Answer:
greatest speed=0.99c
least speed=0.283c
Explanation:
To solve this problem, we have to go to frame of center of mass.
Total available energy fo π + and π - mesons will be difference in their rest energy:
[tex]E_{0,K_{0} }-2E_{0,\pi } =497Mev-2*139.5Mev\\[/tex]
=218 Mev
now we have to assume that both meson have same kinetic energy so each will have K=109 Mev from following equation for kinetic energy we have,
K=(γ-1)[tex]E_{0,\pi }[/tex]
[tex]K=E_{0,\pi}(\frac{1}{\sqrt{1-\beta ^{2} } } -1)\\\frac{1}\sqrt{1-\beta ^{2} }}=\frac{K}{E_{0,\pi}}+1\\ {1-\beta ^{2}=\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta ^{2}=1-\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta = +-\sqrt{\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\\\\beta =+-\sqrt{1-\frac{1}{(\frac{109Mev}{139.5Mev+1)^2}}[/tex]
[tex]u'=+-0.283c[/tex]
note +-=±
To find speed least and greatest speed of meson we would use relativistic velocity addition equations:
[tex]u=\frac{u'+v}{1+\frac{v}{c^{2} } } u'\\u_{max} =\frac{u'_{+} +v_{} }{1+\frac{v}{c^{2} } } u'_{+} \\u_{max} =\frac{0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } 0.828\\ u_{max} =0.99c\\u_{min} =\frac{u'_{-} +v_{} }{1+\frac{v}{c^{2} } } u'_{-}\\u_{min} =\frac{-0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } -0.828c\\u_{min} =0.283c[/tex]
Final answer:
The greatest and least speeds of the π mesons resulting from the decay of a high-speed K0 meson are calculated using the relativistic laws of velocity addition. The maximum speed is (85/86)c, and the minimum speed is (5/14)c.
Explanation:
The question revolves around a high-speed K0 meson decaying into two π mesons (π+ and π-). To determine the greatest and least speeds that the π mesons can have, we employ the relativistic addition of velocities. The formula used in this context is Vx=(Vx'+V)/(1+(Vx'V/c²)), where V is the speed of the K0 meson, Vx' the speed of π mesons in the K0 meson's rest frame, and c is the speed of light.
The maximum speed is achieved when the π meson is emitted in the same direction as the K0 meson's movement, while the minimum speed occurs when the π meson is emitted in the opposite direction. Considering the K0 meson's speed (0.9c) and the π mesons' speed in the rest frame (0.8c), the equations result in the maximum speed being (85/86)c and the minimum speed being (5/14)c respectively.
g If you keep the launch angle fixed, but double the initial launch speed, what happens to the range?
Answer:
Range will become 4 times of initial range
Explanation:
Let the velocity of projection is u
And angle at which projectile is projected is [tex]\Theta[/tex]
And acceleration due to gravity is [tex]g\ m/sec^2[/tex]
So range of projectile is equal to [tex]R=\frac{u^2sin2\Theta }{g}[/tex]........eqn 1
Now in second case it is given that velocity of launching is doubled
So new velocity [tex]u_{new}=2u[/tex]
So new range will be equal to [tex]R_{new}=\frac{(2u)^2sin2\Theta }{g}=\frac{4u^2sin2\Theta }{g}[/tex] .....eqn 2
Now dividing eqn 2 by eqn 1
[tex]\frac{R_{new}}{R}=\frac{4u^2sin2\Theta }{g}\times \frac{g}{u^2sin2\Theta }[/tex]
[tex]R_{new}=4R[/tex]
So if we double the initial launch speed then range will become 4 times
Doubling the initial launch speed while keeping the launch angle fixed quadruples the range of a projectile due to the direct proportionality between the range and the square of the initial speed, considering factors like launch angle, gravity, and air resistance.
If you keep the launch angle fixed, but double the initial launch speed, the range of a projectile will increase. When the initial speed is doubled, the range increases four times because the range is directly proportional to the square of the initial speed. This phenomenon is influenced by factors such as launch angle, gravity, and air resistance.
A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across a 2.25 mm length of the wire? VV = nothing VV SubmitRequest Answer Part B What would the potential difference in part A be if the wire were silver instead of copper, but all else was the same?
Answer:
a) 2.063*10^-4
b) 1.75*10^-4
Explanation:
Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:
[tex]A=\frac{\pi }{4}d^{2} \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\[/tex]
a) The Potential difference across a 2.00 in length of a 14-gauge copper
wire:
L= 2.00 m
From Table Copper Resistivity [tex]p[/tex]= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:
[tex]R=\frac{pL}{A}[/tex]
=0.0165Ω
The Potential difference across the copper wire is:
V=IR
=2.063*10^-4
b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m
The Resistance of the Silver wire is:
[tex]R=\frac{pL}{A}[/tex]
=0.014Ω
The Potential difference across the Silver wire is:
V=IR
=1.75*10^-4
What is the self-inductance of a solenoid 30.0 cm long having 100 turns of wire and a cross-sectional area of 1.00 × 10-4 m2? (μ0 = 4π × 10-7 T • m/A)
Answer:
[tex]L=4.19*10^{-6}H[/tex]
Explanation:
The self-inductance of a solenoid is defined as:
[tex]L=\frac{\mu N^2A}{l}[/tex]
Here [tex]\mu_0[/tex] is the the permeability of free space, N is the number of turns in the solenoid, A is the cross-sectional area os the solenoid and l its length. We replace the given values to get the self-inductance:
[tex]L=\frac{4\pi*10^{-7}\frac{T\cdot m}{A}(100)^2(1*10^{-4}m^2)}{30*10^{-2}m}\\L=4.19*10^{-6}H[/tex]
The self-inductance of the given solenoid would be approximately 1.395x10^-5 H. The calculation involved substituting known variables into the formula for self-inductance and solving.
Explanation:The self-inductance of a solenoid can be calculated using the formula: L = µ₀n²A/l where L is the self-inductance, µ₀ is the permeability of free space, n is the number of turns per unit length, A is the cross-sectional area, and l is the length of the solenoid.
In this scenario, the cross-sectional area (A) = 1.00 × 10⁻⁴ m², length of the solenoid (l) = 30.0 cm = 0.3m, and the number of turns (n) = 100, so n (number of turns per unit length) = 100/0.3 = 333.33 turns/m. µ₀, the permeability of free space is a constant which is given as 4π × 10⁻⁷ T m/A. So, substituting these values into the formula will give:
L (self-inductance) = 4π × 10⁻⁷ T m/A * (333.33 /m)² * 1.00 × 10⁻⁴ m² / 0.3 m = 1.395x10^-5 H, (where H indicates henries, the unit for inductance)
Learn more about self-inductance here:https://brainly.com/question/40927173
#SPJ11
A student throws a set of keys vertically upward to her sorority sister, who is in a window 3.50 m above. The second student catches the keys 1.10 s later. (a) With what initial velocity were the keys thrown? magnitude m/s direction (b) What was the velocity of the keys just before they were caught? magnitude m/s direction
Answer:
a) 8.58 m/s upward
b) -2.211 m/s downward
Explanation:
Let gravitational acceleration g = -9.81m/s2. This is negative because it deceleration the upward motion of the key.
(a)We have the following equation of motion
[tex]s = v_0t + gt^2/2[/tex]
where [tex]v_0[/tex] is the initial upward velocity of the keys, t = 1.1s is the time it takes for the keys to travel a distance of s = 3.5 m
[tex]3.5 = v_01.1 - 9.81*1.1^2/2[/tex]
[tex]3.5 = 1.1v_0 - 5.94 [/tex]
[tex]1.1v_0 = 3.5 + 5.94 = 9.44[/tex]
[tex]v_0 = 9.44 / 1.1 = 8.58 m/s[/tex]
So the keys were thrown initially upward with a speed of 8.58 m/s
(b) If the initial velocity of the key is 8.58 m/s and it is subjected to a deceleration of 9.81m/s2 for 1.1s then the velocity right at the 1.1s instant is
[tex]v = v_0 + gt = 8.58 - 9.81*1.1 = -2.211 m/s[/tex]
So they keys would have a downward speed of 2.211 m/s
A point charge is placed at the center of a spherical Gaussian surface. The electricflux ΦEischangedif(a) a second point charge is placed outside the sphere(b) the point charge is moved outside the sphere(c) the point charge is moved off center, but still inside the original sphere(d) the sphere is replaced by a cube of one-tenth the volume (the original charge remains in thecenter)
Answer:
(b) the point charge is moved outside the sphere
Explanation:
Gauss' Law states that the electric flux of a closed surface is equal to the enclosed charge divided by permittivity of the medium.
[tex]\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}[/tex]
According to this law, any charge outside the surface has no effect at all. Therefore (a) is not correct.
If the point charge is moved off the center, the points on the surface close to the charge will have higher flux and the points further away from the charge will have lesser flux. But as a result, the total flux will not change, because the enclosed charge is the same.
Therefore, (c) and (d) is not correct, because the enclosed charge is unchanged.
A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled 6.00 mm to the bottom of the incline is 3.80 m/s.What is the speed of the block when it is 3.00 m from the top of the incline?
Answer:
Explanation:
Given
Speed of block at bottom is [tex]v=3.8\ m/s[/tex]
Distance traveled [tex]s=6\ m[/tex]
initial velocity is zero
using equation of motion
[tex]v^2-u^2=2as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](3.8)^2-0=2\times a\times 6[/tex]
[tex]a=1.203\ m/s^2[/tex]
when it is 3 m from top then
[tex]v^2-u^2=2as[/tex]
[tex]v^2-0=2\times 1.203\times 3[/tex]
[tex]v=2.68\ m/s[/tex]
The problem is a physics question on kinematics, which requires calculating the speed of a block half-way down a frictionless incline using the kinematic equation for constant acceleration.
Explanation:The student's question involves finding the speed of a block at a certain point as it slides down a frictionless incline. The kinematics of motion on an inclined plane is the focus here. We are given the speed of the block after traveling 6.00 mm and we need to calculate its speed after it has traveled 3.00 m down the incline.
To solve this, we can use the kinematic equation for constant acceleration, which is stated as:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Since the block starts from rest, u = 0. We can find the acceleration by using the given final speed and distance from the first part of the trip (6.00 mm to the bottom). We substitute the acceleration into the kinematic equation again using s as 3.00 m to find the speed at that point.
The electric field near the surface of Earth points downward and has a magnitude of 152 N/C. What is the ratio of the magnitude of the upward electric force on an electron to the magnitude of gravitational force on the electron?
Answer:
[tex]2.7\times 10^{12}[/tex]
Explanation:
We are given that
Electric field =[tex]E=152N/C[/tex]
We have to find the ratio of magnitude of the upward electric force on an electron to the magnitude of gravitational force on the electron.
We know that
F=qE
Charge on an electron,q=[tex]1.6\times 10^{-19}C[/tex]
Using the formula
Upward electric force=[tex]152\times 1.6\times 10^{-19}[/tex] N
Upward electric force=[tex]F_e=2.43\times 10^{-17}[/tex] N
Mass of electron=[tex]m_e=9.1\times 10^{-31} kg[/tex]
[tex]g=9.8m/s^2[/tex]
Gravitational force=[tex]F=mg[/tex]
Using the formula
Gravitational force=[tex]F_g=9.1\times 10^{-31}\times 9.8=8.92\times 10^{-30} N[/tex]
Ratio of Fe to the Fg=[tex]\frac{2.43\times 10^{-17}}{8.92\times 10^{-30}}[/tex]
[tex]\frac{F_e}{F_g}=2.7\times 10^{12}[/tex]
A 1.75 µF capacitor and a 6.00 µF capacitor are connected in series across a 3.00 V battery. How much charge (in µC) is stored on each capacitor?
Answer:
Explanation:
Given
First capacitor magnitude [tex]C=1.75\ \mu F[/tex]
Second capacitor magnitude [tex]C=6.00\ \mu F[/tex]
Voltage of battery [tex]V=3.00\ V[/tex]
Both capacitor are connected in series so net capacitor is given by
[tex]\frac{1}{C_{net}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}[/tex]
[tex]\frac{1}{C_{net}}=\frac{1}{1.75}+\frac{1}{6}[/tex]
[tex]C_{net}=\frac{6\times 1.75}{1.75+6}[/tex]
[tex]C_{net}=1.35\ \mu F[/tex]
So charge Across each capacitor is given by
[tex]Q=C_{net}\times V[/tex]
[tex]Q=1.35\times 10^{-6}\times 3[/tex]
[tex]Q=4.064\ \mu C[/tex]
A 230mg piece of gold is hammered into a sheet measuring 23cm x 17cm. What is the thickness of the sheet in meters? Density of gold is 19.32g/cm3
Answer:
t= 0.0003 mm
Explanation:
Given that
mass ,m = 230 mg
m = 0.23 g
Area ,A= 23 x 17 cm²
A= 391 cm²
Density ,ρ = 19.32 g/cm³
Lets take thinness of the sheet = t cm
We know that
Mass = Density x Volume
m = ρ A t
Now by putting the values in the above equation we get
0.23 = 19.32 x 391 x t
[tex]t=\dfrac{0.23}{19.32\times 391}\ cm[/tex]
t=0.000030 cm
t= 0.0003 mm
That is why the thickness of the sheet will be 0.0003 mm.
During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive force of 115 N to the tire's rim. The mass of the wheel is 1.70 kg and, for the purpose of this problem, assume that all of this mass is concentrated on the outside radius of the wheel. The diameter of the wheel is 50.0 cm. A chain passes over a sprocket that has a diameter of 9.50 cm. In order for the wheel to have an angular acceleration of 4.90 rad/s2, what force, in Newtons, must be applied to the chain? (Enter the magnitude only.) N
Answer:
616.223684211 N
Explanation:
[tex]F_r[/tex] = Resistive force on the wheel = 115 N
F = Force acting on sprocket
[tex]r_2[/tex] = Radius of sprocket = 4.75 cm
[tex]r_1[/tex] = Radius of wheel = 25 cm
Moment of inertia is given by
[tex]I=mr_1^2\\\Rightarrow I=1.7\times 0.25^2\\\Rightarrow I=0.10625\ kgm^2[/tex]
Torque
[tex]\tau=I\alpha\\\Rightarrow \tau=0.10625\times 4.9\\\Rightarrow \tau=0.520625\ Nm[/tex]
Torque is given by
[tex]\tau=Fr_2-F_rr_1\\\Rightarrow F=\dfrac{\tau+F_rr_1}{r_2}\\\Rightarrow F=\dfrac{0.520625+115\times 0.25}{0.0475}\\\Rightarrow F=616.223684211\ N[/tex]
The force on the chain is 616.223684211 N
A 6 kg bowling ball moves with a speed of 3 m/s. How fast does a 7 kg bowling ball need to move so that it has the same kinetic energy?
Answer: 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy.
Explanation:
Kinetic energy is the energy possessed by a body by virtue of its motion.
[tex]K.E=\frac{1}{2}mv^2[/tex]
m = mass of object
v= velocity of the object
[tex]K.E=\frac{1}{2}\times 6kg\times (3m/s)^2=27Joules[/tex]
b) for a 7 kg bowl to have kinetic energy of 27 Joules:
[tex]27J=\frac{1}{2}\times 7kg\times v^2[/tex]
[tex]v^2=7.7[/tex]
[tex]v=2.8m/s[/tex]
Thus 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy
falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.040 s. The average force exerted on him by the ground is 18 000 N, where the upward direction is taken to be the positive direction. From what height did the student fall
Answer:
y = 2,645 10⁴ / m²
m=80 kg, y = 4.13 m
Explanation:
We must solve this problem in two parts, one when it is in free fall and another for the collision with the floor
Let's start by analyzing the crash with the floor,
Initial instant When it arrives but if you start to stop
p₀ = m v
Final moment. When he stopped
[tex]p_{f}[/tex] = 0
The momentum is related to the moment by
I = Δp = p_{f} –p₀
F t = 0 - mv
v = -F t / m
Let's calculate
v = -18000 0.040 / m
v = -720 / m
The sign indicates that the speed goes down
Now we use energy conservation at two points
Lowest point. Just before crashing
Em₀ = K = ½ m v²
Highest point. From where it began to fall
Em_{f} = U = m g y
Energy is conserved in the fall
Em₀ = Em_{f}
½ m v² = m g y
y = ½ v² / g
y = ½ (720 / m)² /9.8
y = 2,645 10⁴ / m²
For an explicit height value, the object's mass must be known, suppose the masses are m = 80 kg
y = 2,645 10⁴/80²
y = 4.13 m
When a tractor moves with uniform velocity its heavier wheel rotates slowly than its lighter wheel why?
Answer:
Because of Moment of inertia.
Explanation:
Larger wheels have larger moment of inertia,when they rotate rotational energy is stored in spinnning of motion so they rotate slowly while lighter wheels move faster comparitively because they use smaller moment of inertia.The thing is to keeping tyre in contact with road,when a vehicla hits a
jolt it is difficult with heavy tyre to be in contact with road. When vehicle loses contact the driver will lose steering control which result in giving a way to sliding friction.
The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by various atmospheric processes, including lightning. What is the excess charge on the surface of the earth?
Answer:
The excess charge on earth's surface was calculated to be 4.56 × 10⁵ C
Explanation:
Using the formula for an electric field;
E = kQ/r²
k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²
E = 100N/C
r = radius of the earth = 6400 km = 6400000m
Q = Er²/k = 100 × (6400000)²/(8.99 × 10⁹)
Q = 455617.4 C = 4.56 × 10⁵ C
Hope this helps!!!
The excess charge on the surface of the earth is 4.55×10⁵C
To calculate the charge on the surface of the earth, we apply the equation of electric field strength, that is:
[tex]E = k\frac{Q}{r^2}[/tex]
here, E is the electric field strength = 100N/C (given)
k = 9×10⁹ Nm²/C²
Q is the charge, and
r = distance = radius of earth = 6.4×10⁶ m
Now,
[tex]Q = \frac{r^2E}{k}\\ \\=\frac{(6.4*10^6)^2*100}{9*10^9}\\\\Q=4.55*10^5 C[/tex] is the charge acquired on the surface of the earth.
Learn more:
https://brainly.com/question/9774180?referrer=searchResults
A 2-ft3 tank contains a gas at 2 atm(g) and 60 oF. This tank is connected to a second tank containing the same gas at atmospheric pressure and 60 oF. The two tanks are connected and allowed to reach equilibrium. The final conditions are measured to be 1 atm(g) and 60oF. What is the volume of the second tank
Answer : The volume of the second tank is, [tex]4ft^3[/tex]
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex]
or,
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 2 atm
[tex]P_2[/tex] = final pressure of gas = 1 atm
[tex]V_1[/tex] = initial volume of gas = [tex]2ft^3[/tex]
[tex]V_2[/tex] = final volume of gas = ?
Now put all the given values in the above equation, we get:
[tex]2atm\times 2ft^3=1atm\times V_2[/tex]
[tex]V_2=4ft^3[/tex]
Therefore, the volume of the second tank is, [tex]4ft^3[/tex]
The period of a carrier wave is T=0.005 seconds. Determine the frequency and wavelength of the carrier wave.
Answer:
f = 200[Hz]; L=1500[km]
Explanation:
We know that frequency is the reciprocal of the period, therefore.
[tex]f=\frac{1}{T} \\f=1/0.005\\f=200[Hz][/tex]
And the wavelength is determined using the following equation.
[tex]L=\frac{c}{f} \\where:\\c = light speed = 3*10^{8}[m/s]\\ f = frecuency [Hz]\\L = wavelength [m]\\L= 3*10^{8}/ 200\\ L = 1500000[m] = 1500[km][/tex]
Final answer:
The frequency of a carrier wave with a period of 0.005 seconds is calculated to be 200 Hz. The wavelength cannot be determined without additional information on the wave's speed.
Explanation:
The question involves calculating the frequency and wavelength of a carrier wave given its period (T=0.005 seconds). The frequency (f) of a wave is the reciprocal of its period, defined by the equation f = 1/T. Therefore, for a period of 0.005 seconds, the frequency would be f = 1/0.005 Hz, which equals 200 Hz. To determine the wavelength (λ), we would need the speed (v) of the wave, which is typically the speed of light (c) for electromagnetic waves, but since the speed and specific type of wave are not provided, we can't calculate the wavelength directly from the given information here.
a. How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter to produce an electric field of 1350 N/C just outside the surface of the sphere.b. What is the electric field at a point 10.0 cm outside the surface of the sphere?
Answer:
Explanation:
The electric field outside the sphere is given as,
E = k Q /r²
here Q = n x 1.6 x 10⁻¹⁹ C
where n is the number of electons
if the dimeter of sphere d= 25 cm= 0.25 m
then the radius r = 0.125 m
we get
n= E r²/ k x 1.6 x 10⁻¹⁹ C
n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)
n = 14664731646
A very large sheet of a conductor carries a uniform charge density of 4.00 pC/mm2 on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Answer:
451977.40113 N/C
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
[tex]\sigma[/tex] = Surface charge density = [tex]4\ pc/mm^2[/tex]
Electric field near the surface of a charged conductor is given by
[tex]E=\dfrac{\sigma}{\epsilon_0}\\\Rightarrow E=\dfrac{4\times 10^{-12}\times 10^{6}}{8.85\times 10^{-12}}\\\Rightarrow E=451977.40113\ N/C[/tex]
The electric field is 451977.40113 N/C
If the separation between the first and the second minima of a single-slit diffraction pattern is 6.0 mm, what is the distance between the screen and the slit?
Incomplete question.The complete question is here
If the separation between the first and the second minima of a single-slit diffraction pattern is 6.0 mm, what is the distance between the screen and the slit? The light wavelength is 500 nm and the slit width is 0.16 mm.
Answer:
The distance between first and second minimum is 1.92m
Explanation:
Given data
Wavelength of light λ=500nm
Slit width D=0.16 mm
The distance between first and second minima is 6.0 mm
To find
Distance L between the screen and the slit
Solution
As we know that
Yn=(nλL)/D
Where is L is distance between the slit and screen
D is slit width
n is order of minimum
Yn is distance of nth minimum from center maximum
λ is wavelength of light
The Distance between first and second minimum is given as:
Y₂-Y₁=6 mm
(2λL)/D-(1λL)/D=6 mm
(2λL-1λL)/D=6 mm
(λL/D)=6 mm
[tex]L=\frac{6mm(0.16mm)}{500nm}\\ L=1.92m[/tex]
The distance between first and second minimum is 1.92m
Final answer:
The screen distance is 0.19 meters, calculated using the separation between minima, slit width, and light wavelength in single-slit diffraction.
Explanation:
Solve for the distance between the screen and the slit (L).
1. Given Values:
Separation between minima (Δx) = 6.0 mm (convert to meters: 0.0060 m)
Wavelength of light (λ) = 500 nm (convert to meters: 500 x 10⁻⁹ m)
Slit width (a) = 0.16 mm (convert to meters: 0.16 x 10⁻³ m)
2. Relate Separation and Screen Distance:
We can use the relationship between the angular separation (Δθ) and the linear separation (Δx) on the screen:
Δθ ≈ Δx / L
3. Relate Separation and Minimum Position:
We know the separation (Δθ) is related to the difference between the positions of the first (m = 1) and second (m = 2) minima:
Δθ = θ₂ - θ₁ = (λ / a)
4. Combine Equations:
Substitute the equation for Δθ from step 3 into the equation from step 2:
(λ / a) ≈ Δx / L
5. Solve for Screen Distance (L):
Now we can arrange the equation to solve for L:
L ≈ Δx * a / λ
6. Plug in the Values:
L ≈ (0.0060 m) * (0.16 x 10⁻³ m) / (500 x 10⁻⁹ m)
7. Calculate and Round:
L ≈ 0.192 m (round to two significant figures)
Therefore, the distance between the screen and the slit is approximately 0.19 meters.
A 2-m telescope can collect a given amount of light in 1 hour. Under the same observing conditions, how much time would be required for a 6-m telescope to perform the same task? A 12-m telescope?
Answer
given,
diameter of telescope = 2 m
time to collect area = 1 hour
Diameter of the another telescope = 6 m
The light collected by the telescope is directly proportional to the area of its primary mirror.
Area of the mirror is directly proportional to the square of diameter.
so,
6 m diameter telescope will carry (6/2)² = 9 times more light than 2 m telescope.
Time taken to collect light = 60/9 = 6.67 minutes.
now, For 12 m telescope
12 m diameter telescope will carry (12/2)² = 36 times more time than 2 m telescope.
Time taken by 12 m telescope to collect light= 60/36 = 1.7 minutes.
To find the time required for a 6-m telescope and a 12-m telescope to perform the same task as a 2-m telescope, we can use a formula and rearrange it to solve for the time taken.
Explanation:To find out how much time would be required for a 6-m telescope and a 12-m telescope to perform the same task as a 2-m telescope, we can use the formula:
(light collected by telescope 1) × (time taken by telescope 1) = (light collected by telescope 2) × (time taken by telescope 2)
Let's solve for the time taken by the 6-m telescope and the 12-m telescope:
For the 6-m telescope: (light collected by 2-m telescope) × (1 hour) = (light collected by 6-m telescope) × (time taken by 6-m telescope)For the 12-m telescope: (light collected by 2-m telescope) × (1 hour) = (light collected by 12-m telescope) × (time taken by 12-m telescope)By rearranging the equation and solving for the time, we can find the answer.
Learn more about Time required for larger telescopes here:https://brainly.com/question/32814773
#SPJ3
In the SI system of units, dynamic viscosity of water μ at temperature T (K) can be computed from μ=A10B/(T-C), where A=2.4×10-5, B=250 K and C=140 K. (6 points) (a) Determine the dimensions of A. (b) Determine the kinematic viscosity of water at 20°C. Express your results in both SI and BG units.
Answer:
0.00000103149529075 m²/s
0.0103149529075 stokes
Explanation:
A = [tex]2.4\times 10^-5[/tex]
B = 250 K
C = 140 K
T = 20°C
[tex]\rho[/tex] = Density of water = 998 kg/m³
Viscosity is given by
[tex]\mu=A\times 10^{\dfrac{B}{T-C}}\\\Rightarrow \mu=2.4\times 10^{-5}\times 10^{\dfrac{250}{273.15+20-140}}\\\Rightarrow \mu=0.00102943230017\ Pas[/tex]
Kinematic viscosity is given by
[tex]\nu=\dfrac{\mu}{\rho}\\\Rightarrow \nu=\dfrac{0.00102943230017}{998}\\\Rightarrow \nu=0.00000103149529075\ m^2/s[/tex]
The kinematic viscosity is 0.00000103149529075 m²/s
In BG units [tex]0.00000103149529075\times 10^4=0.0103149529075\ stokes[/tex]
In the SI system of units, the dynamic viscosity of water can be computed using the equation μ=A10B/(T-C). The dimensions of A are (m⋅s²)/kg. To find the kinematic viscosity of water at 20°C, we can use the formula ν=μ/ρ, where μ is the dynamic viscosity and ρ is the density of water. The kinematic viscosity of water at 20°C is approximately 0.0001 m²/s in SI units and 1 cm²/s in BG units.
Explanation:(a) Dimensions of A:
To determine the dimensions of A, we need to rearrange the equation and analyze the units on both sides. We know that viscosity has units of kilograms per meter per second (kg/m/s). Plugging in the given values for B and C, we have μ=A10B/(T-C). The dimensions of 10B/(T-C) are (dimensionless). Therefore, the dimensions of A must be kg/m/s multiplied by the inverse of the dimensions of 10B/(T-C), which is 1/(kg/m/s) or (m⋅s²)/kg.
(b) Kinematic viscosity at 20°C:
To find the kinematic viscosity of water at 20°C, we can use the formula ν=μ/ρ, where μ is the dynamic viscosity and ρ is the density of water. The density of water at 20°C is approximately 998 kg/m³. Plugging in the given values for A and C, we can calculate μ using the equation μ=A10B/(T-C). Substituting the values into the formula ν=μ/ρ, we get ν≈μ/ρ≈(A10B/(T-C))/ρ.
In the SI unit, ν≈(2.4×10-5×10250 K/(293 K-140 K))/998 kg/m³≈0.0001 m²/s.
In the BG unit, ν≈0.0001 m²/s×104 cm²/1 m²≈1 cm²/s.
You are on the ground and observing a jet traveling in air at a constant height of 4500 m with a speed of 650 m/s. How long after the spaceship has passed directly overhead will the shock wave reach you
Answer:
[tex]t=11.1s[/tex]
Explanation:
Given data
Jet Speed v=650 m/s=1.89504 mach
height h=4500 m
Speed of the Sound V=343 m/s
To find
time t
Solution
As we know that
[tex]Vs_{Velocity}=d_{distance} /t_{time} \\d=Vs*t[/tex]
where
[tex]Vs=1.895*V[/tex]
Also from trigonometric properties we know that
[tex]tan\alpha =\frac{Perpendicular}{Base}[/tex]
We have use height h as perpendicular and distance d is base
So
[tex]tan\alpha =h/d\\tan\alpha=h*(1/Vs*t)\\t=\frac{h}{tan\alpha } \frac{1}{Vs}\\[/tex]
First we need to angle α
Since
[tex]Sin\alpha =V/Vs[/tex]
[tex]Sin\alpha =\frac{v}{1.895v}\\\alpha =32^{o}[/tex]
Substitute given values and angle to find
[tex]t=\frac{h}{tan\alpha } \frac{1}{Vs}\\t=\frac{4500m}{tan(32) } \frac{1}{1.895*343m/s}\\t=11.1s[/tex]
Final answer:
To calculate the time for a shock wave to reach an observer on the ground, divide the altitude of the jet by the speed of sound. For a jet at 4500 m and a sound speed of 343 m/s, it will take approximately 13.12 seconds for the shock wave to reach the ground.
Explanation:
The question pertains to the time it will take for the shock wave from a jet traveling at a constant speed and altitude to reach an observer on the ground.
To calculate this, one has to consider the speed of sound and the altitude of the jet from the ground. Using Pythagoras' theorem, one can determine the distance the sound has to travel to reach the observer. Then divide the distance by the speed of sound to find the time for the shock wave to reach the ground.
Let's denote the speed of sound as v and the altitude of the jet as h. The time t it takes for the shock wave to reach the observer can be calculated using the formula:
t = h / v
As given, the jet is flying at an altitude of 4500 m (4.5 km) and the speed of sound is generally taken to be 343 m/s. Plugging these values into the formula:
t = 4500 m / 343 m/s
t approx 13.12 seconds
Therefore, it will take approximately 13.12 seconds for the shock wave to reach the observer on the ground.
Calculate the speed of an electron (in m/s) after it accelerates from rest through a potential difference of 160 V.
Answer:
v = 7.5*10⁶ m/s
Explanation:
While accelerating through a potential difference of 160 V, the electron undergoes a change in the electric potential energy, as follows:
ΔUe = q*ΔV = (-e)*ΔV = (-1.6*10⁻¹⁹ C) * 160 V = -2.56*10⁻¹⁷ J (1)
Due to the principle of conservation of energy, in absence of non-conservative forces, this change in potential energy must be equal to the change in kinetic energy, ΔK:
ΔK = Kf -K₀
As the electron accelerates from rest, K₀ =0.
⇒ΔK =Kf = [tex]\frac{1}{2}*me*vf^{2}[/tex] (2)
From (1) and (2):
ΔK = -ΔUe = 2.56*10⁻¹⁷ J = [tex]\frac{1}{2}*me*vf^{2}[/tex]
where me = mass of the electron = 9.1*10⁻³¹ kg.
Solving for vf:
[tex]vf =\sqrt{\frac{2*(2.56e-17J)}{9.1e-31kg} } =7.5e6 m/s[/tex]
⇒ vf = 7.5*10⁶ m/s
Final answer:
The speed of an electron after accelerating from rest through a potential difference of 160 V can be calculated using the conservation of energy principle, resulting in a velocity of approximately 5.93 × 10^6 m/s.
Explanation:
To calculate the speed of an electron after it accelerates from rest through a potential difference, we use the concept of conservation of energy where the electrical potential energy converted into kinetic energy is expressed as qAV = ½mv². Here, q is the charge of the electron (-1.60 × 10^-19 C), V is the potential difference (160 V), m is the mass of the electron (9.11 × 10^-13 kg), and v is the velocity of the electron we need to find. Rearranging the equation to solve for v and substituting the given values gives:
v = sqrt(2 × q × V / m) = √(2 × (-1.60 × 10^-19 C) × 160 V / (9.11 × 10^-13kg))
Performing the calculation yields a velocity of approximately 5.93 × 10^6 m/s.
The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s. A seismograph records the arrival of the transverse waves 56.4 s after that of the longitudinal waves. How far away was the earthquake? Answer in units of km.
Answer:
[tex]d=691.71km[/tex]
Explanation:
The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:
[tex]t=\frac{d}{v_t}-\frac{d}{v_l}[/tex]
Here d is the distance at which the earthquake take place and [tex]v_t, v_l[/tex] is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:
[tex]t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km[/tex]
A hydrogen atom has a single proton at its center and a single electron at a distance of approximately 0.0539 nm from the proton. a. What is the electric potential energy in joules?b. What is the significance of the sign of the answer?
To solve this problem we will apply the concepts related to electrostatic energy, defined as,
[tex]U_e = \frac{kq_1q_2}{d^2}[/tex]
Here,
k = Coulomb's constant
[tex]q_{1,2}[/tex] = Charge of each object (electron and proton, at this case the same)=
d = Distance
Replacing with our values we have that
[tex]U_e = \frac{(9*10^9Nm^2)(1.6*10^{-19}C)(-1.6*10^{-19}C)}{0.0539*10^{-9}m}[/tex]
[tex]U_e = -4.27*10^{-18}J[/tex]
Therefore for the Part A the answer is [tex]-4.27*10^{-18}J[/tex] and por the Part B the sign indicates that the force between the proton and electron is attractive
A point charge gives rise to an electric field with magnitude 2 N/C at a distance of 4 m. If the distance is increased to 20 m, then what will be the new magnitude of the electric field?
Answer:
0.08 N/C
Explanation:
Electric Field: This is defined as the force per unit charge exerted at a point. The expression for electric field is given as,
E = Kq/r².............................. Equation 1
Where E = Electric Field, q = Charge, k = proportionality constant, r = distance.
making q the subject of the equation,
q = Er²/k............................... Equation 2
Given: E = 2 N/C, r = 4 m,
Substitute into equation 2
q = 2(4)²/k
q = 32/k C.
When r is increased to 20 m,
E = k(32/k)/20²
E = 32/400
E = 0.08 N/C.
Hence the electric Field = 0.08 N/C
The initial electric field from a charge is 2 N/C at 4m distance. When the distance increases fivefold, the strength of the electric field decreases by the square of this factor, resulting in a new electric field magnitude of 0.08 N/C.
Explanation:This question is about the relationship between an electric field and its distance from a point charge. According to the formula for the magnitude of the electric field generated by a point charge, which is E = kQ / r^2, where E is the electric field strength, k is Coulomb's constant, Q is the charge, and r is the distance from the charge, the electric field is inversely proportional to the square of the distance.
At a distance of 4m, the electric field's magnitude is 2 N/C. If you increase the distance to 20m (which is 5 times more than the initial distance), the new electric field magnitude will be the initial magnitude divided by the square of this factor, i.e., 2 N/C divided by 5^2 = 2 N/C / 25, which equals to 0.08 N/C.
Learn more about Electric Field here:https://brainly.com/question/33547143
#SPJ3
An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed of the plane (its speed in still air) is 320.0 km/h (about 200 mi/h), in which direction should the pilot head? (b) What is the speed of the plane over the ground? Draw a vector diagram.
Answer:
a) 75.5 degree relative to the North in north-west direction
b) 309.84 km/h
Explanation:
a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it
So the pilot should head to the West-North direction at an angle of
[tex]cos(\alpha) = 80/320 = 0.25[/tex]
[tex]\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0[/tex] relative to the North-bound.
b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is
[tex]320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h[/tex]
Answer:
a) Ф=14°, north of west
b) 310 km/h
Explanation:
we have,
[tex]v_{p/G}[/tex], velocity of plane relative to the ground(west)
[tex]v_{p/A}[/tex],velocity of plane relative to the air(320 km/h)
[tex]v_{A/G}[/tex],velocity of air relative to the ground(80 km/h, due south).
[tex]v_{p/G}[/tex]=[tex]v_{p/A}[/tex]+[tex]v_{A/G}[/tex].........(1)
a) sin(Ф)=[tex]\frac{v_{A/G}}{v_{p/A}}[/tex]
=[tex]\frac{80km/h}{320km/h}[/tex]
Ф=14°, north of west
b) using Pythagorean theorem
[tex]v_{p/G}=\sqrt{v^2_{p/A}+v^2_{A/G}}[/tex]
[tex]v_{p/G}[/tex]=310 km/h
note:
diagram is attached
You throw a rock upward. The rock is moving upward. but it is slowing down. If we define the ground as the origin, theposition of the rock is ____ and the velocity of the rock is ____ .A. positive, positiveB. positive, negativeC. negative, positiveD. negative. negative
Answer:
.A. positive, positive.
Explanation:
When we throw a rock upward , it will decelerate due to gravitation . It will have acceleration in downward direction or - ve acceleration in upward direction.
If we define the ground as the origin and upward direction as positive , anything in upward direction will be positive and in downward direction will be negative . For example , in the case described above , acceleration of rock thrown upward is negative because is in downward direction .
Its position is in upward direction , so its position is positive with respect to ground.
It is going in upward direction . So velocity too will be positive.
A sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz. What is the maximum speed of the needle?
Answer:
v = 0.2035 m/s
Explanation:
given,
Amplitude,A = 0.0127 m
Frequency, f = 2.55 Hz
maximum speed of the needle = ?
we know,
x = A cos ω t
velocity of the motion
v = - A ω sin ω t
for maximum speed
sin ω t = 1
v = - A ω
ω = 2 π f
now,
v = A 2 π f
v = 0.0127 x 2 π x 2.55
v = 0.2035 m/s
Hence, the maximum speed of the needle is equal to 0.2035 m/s
Nuclear reactors use fuel rods to heat water and generate steam. Is this process endothermic or exothermic?
Explanation:
Exothermic reaction is defined as the reaction in which release of heat takes place. This also means that in an exothermic reaction, bond energies of reactants is less than the bond energies of products.
Hence, difference between the energies between the reactants and products releases as heat and therefore, enthalpy of the system will decrease.
Whereas in an endothermic reaction, heat is supplied from outside and absorbed by the reactant molecules. Hence, enthalpy of the system increases.
As water acts as a coolent and when fuel rods in a nuclear reactor are immersed in it then heat created by coolent is absorbed by water and then it changes into steam.
Since, absorption of heat occurs in the nuclear reactor. Therefore, it is an endothermic reaction.
Thus, we can conclude that nuclear reactors use fuel rods to heat water and generate steam. This process is endothermic.
Answer:
The heat produced by the nuclear reactor is an exothermic process while the heat absorbed by the water to convert into steam is an endothermic process.
Explanation:
Nuclear reactor being the heat of a nuclear power plant uses the radioactive uranium fuel to generate the heat by the process of nuclear fission in a controlled manner.
The processing of uranium is carried out into small ceramic pellets which are stacked together into sealed metal tubes known as fuel rods.
Usually more than 200 such rods are bunched together leading to the formation of a fuel assembly.
The core of the reactor is often made up of a couple hundred assemblies, according to its power level.
Inside the reactor vessel, these fuel rods are immersed into water which serve as both a coolant and moderator. The moderator helps slow down the neutrons produced by fission to sustain the chain reaction.