 Newtons, assuming there is no friction.
The formula to calculate work (W) is:
W = F * d
where W is work, F is the force applied, and d is the distance covered.
Let's plug in the values provided:
F = 22 N
d = 3.9 m
Thus, the work done is:
W = 22 N * 3.9 m
W = 85.8 J
Therefore, the correct answer is approximately 86 J.
The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determine the magnitude and the direction of the non-conservative force acting on the box as it slides.
Answer:
F = - 59.375 N
Explanation:
GIVEN DATA:
Initial velocity = 11 m/s
final velocity = 1.5 m/s
let force be F
work done = mass* F = 4*F
we know that
Change in kinetic energy = work done
kinetic energy = [tex]= \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})[/tex]
kinetic energy = [tex]= \frac{1}{2}*4*(1.5^{2}-11^{2})[/tex] = -237.5 kg m/s2
-237.5 = 4*F
F = - 59.375 N
The magnitude of the non-conservative force acting on the box is 236 J and it acts horizontally.
Explanation:The magnitude and direction of the non-conservative force acting on the box can be determined using the work-energy theorem. The work done by the non-conservative force is equal to the change in kinetic energy of the box. From the given information, the initial kinetic energy of the box is 0.5 * 4.0 kg * (11 m/s)^2 = 242 J and the final kinetic energy is 0.5 * 4.0 kg * (1.5 m/s)^2 = 6 J. Therefore, the work done by the non-conservative force is 242 J - 6 J = 236 J. Since the box moves horizontally, the non-conservative force acts horizontally as well.
Learn more about Non-conservative force here:https://brainly.com/question/199265
#SPJ3
A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, equal to that of the frame. A uniform magnetic field is turned on perpendicular to the plane of the coil. This field changes at a constant rate from 20 to 80 mT in a time of 20 ms. What is the magnitude of the induced emf in the coil at the instant the magnetic field has a magnitude of 50 mT?
Answer:
Approximately 18 volts when the magnetic field strength increases from [tex]\rm 20\; mT[/tex] to [tex]\rm 80\;mT[/tex] at a constant rate.
Explanation:
By the Faraday's Law of Induction, the EMF [tex]\epsilon[/tex] that a changing magnetic flux induces in a coil is:
[tex]\displaystyle \epsilon = N \cdot \frac{d\phi}{dt}[/tex],
where
[tex]N[/tex] is the number of turns in the coil, and[tex]\displaystyle \frac{d\phi}{dt}[/tex] is the rate of change in magnetic flux through this coil.However, for a coil the magnetic flux [tex]\phi[/tex] is equal to
[tex]\phi = B \cdot A\cdot \cos{\theta}[/tex],
where
[tex]B[/tex] is the magnetic field strength at the coil, and[tex]A\cdot \cos{\theta}[/tex] is the area of the coil perpendicular to the magnetic field.For this coil, the magnetic field is perpendicular to coil, so [tex]\theta = 0[/tex] and [tex]A\cdot \cos{\theta} = A[/tex]. The area of this circular coil is equal to [tex]\pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}[/tex].
[tex]A\cdot \cos{\theta} = A[/tex] doesn't change, so the rate of change in the magnetic flux [tex]\phi[/tex] through the coil depends only on the rate of change in the magnetic field strength [tex]B[/tex]. The size of the magnetic field at the instant that [tex]B = \rm 50\; mT[/tex] will not matter as long as the rate of change in [tex]B[/tex] is constant.
[tex]\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}[/tex].
As a result,
[tex]\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V[/tex].
Using Faraday's law, the magnitude of the induced emf in the coil at the instant the magnetic field is 50 mT is found to be 18 volts. The negative sign indicates the emf opposes the change in flux according to Lenz's law.
Explanation:The question is about the calculation of the magnitude of induced emf in the coil at the instant the magnetic field has a magnitude of 50 mT. To approach this, we can use Faraday's law of electromagnetic induction. It states that the induced emf is equal to the rate of change of the magnetic flux.
First, let us recall the expression of the magnetic flux (Φ): Φ = B * A, where B is the magnetic field's magnitude and A is the area through which it passes. For a circular coil, A = π * (radius)^2. Applying the given radius 0.08 m, we calculate A = 0.02 m².
Given that the magnetic field changes linearly with time from 20 mT to 80 mT in 20 ms, we can determine the rate of change of the magnetic field which is (80 mT - 20 mT) / 20 ms = 3 T/s. Therefore, the rate of change of Flux is dΦ/dt = (B * A)/dt = 3 T/s * 0.02 m² = 0.06 Wb/s.
Finally, as per Faraday's law, the induced emf = -NdΦ/dt, where N is the number of turns in the coil. Therefore, for N = 300 turns, emf = -300 * 0.06 Wb/s = -18 V. The negative sign indicates the emf would oppose the change in flux according to Lenz's law.
Learn more about Induced emf here:https://brainly.com/question/36650405
#SPJ3
A uniform solid sphere of mass M and radius R rotates with an angular speed ω about an axis through its center. A uniform solid cylinder of mass M, radius R, and length 2R rotates through an axis running through the central axis of the cylinder. What must be the angular speed of the cylinder so it will have the same rotational kinetic energy as the sphere?
Answer:
[tex]\omega' = 0.89\omega[/tex]
Explanation:
Rotational inertia of uniform solid sphere is given as
[tex]I = \frac{2}{5}MR^2[/tex]
now we have its angular speed given as
angular speed = [tex]\omega[/tex]
now we have its final rotational kinetic energy as
[tex]KE = \frac{1}{2}(\frac{2}{5}MR^2)\omega^2[/tex]
now the rotational inertia of solid cylinder about its axis is given by
[tex]I = \frac{1}{2}MR^2[/tex]
now let say its angular speed is given as
angular speed = [tex]\omega'[/tex]
now its rotational kinetic energy is given by
[tex]KE = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2[/tex]
now if rotational kinetic energy of solid sphere is same as rotational kinetic energy of solid sphere then
[tex]\frac{1}{2}(\frac{2}{5}MR^2)\omega^2 = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2[/tex]
[tex]\frac{2}{5}\omega^2 = \frac{1}{2}\omega'^2[/tex]
[tex]\omega' = 0.89\omega[/tex]
Answer:
w_cyl = ±√(4/5) ω
Explanation:
Kinetic energy
E = (1/2)Iw²
where I is the moment of inertia and w the angular frequency of rotation.
The moment of inertia of a solid sphere of mass M and radius R is:
I = (2/5)MR²,
Solid cylinder is of mass M and radius R
I = (1/2)MR²
Equate the energies through
(1/2)×(2/5) M R²× (w_sphere)² = (1/2)× (1/2) MR² × (w_cyl)²
(w_cyl)² = (4/5)(w_sphere)²
w_cyl = ±√(4/5) ω
The energy of rotation is independent of the direction of rotation
A ball is thrown straight down with a speed of Va2.00 from a building 40.00 meters high. How much S time passes before the ball reaches the ground? a.) -3.07s b.) - 2.46 s c.) =2.66s d.) =9.89s e.) None of the above.A ball is thrown straight down with a speed of Va2.00 from a building 40.00 meters high. How much S time passes before the ball reaches the ground? a.) -3.07s b.) - 2.46 s c.) =2.66s d.) =9.89s e.) None of the above.
Answer:
Option C is the correct answer.
Explanation:
Considering vertical motion of ball:-
Initial velocity, u = 2 m/s
Acceleration , a = 9.81 m/s²
Displacement, s = 40 m
We have equation of motion s= ut + 0.5 at²
Substituting
s= ut + 0.5 at²
40 = 2 x t + 0.5 x 9.81 x t²
4.9t² + 2t - 40 = 0
t = 2.66 s or t = -3.06 s
So, time is 2.66 s.
Option C is the correct answer.
At what frequency will a 31.0 mH inductor have a reactance of 637.0 Ω?
Answer:
3272.4 Hz
Explanation:
L = 31 mH
XL = 637 ohm
XL = 2 π f L
f = XL / (2 π L)
f = 637 / ( 2 x 3.14 x 31 x 10^-3)
f = 3272.4 Hz
Emperor penguins are known to dive down to a depth of about 530 meter to hunt. If atmospheric pressure is 1.013 x 10^5 Pa, and the density of seawater is about 1025 kg/m^3, what is the absolute pressure a penguin experiences at that depth?
Answer:
5.4 x 10⁶ Pa
Explanation:
h = depth to which penguins dive under seawater = 530 m
P₀ = Atmospheric pressure = 1.013 x 10⁵ pa
ρ = density of seawater = 1025 kg/m³
P = absolute pressure experienced by penguin at that depth
Absolute pressure is given as
P = P₀ + ρgh
Inserting the values
P = 1.013 x 10⁵ + (1025) (9.8) (530)
P = 5.4 x 10⁶ Pa
A 12.0 kg block rests on an inclined plane. The plane makes an angle of 31.0° with the horizontal, and the coefficient of friction between the block and the plane is 0.158. The 12.0 kg block is tied to a second block (mass = 38.0 kg) which hangs over the end of the inclined plane after the rope passes over an ideal pulley. (a) What is the acceleration of each of the two blocks, and (b) what is the tension in the rope?
Answer:
The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.
Explanation:
Given that,
Mass = 12.0 kg
Angle = 31.0°
Friction coefficient = 0.158
Mass of second block = 38.0 kg
Using formula of frictional force
[tex]f_{\mu} = \mu N[/tex]....(I)
Where, N = normal force
[tex]N = mg\cos\theta[/tex]
Put the value of N into the formula
[tex]N =12\times9.8\times\cos 31^{\circ}[/tex]
[tex]N=100.80\ N[/tex]
Put the value of N in equation (I)
[tex]f_{mu}=0.158\times100.80[/tex]
[tex]f_{mu}=15.9264\ N[/tex]
Now, Weight of second block
[tex]W = mg[/tex]
[tex]W=38.0\times9.8[/tex]
[tex]W=372.4\ N[/tex]
The horizontal force is
[tex]F = mg\sintheta[/tex]
[tex]F=12\times9.8\times\sin 31^{\circ}[/tex]
[tex]F=60.5684\ N[/tex]....(II)
(I). We need to calculate the acceleration
[tex]a=m_{2}g-\dfrac{f_{\mu}+mg\sin\theta}{m_{1}+m_{2}}[/tex]
[tex]a=\dfrac{372.4-(15.9264+60.5684)}{12+38}[/tex]
[tex]a=5.92\ m/s^2[/tex]
(II). We need to calculate the tension in the rope
[tex]m_{2}g-T=m_{2}a[/tex]
[tex]-T=38\times5.92-38\times9.8[/tex]
[tex]T=147.44\ N[/tex]
Hence, The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.
A large container, 120 cm deep is filled with water. If a small hole is punched in its side 77.0 cm from the top, at what initial speed will the water flow from the hole? Please use a equation and explain every step
Answer:
The water will flow at a speed of 3,884 m/s
Explanation:
Torricelli's equation
v = [tex]\sqrt{2gh}[/tex]
*v = liquid velocity at the exit of the hole
g = gravity acceleration
h = distance from the surface of the liquid to the center of the hole.
v = [tex]\sqrt{2*9,8m/s^2*0,77m}[/tex] = 3,884 m/s
We have that for the Question"A large container, 120 cm deep is filled with water. If a small hole is punched in its side 77.0 cm from the top, at what initial speed will the water flow from the hole" it can be said that initial speed the water flow from the hole is
v=3.88m/s
From the question we are told
A large container, 120 cm deep is filled with water. If a small hole is punched in its side 77.0 cm from the top, at what initial speed will the water flow from the hole? Please use a equation and explain every step
Generally the equation for the water flow speed is mathematically given as
[tex]v=\sqrt(2gh)\\\\v=\sqrt{2*9.8*0.77}[/tex]
v=3.88m/s
Therefore
initial speed the water flow from the hole is
v=3.88m/s
For more information on this visit
https://brainly.com/question/23379286
A 66.0-kg boy and his 45.0-kg sister, both wearing roller blades, face each other at rest. The girl pushes the boy hard, sending him backward with a velocity 2.80 m/s toward the west. Ignore friction. (a) Describe the subsequent motion of the girl.
Answer:
the girl moves towards right with a velocity of 4.1m/s
Explanation:
Since the system is isolated the momentum of the system is conserved
Initial momentum = Final Momentum
Since initially the system is at rest thus [tex]\overrightarrow{p_{i}}=0[/tex]
Now the final momentum of boy = [tex]m_{boy}×velocity[/tex]
[tex]\overrightarrow{p_{boy}}=66.0\times -2.80m/s\\\\\overrightarrow{p_{boy}}=-184.8kgm/s[/tex]
Now for girl let the velocity = u hence her moumentum is 45[tex]\times u[/tex]
Thus equating final momentum to zero we have
[tex]-184.8kgm/s[/tex]+[tex]45\times u = 0[/tex]
[tex]u=\frac{184.8}{45}m/s[/tex]
hence [tex]u=4.1m/s[/tex]
Thus the girl moves towards right with a velocity of 4.1m/s
When the girl pushes her brother on roller blades, causing him to move backwards, the principle of conservation of momentum dictates that the girl will move in the opposite direction with a velocity of 4.16 m/s towards the east.
Explanation:The scenario presented involves the conservation of momentum, which is a fundamental concept in physics. When the girl pushes her brother on roller blades, and he moves backward at a velocity of 2.80 m/s, by the principle of conservation of momentum, the girl will move in the opposite direction.
Since no external forces are acting on the system (assuming friction is ignored), the total momentum before and after the push remains constant. The combined momentum of the boy and girl before the push is zero because they are both initially at rest. After the push, the momentum of the boy is his mass multiplied by his velocity (66.0 kg × 2.80 m/s).
To find the subsequent motion of the girl, let's calculate:
Initial total momentum = 0 kg·m/s (since both are at rest)Final total momentum = Initial total momentum (due to conservation of momentum)Final momentum of the boy = (mass of boy) × (velocity of boy) = 66.0 kg × (-2.80 m/s)Final momentum of the girl = (mass of girl) × (velocity of girl)0 = (66.0 kg × -2.80 m/s) + (45.0 kg × velocity of girl)Velocity of girl = - (66.0 kg × -2.80 m/s)/45.0 kgVelocity of girl = 4.16 m/s (towards the east)Therefore, the girl will move towards the east at a velocity of 4.16 m/s as a result of the push.
Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. How does the molecular mass of A compare to that of B? (a) It is twice that of B. (b) It is one half that of B (c) It is four times that of B (d) It is one fourth that of B. (e) It is 1.4 times that of B.
Answer:
option (d)
Explanation:
The relation between the rms velocity and the molecular mass is given by
v proportional to \frac{1}{\sqrt{M}} keeping the temperature constant
So for two gases
[tex]\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}[/tex]
[tex]\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}[/tex]
[tex]{\frac{M_{B}}{M_{A}}} = 4[/tex]
[tex]{\frac{M_{B}}{4}} = M_{A}[/tex]
If the RMS speed of the molecules of gas A is twice that of gas B then the molecular mass of A compared to that of B is one-fourth that of B.
How does the molecular mass of A compare to that of B?We know that the RMS velocity of the molecule is given as,
[tex]V = \dfrac{1}{\sqrt{M}}[/tex]
Given to us
RMS speed of the molecules of gas A is twice that of gas B, therefore, [tex]V_A = 2 V_B[/tex]
Substitute the value of RMS in the equation [tex]V_A = 2 V_B[/tex],
[tex]\dfrac{1}{\sqrt{M_A}} = \dfrac{2}{\sqrt{M_B}}\\\\\\\sqrt{\dfrac{M_B}{M_A}} = 2\\\\\\\dfrac{M_B}{M_A} = 4[/tex]
Hence, If the RMS speed of the molecules of gas A is twice that of gas B then the molecular mass of A compared to that of B is one-fourth that of B.
Learn more about RMS speed:
https://brainly.com/question/15514391
Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm. What is the width of the slit?
Answer:
The width of the slit is 0.167 mm
Explanation:
Wavelength of light, [tex]\lambda=608\ nm=608\times 10^{-9}\ m[/tex]
Distance from screen to slit, D = 88.5 cm = 0.885 m
The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m
We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :
[tex]y=\dfrac{mD\lambda}{a}[/tex]
where
a = width of the slit
[tex]a=\dfrac{mD\lambda}{y}[/tex]
[tex]a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}[/tex]
a = 0.000167 m
[tex]a=1.67\times 10^{-4}\ m[/tex]
a = 0.167 mm
So, the width of the slit is 0.167 mm. Hence, this is the required solution.
What is the average distance between the parabola yequals=77x(1616minus−x) and the x-axis on the interval left bracket 0 comma 16 right bracket[0,16]?
Solution:
To calculate the average distance between the given parabola and the x-axis
y = 77x(1616 - x)
x ∈ [0, 16]
avg distance = [tex]\int_{0}^{16}\frac{77x(1616 - x)dx}{\int_{0}^{16} x dx}[/tex]
= [tex]2\int_{0}^{16}(\frac{(124432x - 77x^{2})dx}{[x^{2}]_{0}^{16}}[/tex]
=[tex] 2\int_{0}^{16}\frac{\frac{124432x^{2}}{2}- \frac{77x^{3}}{3}}{[x^{2}]_{0}^{16}}[/tex]
= [tex]16^{2}[\frac{62216 - 25.67\times 16}{16^{2}}][/tex]
avg distance = 61805 unit
A 4.0-V battery that can store 300.0 J of energy is connected to a resistor. How much electrical charge must flow between the battery's termiñals to completely drain the battery if it is fully charged? A) 1200 C B) 0.05 C C) 0.01 C D) 75 C
Answer:
Electrical charge, q = 75 C
Explanation:
It is given that,
Voltage, V = 4 V
Energy stored, E = 300 J
Energy stored in the battery is given by :
[tex]E=q\times V[/tex]
q is the electrical charge
[tex]q=\dfrac{E}{V}[/tex]
[tex]q=\dfrac{300\ J}{4\ V}[/tex]
q = 75 C
So, the electrical charge of 75 C must flow between the battery's terminals to completely drain the battery if it is fully charged. Hence, this is the required solution.
A 4.80 Kg watermelon is dropped from rest from the roof of an 18.0 m building. Calculate the work done by gravity on the watermelon from the roof to the ground.
Answer:
Work, W = 846.72 Joules
Explanation:
Given that,
Mass of the watermelon, m = 4.8 kg
It is dropped from rest from the roof of 18 m building. We need to find the work done by the gravity on the watermelon from the roof to the ground. It is same as gravitational potential energy i.e.
W = mgh
[tex]W=4.8\ kg\times 9.8\ m/s^2\times 18\ m[/tex]
W = 846.72 Joules
So, the work done by the gravity on the watermelon is 846.72 Joules. Hence, this is the required solution.
An electric teapot has a heating element that operates at 120 V and with a current of 2.00 A. Assuming the water absorbs all the energy delivered to the heating element, calculate the time interval (in s) during which the temperature of 0.891 kg of water rises from room temperature (23.0°C) to the boiling point. (The specific heat f water is 4,186 J/(kg°C).)
Answer:
1196.62 sec
Explanation:
V = electric potential difference at which teapot operates = 120 volts
i = current = 2.00 A
t = time of operation
m = mass of water = 0.891 kg
T₀ = initial temperature = 23.0 °C
T = final temperature = 100 °C
c = specific heat of water = 4186 J/(Kg °C)
Using conservation of energy
V i t = m c (T - T₀)
(120) (2.00) t = (0.891) (4186) (100 - 23.0)
t = 1196.62 sec
An iron container has a mass of 200 g and contains 50 g of water @ 40°C. 50 g of ice @ -6°C are poured. Calculate the equilibrium temperature and describe the final composition.
Answer:
final equilibrium temperature of the system is ZERO degree Celcius
Explanation:
Hear heat given by water + iron = heat absorbed by ice
so here first we will calculate the heat given by water + iron
[tex]Q_1 = m_1s_2\Delta T_1 + m_2 s_2 \Delta T_1[/tex]
[tex]Q_1 = (200)(0.450)(40 - T) + (50)(4.186)(40 - T)[/tex]
now the heat absorbed by ice so that it will melt and come to the final temperature
[tex]Q_2 = m s \Delta T + mL + m s_{water}\Delta T'[/tex]
[tex]Q_2 = 50(2.09)(0 + 6) + 50(335) + 50(4.186)(T - 0)[/tex]
now we will have
[tex]17377 + 209.3T = 3600 - 90T + 8372 - 209.3T[/tex]
[tex]17377 + 209.3T + 90T + 209.3T = 11972[/tex]
[tex]T = -10.6[/tex]
since it is coming out negative which is not possible so here the ice will not completely melt
so final equilibrium temperature of the system is ZERO degree Celcius
Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 kg that is traveling horizontally at 10.9 m/s . Olaf's mass is 70.2 kg. (a) If Olaf catches the ball, with what speed vf do Olaf and the ball move afterward? Express your answer numerically in meters per second. (b) If the ball hits Olaf and bounces off his chest horizontally at 8.10 m/s in the opposite direction, what is his speed vf after the collision? Express your answer numerically in meters per second.
Explanation:
Momentum is conserved.
a) In the first scenario, Olaf and the ball have the same final velocity.
mu = (M + m) v
(0.400 kg) (10.9 m/s) = (70.2 kg + 0.400 kg) v
v = 0.0618 m/s
b) In the second scenario, the ball has a final velocity of 8.10 m/s in the opposite direction.
mu = mv + MV
(0.400 kg) (10.9 m/s) = (0.400 kg) (-8.10 m/s) + (70.2 kg) v
v = 0.108 m/s
a) After Olaf catches the ball, Olaf and the ball will move at a speed of 0.062 m/s.
b) The speed of Olaf after the ball bounces off his chest is 0.11 m/s.
a) We can find the speed of Olaf and the ball by conservation of linear momentum.
[tex] p_{i} = p_{f} [/tex]
[tex] m_{o}v_{i_{o}} + m_{b}v_{i_{b}} = m_{o}v_{f_{o}} + m_{b}v_{f_{b}} [/tex]
Where:
[tex] m_{o}[/tex]: is the mass of Olaf = 70.2 kg
[tex] m_{b}[/tex]: is the mass of the ball = 0.400 kg
[tex] v_{i_{o}}[/tex]: is the intial speed of Olaf = 0 (he is at rest)
[tex]v_{i_{b}}[/tex]: is the initial speed of the ball = 10.9 m/s
[tex] v_{f_{o}}[/tex] and [tex]v_{f_{b}} [/tex]: are the final speed of Olaf and the final speed of the ball, respetively.
Since Olaf catches the ball, we have that [tex] v_{f_{o}} = v_{f_{b}} = v [/tex], so:
[tex] m_{o}v_{i_{o}} + m_{b}v_{i_{b}} = v(m_{o} + m_{b}) [/tex]
We will take the direction of motion of the ball to the right side, and this will be the positive x-direction.
By solving for "v" we have:
[tex] v = \frac{m_{o}v_{i_{o}} + m_{b}v_{i_{b}}}{m_{o} + m_{b}} = \frac{70.2 kg*0 + 0.400 kg*10.9 m/s}{70.2 kg + 0.400 kg} = 0.062 m/s [/tex]
Hence, Olaf and the ball will move at a speed of 0.062 m/s.
b) The final speed of Olaf after the collision can be calculated, again with conservation of linear momentum.
[tex] m_{o}v_{i_{o}} + m_{b}v_{i_{b}} = m_{o}v_{f_{o}} + m_{b}v_{f_{b}} [/tex]
In this case, since the ball hits Olaf and bounces off his chest, we have that [tex] v_{f_{o}} \neq v_{f_{b}}[/tex]
[tex] 0.400 kg*10.9 m/s = 70.2 kg*v_{f_{o}} + 0.400 kg*(-8.10 m/s) [/tex]
The minus sign of the speed of the ball is because it moves to the negative direction of motion after the collision.
[tex] v_{f_{o}} = \frac{0.400 kg*10.9 m/s + 0.400 kg*8.10 m/s}{70.2 kg} = 0.11 m/s [/tex]
Therefore, the speed of Olaf after the collision will be 0.11 m/s in the positive x-direction.
You can see another example of conservation of linear momentum here: https://brainly.com/question/22257322?referrer=searchResults
I hope it helps you!
The mass of mars is 6.38x10^23 kg and its radius is 3.38 x10^6m. Mars rotates on its axis with a period of 1.026 days.(G=6.67x10^-11 Nm^2/kg^2). calculate the orbital speed for a satellite at an altitude of 1.62x10^6 m.
Answer:
v = 2917.35 m/s
Explanation:
let Fc be the centripetal force avting on the satelite , Fg is the gravitational force between mars and the satelite, m is the mass of the satelite and M is the mass of mars.
at any point in the orbit the forces acting on the satelite are balanced such that:
Fc = Fg
mv^2/r = GmM/r^2
v^2 = GM/r
v = \sqrt{GM/r}
= \sqrt{(6.6708×10^-11)(6.38×10^23)/(3.38×10^6 + 1.62×10^6)}
= 2917.35 m/s
Therefore, the orbital velocity of the satelite orbiting mars is 2917.35 m/s.
A proton in a certain particle accelerator has a kinetic energy that is equal to its rest energy. What is the TOTAL energy of the proton as measured by a physicist working with the accelerator? (c = 3.00 × 108 m/s, mproton = 1.67 × 10-27 kg)
Answer:
Total energy, [tex]TE=3.006\times 10^{-10}\ J[/tex]
Explanation:
It is given that, a proton in a certain particle accelerator has a kinetic energy that is equal to its rest energy. Let KE is the kinetic energy of the proton and E₀ is its rest energy. So,
[tex]KE=E_o[/tex]
The total energy of the proton is equal to the sun of kinetic energy and the rest mass energy.
[tex]TE=KE+E_o[/tex]
[tex]TE=2E_o[/tex]
[tex]TE=2m_{proton}c^2[/tex]
[tex]TE=2\times 1.67\times 10^{-27}\ kg\times (3\times 10^8\ m/s)^2[/tex]
[tex]TE=3.006\times 10^{-10}\ J[/tex]
So, the total energy of the proton as measured by a physicist working with the accelerator is [tex]3.006\times 10^{-10}\ J[/tex]
Beats are the difference in (a) Frequency (b) Amplitude (c) Intensity (d) None
Answer:
Beats are the difference in frequency.
(a) is correct option.
Explanation:
Beat :
Beat is the difference of the frequency of two waves.
The difference in frequency is equal to the number of beat per second.
Amplitude :
Amplitude of the wave is the maximum displacement.
Frequency :
Frequency is the number oscillations of wave in per second.
Intensity :
Intensity is the power per unit area.
Hence, Beats are the difference in frequency.
An astronaut aboard the International Space Station, which is orbiting at an altitude of 4.00 x 105 m above the Earth's surface, has a gravitational potential energy of 2.94 x 106 J. What is the weight of the astronaut when he returns to the Earth's surface
Answer:
The weight of the astronaut is 0.4802 N.
Explanation:
Gravitational potential energy, [tex]U=2.94\times 10^6\ J[/tex]
Distance above earth, [tex]d=4\times 10^5\ m[/tex]
The gravitational potential energy is given by :
[tex]U=\dfrac{GMm}{R}[/tex]
G is universal gravitational constant
M is the mass of Earth, [tex]M=5.97\times 10^{24}\ kg[/tex]
m is mass of astronaut
R is the radius of earth, R = R + d
[tex]R=6.37\times 10^6\ m+4\times 10^5\ m=6770000\ m[/tex]
[tex]m=\dfrac{U(R+d)^2}{GM}[/tex]
[tex]m=\dfrac{2.94\times 10^6\ J\times (6770000\ m)}{6.67\times 10^{-11}\times 5.97\times 10^{24}\ kg}[/tex]
m = 0.049 kg
The weight of the astronaut is given by :
W = mg
[tex]W=0.049\ kg\times 9.8\ m/s^2[/tex]
W = 0.4802 N
So, the weight of the astronaut when he returns to the earth surface is 0.4802 N. Hence, this is the required solution.
Final answer:
The weight of the astronaut when he returns to the Earth's surface is approximately 73.32 N.
Explanation:
To calculate the weight of the astronaut when he returns to the Earth's surface, we can use the formula for gravitational potential energy:
PE = mgh
where PE is the gravitational potential energy, m is the mass of the astronaut, g is the acceleration due to gravity, and h is the altitude of the astronaut.
Given that the gravitational potential energy is 2.94 x 10^6 J and the altitude is 4.00 x 10^5 m, we can rearrange the formula to solve for m:
m = PE / (gh)
Substituting the values, we get:
m = (2.94 x 10^6 J) / ((9.80 m/s^2) * (4.00 x 10^5 m))
Calculating this, we find that the mass of the astronaut is approximately 7.49 kg.
Now, to find the weight of the astronaut when he returns to the Earth's surface, we can use the formula:
Weight = mg
Substituting the mass we just calculated, we get:
Weight = (7.49 kg) * (9.80 m/s^2)
Calculating this, we find that the weight of the astronaut when he returns to the Earth's surface is approximately 73.32 N.
A satellite is in a circular orbit around the Earth at an altitude of 3.18x10 m. Find the period and th orbital speed of the satellite? A. T= 2.94 h, B. T= 3.23 h, v 5610 m/s C. T= 1.75 h, v = 5920 m/s D. T 1.12 h, v 4980 m/s E. T 2.58 h, v 6460 m/s
Answer:
112.17 m/s
56.427 years
Explanation:
h = 3.18 x 10^10 m
R = 6.4 x 10^6 m
r = R + h = 3.18064 x 10^10 m
M = 6 x 10^24 kg
The formula for the orbital velocity is given by
[tex]v = \sqrt{\frac{G M }{r}}[/tex]
[tex]v = \sqrt{\frac{6.67 \times 10^{-11}\times 6\times 10^{24} }{3.18064\times 10^{10}}}[/tex]
v = 112.17 m/s
Orbital period, T = 2 x 3.14 x 3.18064 x 10^10 / 112.17
T = 0.178 x 10^10 s
T = 56.427 years
A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the centerline at a point 1.15 m behind the front axle. Find the force exerted by the ground on each wheel.
To find the force exerted by the ground on each wheel of the automobile, we can analyze the forces acting on the car and use the principle of equilibrium. By considering the weight of the car and the distribution of weight between the front and rear axles, we can determine the force exerted by the ground on each wheel.
Explanation:The force exerted by the ground on each wheel of the automobile can be determined by considering the forces acting on the car. Since the car is in equilibrium, the sum of the vertical forces must be zero. The weight of the car is distributed between the front and rear axles according to their distances from the center of mass. Using this information, we can calculate the force exerted by the ground on each wheel.
First, we find the weight of the car by multiplying its mass by the acceleration due to gravity: W = (mass of the car) x (acceleration due to gravity). In this case, the acceleration due to gravity is 9.8 m/s².
Next, we find the force exerted by the ground on the rear wheels. Since the car is in equilibrium, the sum of the torques about any point must be zero. Taking the point where the front wheels contact the ground as the pivot point, we can set up an equation using the weight of the car, the distances between the front and rear axles and the center of mass, and the force exerted by the ground on the rear wheels. Solving this equation allows us to find the force exerted by the ground on the rear wheels, and since the front and rear wheels share the weight of the car equally, the force exerted by the ground on each wheel is half of this value.
Learn more about the force here:brainly.com/question/30526425
#SPJ12
A rectangular key was used in a pulley connected to a line shaft with a power of 7.46 kW at a speed of 1200 rpm. If the shearing stress of the shaft and key are 30 MPa and 240 MPa, respectively what is the length of the rectangular key if the width is one fourth that of the shaft diameter?
Answer:
The length of the rectangular key is 0.4244 m
Explanation:
Given that,
Power = 7.46 kW
Speed = 1200 rpm
Shearing stress of shaft = 30 MPa
Mini shearing stress of key = 240 MPa
We need to calculate the torque
Using formula of power
[tex]P=\dfrac{2\pi NT}{60}[/tex]
Where, P = power
N = number of turns
Put the value into the formula
[tex]7.46\times10^{3}=\dfrac{2\pi\times1200\times T}{60}[/tex]
[tex]T=\dfrac{7.46\times10^{3}\times60}{2\pi\times1200}[/tex]
[tex]T=59.36\ N-m[/tex]
We need to calculate the distance
[tex]\tau_{max}=\dfrac{16T}{\pi d^3}[/tex]
[tex]d^3=\dfrac{16\times59.36}{\pi\times30}[/tex]
[tex]d=(10.077)^{\dfrac{1}{3}}[/tex]
[tex]d=2.159\ m[/tex]
Width of key is one fourth of the shaft diameter
[tex]W=\dfrac{1}{4}\times2.159[/tex]
[tex]W=0.53975\ m[/tex]
The shear stress induced in key
[tex]\tau_{max}=\dfrac{F}{Wl}[/tex]
[tex]\tau_{max}=\dfrac{\dfrac{T}{\dfrac{d}{2}}}{wl}[/tex]
[tex]\tau_{max}=\dfrac{2T}{dWl}[/tex]
[tex]240=\dfrac{2\times59.36}{2.159\times0.53975\times l}[/tex]
[tex]l=\dfrac{2\times59.36}{2.159\times0.53975\times240}[/tex]
[tex]l=0.4244\ m[/tex]
Hence, The length of the rectangular key is 0.4244 m
An electron and a proton are each placed at rest in a uniform electric field of magnitude 554 N/C. Calculate the speed of each particle 52.0 ns after being released. electron m/s proton m/s Need Help?
Explanation:
It is given that,
Electric field, E = 554 N/C
Time, [tex]t=52\ ns=52\times 10^{-9}\ s[/tex]
Electric force, F = qE
For both electron and proton, [tex]F=1.6\times 10^{-19}\ C\times 554\ N/C[/tex]
[tex]F=8.86\times 10^{-17}\ N[/tex]
For electron, [tex]F=m_ea_e[/tex]
[tex]a_e=\dfrac{F}{m_e}[/tex]
[tex]a_e=\dfrac{8.86\times 10^{-17}\ N}{9.1\times 10^{-31}\ kg}[/tex]
[tex]a_e=9.73\times 10^{13}\ m/s^2[/tex]
Using first equation of motion as :
[tex]v=u+at[/tex]
u = 0
[tex]v=9.73\times 10^{13}\ m/s^2\times 52\times 10^{-9}\ s[/tex]
v = 5059600 m/s
or
v = 5.05 × 10⁶ m/s
For proton :
[tex]F=m_pa_p[/tex]
[tex]a_p=\dfrac{F}{m_e}[/tex]
[tex]a_p=\dfrac{8.86\times 10^{-17}\ N}{1.67\times 10^{-27}\ kg}[/tex]
[tex]a_p=5.3\times 10^{10}\ m/s^2[/tex]
Using first equation of motion as :
[tex]v=u+at[/tex]
u = 0
[tex]v=5.3\times 10^{10}\ m/s^2\times 52\times 10^{-9}\ s[/tex]
v = 2756 m/s
Hence, this is the required solution.
An electron moving with a velocity v = 5.0 × 10^7 m/s i enters a region of space where perpendicular electric and a magnetic fields are present. The electric field is E = 104 V/m j. What magnetic field will allow the electron to go through the region without being deflected?
Answer:
2.08 x 10⁻⁶ T
Explanation:
[tex]v[/tex] = velocity of electron = 5.0 x 10⁷ m/s
q = charge on electron
B = magnetic field = ?
E = electric field = 104 V/m
Magnetic force on the electron is given as
[tex]F_{B} = qvB[/tex]
Electric force on the electron is given as
[tex]F_{E} = qE[/tex]
For the electron to pass without being deflected, we must have
[tex]F_{B} = F_{E}[/tex]
[tex]qvB = qE[/tex]
[tex]vB = E[/tex]
(5.0 x 10⁷) B = 104
B = 2.08 x 10⁻⁶ T
A parallel-plate capacitor consists of plates of area 1.5 x 10^-4 m^2 separated by 2.0 mm The capacitor is connected to a 12-V battery. How much energy is stored in the capacitor? (a) 4.8 x 10^-11 J (b) 2.8 X 10^-11 J (c) 1.8 x 10^-11 J d) 4.8 X 10^-12 J
Answer:
4.78 x 10^-11 J
Explanation:
A = 1.5 x 10^-4 m^2
d = 2 mm = 2 x 10^-3 m
V = 12 V
Let C be the capacitance of the capacitor
C = ε0 A / d
C = (8.854 x 10^-12 x 1.5 x 10^-4) / (2 x 10^-3)
C = 6.64 x 10^-13 F
Energy stored, U = 1/2 CV^2
U = 0.5 x 6.64 x 10^-13 x 12 x 12
U = 4.78 x 10^-11 J
The energy stored in a parallel-plate capacitor connected to a 12-V battery, with plate area of 1.5 x 10^-4 m^2 and separation of 2.0 mm, is calculated using the formula U = (1/2)CV^2 and is found to be 4.8 x 10^-11 J.
The capacitance C of a parallel-plate capacitor is given by the formula C = ε_0 * A / d, where ε_0 is the vacuum permittivity (ε_0 = 8.85 x 10^-12 F/m), A is the area of the plates, and d is the separation between the plates.
Given that the area A is 1.5 x 10^-4 m^2, the separation d is 2.0 mm = 2.0 x 10^-3 m, and the voltage V is 12 V, we can plug in these values to first determine capacitance and then calculate the energy stored.
First, calculate the capacitance:
C = ε_0 * A / d = (8.85 x 10^-12 F/m)(1.5 x 10^-4 m^2) / (2.0 x 10^-3 m) = 6.6 x 10^-12 F
Next, calculate the energy stored:
U = (1/2)CV^2 = (1/2)(6.6 x 10^-12 F)(12 V)^2 = 4.8 x 10^-11 J
Therefore, the energy stored in the capacitor is 4.8 x 10^-11 J, which corresponds to option (a).