Answer:
He becomes a croco-dile food
Explanation:
From the question we are told that
The height is h = 2.0 m
The angle is [tex]\theta = 20^o[/tex]
The distance is [tex]w = 10m[/tex]
The speed is [tex]u = 11 m/s[/tex]
The coefficient of static friction is [tex]\mu = 0.02[/tex]
At equilibrium the forces acting on the motorcycle are mathematically represented as
[tex]ma = mgsin \theta + F_f[/tex]
where [tex]F_f[/tex] is the frictional force mathematically represented as
[tex]F_f =\mu F_x =\mu mgcos \theta[/tex]
where [tex]F_x[/tex] is the horizontal component of the force
substituting into the equation
[tex]ma = mgsin \theta + \mu mg cos \theta[/tex]
[tex]ma =mg (sin \theta + \mu cos \theta )[/tex]
making a the subject of the formula
[tex]a = g(sin \theta = \mu cos \theta )[/tex]
substituting values
[tex]a = 9.8 (sin(20) + (0.02 ) cos (20 ))[/tex]
[tex]= 3.54 m/s^2[/tex]
Applying " SOHCAHTOA" rule we mathematically evaluate that length of the ramp as
[tex]sin \theta = \frac{h}{l}[/tex]
making [tex]l[/tex] the subject
[tex]l = \frac{h}{sin \theta }[/tex]
substituting values
[tex]l = \frac{2}{sin (20)}[/tex]
[tex]l = 5.85m[/tex]
Apply Newton equation of motion we can mathematically evaluate the final velocity at the end of the ramp as
[tex]v^2 =u^2 + 2 (-a)l[/tex]
The negative a means it is moving against gravity
substituting values
[tex]v^2 = (11)^2 - 2(3.54) (5.85)[/tex]
[tex]v= \sqrt{79.582}[/tex]
[tex]= 8.92m/s[/tex]
The initial velocity at the beginning of the pool (end of ramp) is composed of two component which is
Initial velocity along the x-axis which is mathematically evaluated as
[tex]v_x = vcos 20^o[/tex]
substituting values
[tex]v_x = 8.92 * cos (20)[/tex]
[tex]= 8.38 m/s[/tex]
Initial velocity along the y-axis which is mathematically evaluated as
[tex]v_y = vsin\theta[/tex]
substituting values
[tex]v_y = 8.90 sin (20)[/tex]
[tex]= 3.05 m/s[/tex]
Now the motion through the pool in the vertical direction can mathematically modeled as
[tex]y = y_o + u_yt + \frac{1}{2} a_y t^2[/tex]
where [tex]y_o[/tex] is the initial height,
[tex]u_y[/tex] is the initial velocity in the y-axis
[tex]a_y[/tex] is the initial acceleration in the y axis with a constant value of ([tex]g = 9.8 m/s^2[/tex])
at the y= 0 which is when the height above ground is zero
Substituting values
[tex]0 = 2 + (3.05)t - 0.5 (9.8)t^2[/tex]
The negative sign is because the acceleration is moving against the motion
[tex]-(4.9)t^2 + (2.79)t + 2m = 0[/tex]
Solving using quadratic formula
[tex]\frac{-b \pm \sqrt{b^2 -4ac} }{2a}[/tex]
substituting values
[tex]\frac{-3.05 \pm \sqrt{(3.05)^2 - 4(-4.9) * 2} }{2 *( -4.9)}[/tex]
[tex]t = \frac{-3.05 + 6.9}{-9.8} \ or t = \frac{-3.05 - 6.9}{-9.8}[/tex]
[tex]t = -0.39s \ or \ t = 1.02s[/tex]
since in this case time cannot be negative
[tex]t = 1.02s[/tex]
At this time the position the motorcycle along the x-axis is mathematically evaluated as
[tex]x = u_x t[/tex]
x [tex]=8.38 *1.02[/tex]
[tex]x =8.54m[/tex]
So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool
If the horizontal distance x covered by the motorcyclist is greater than or equal to 10 meters, he survives; otherwise, he does not, From the given calculations, x = 8.21 m, so he becomes crocodile food.
According to question, the height of motorcycle daredevil is h = 2.0 m, the angle of the ramp is θ = 20°, the distance of the wide pool is w = 10 m, the speed of the motorcycle daredevil is u = 11 m/s, and the coefficient of static friction between the ramp and the tyres of the motorcycle daredevil is μ = 0.02.
At equilibrium, the forces acting on the motorcycle are equated as:
ma = mgsinθ + f
here, m = mass of the motorcycle, a is the acceleration of the motorcycle, g is the acceelration due to gravity (g = 9.8 m/s²), and f is the frictional force between the ramp and the tyres of the motorcycle daredevil mathematically represented using normal reaction N as
f =μ N = μ mgcosθ
substituting into the equation
ma = mg sinθ + μ mg cosθ
ma =mg (sinθ + μ cosθ)
⇒ a = g(sinθ + μ cosθ)
⇒ a = 9.8 (sin(20°) + (0.02) cos (20°))
⇒ a = 3.54 m/s²
Now, for the ramp, we can say that:
[tex]sin \theta = \frac{h}{l}[/tex]
[tex]\therefore l = \frac{h}{sin \theta}[/tex]
substituting the above values, we get:
[tex]l = \frac{2}{sin20^{\circ}}[/tex]
or, l = 5.85m
Using Newton equation of motion, the final velocity of the motorcycle at the end of the ramp, can be given as:
[tex]v^2 =u^2 + 2 (-a)l[/tex]
As the motorcycle is moving against gravity, the acceleration (a) is taken negative. From the above values, we get:
[tex]v^2 = (11)^2 - 2(3.54) (5.85)[/tex]
⇒ [tex]v= \sqrt{79.582} \hspace{0.8 mm} m/s[/tex]
or, v = 8.92m/s
The motion of the motorcycle is a projectile motion. The initial velocity at the beginning of the pool (end of ramp) is composed of two component which are the x, and the y components respectively.
Initial velocity along the x-axis which is [tex]v_x[/tex] = v cos 20°
substituting values, we get:
[tex]v_x[/tex] = 8.92 cos (20°)
or, v = 8.38 m/s
Initial velocity along the y-axis which [tex]v_y[/tex] = v sin 20°
substituting values, we get:
[tex]v_y[/tex] = 8.90 sin (20)
[tex]v_y[/tex] = 3.05 m/s
Now the motion through the pool in the vertical direction can mathematically modeled as
[tex]y = y_o + u_yt + \frac{1}{2}a_y t^2[/tex]
where y₀ is the initial height, [tex]u_y[/tex] is the initial velocity in the y-axis, and [tex]a_y[/tex] is the initial acceleration in the y-axis with a constant value of (g = 9.8 m/s^2)
At y = 0
0 = 2 + (3.05) t - 0.5 (9.8) t²
The negative sign is because the acceleration is moving against the motion:
- (4.9) t² + (2.79) t + 2m = 0
To solve for t, we can use the quadratic formula:
[tex]t = \frac{- b \hspace{0.5} \pm \sqrt{(b^2 - 4ac)}}{2a}[/tex]
In this case, a = 4.9, b = -2.79, and c = -2.
Plugging in these values, we get:
[tex]t = \frac{(2.79 \pm \sqrt{(-2.79)^2 - 4(4.9)(-2)}}{2(4.9)}[/tex]
Solving for both possible values of t, we get:
t ≈ [tex]\frac{(2.79 + 6.858)}{9.8}[/tex] ≈ 0.98 seconds
t ≈ [tex]\frac{(2.79 - 6.858)}{9.8}[/tex] ≈ -0.41 seconds
Since time cannot be negative,
∴ t ≈ 0.98 seconds
Hence displacement along the x-coordinate will be:
x = [tex]u_x[/tex] t
⇒ x =8.38 × 0.98 m
x = 8.21 m
So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool.
A light source shines light consisting of two wavelengths, λ1 = 540 nm (green) and λ2 = 450 nm (blue), on two slits separated by 0.180 mm. The two overlapping interference patterns, one from each wavelength, are observed on a screen 1.53 m from the slits. What is the minimum distance (in cm) from the center of the screen to a point where a bright fringe of the green light coincides with a bright fringe of the blue light?
Answer:
The maximun distance is [tex]z_1 = z_2 = 0.0138m[/tex]
Explanation:
From the question we are told that
The wavelength are [tex]\lambda _ 1 = 540nm (green) = 540 *10^{-9}m[/tex]
[tex]\lambda_2 = 450nm(blue) = 450 *10^{-9}m[/tex]
The distance of seperation of the two slit is [tex]d = 0.180mm = 0.180 *10^{-3}m[/tex]
The distance from the screen is [tex]D = 1.53m[/tex]
Generally the distance of the bright fringe to the center of the screen is mathematically represented as
[tex]z = \frac{m \lambda D}{d}[/tex]
Where m is the order of the fringe
For the first wavelength we have
[tex]z_1 = \frac{m_1 (549 *10^{-9} * (1.53))}{0.180*10^{-3}}[/tex]
[tex]z_1=0.00459m_1 m[/tex]
[tex]z_1= 4.6*10^{-3}m_1 m ----(1)[/tex]
For the second wavelength we have
[tex]z_2 = m_2 \frac{450*10^{-9} * 1.53 }{0.180*10^{-3}}[/tex]
[tex]z_2 = 0.003825m_2[/tex]
[tex]z_2 = 3.825 *10^{-3} m_2 m[/tex] ----(2)
From the question we are told that the two sides coincides with one another so
[tex]zy_1 =z_2[/tex]
[tex]4.6*10^{-3}m_1 m = 3.825 *10^{-3} m_2 m[/tex]
[tex]\frac{m_1}{m_2} = \frac{3.825 *10^{-3}}{4.6*10^{-3}}[/tex]
Hence for this equation to be solved
[tex]m_1 = 3[/tex]
and [tex]m_2 = 4[/tex]
Substituting this into the equation
[tex]z_1 = z_2 = 3 * 4.6*10^{-3} = 4* 3.825*10^{-3}[/tex]
Hence [tex]z_1 = z_2 = 0.0138m[/tex]
The minimum distance from the center of the screen where green and blue bright fringes coincide is calculated using principles of double-slit interference. Applying these principles and trigonometry, we find the minimum distance from the center of the screen to this point.
Explanation:To find the minimum distance from the center of the screen to a point where a bright fringe of the green light coincides with a bright fringe of the blue light, we can use the formula for double slit interference:
d sin θ = mλ
where d is the separation of the slits, θ is the angle from the central bright fringe, m is the order of fringe, and λ is the wavelength of light.
For the green and blue lights to interfere constructively on the screen for the first time, their path length difference must be equal to their respective wavelengths. Therefore, the order of fringe for the green light will be different from the blue light. The minimum distance can be found when the fringe order for the green light (m1) is 1 and for the blue light (m2) is, by proportion of the wavelengths, rounded to the nearest whole number. Solving the equation for each colour and setting them equal will give the minimum distance.
Finally, the distance from the center to the fringe can be calculated using the trigonometric relationship:
L tan θ, where L is the distance from the slits to the screen.
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Two astronauts are 1.40 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via electromagnetic waves. The time it takes for sound waves to travel at 328 m/s through the air between the astronauts equals the time it takes for the electromagnetic waves to travel to the earth. How far away from the earth is the spaceship?
Answer:
The spaceship is placed at a distance of [tex]1.28\times 10^6\[/tex] meters from the Earth.
Explanation:
Distance between two astronauts, d = 1.4 m
The time it takes for sound waves to travel at 328 m/s through the air between the astronauts equals the time it takes for the electromagnetic waves to travel to the earth.
For sound, time taken is given by :
[tex]t=\dfrac{1.4}{328}\ ........(1)[/tex]
For electromagnetic wave, time taken to travel the Earth is :
[tex]t=\dfrac{d}{3\times 10^8}\ ........(2)[/tex]
d is the distance from the Earth to the spaceship.
As time are same :
[tex]\dfrac{d}{3\times 10^8}=\dfrac{1.4}{328}[/tex]
[tex]d=\dfrac{1.4}{328}\times 3\times 10^8\\\\d=1.28\times 10^6\ m[/tex]
So, the spaceship is placed at a distance of [tex]1.28\times 10^6[/tex] meters from the Earth.
Waves from two slits are in phase at the slits and travel to a distant screen to produce the second minimum of the interference pattern. The difference in the distance traveled by the wave is: Group of answer choices three wavelengths three halves of a wavelength half a wavelength a wavelength two wavelengths
Answer:
three halves of a wavelength
Explanation:because a
path difference between waves at minimum is (2n+1)*wavelength /2 therefore at second minimum it is for n=1=1.5 times a wavelength. or three halves a wavelength
The path difference in the distance traveled by the wave is three halves of a wavelength. Therefore, option (B) is correct.
What is interference?Interference can be defined as a phenomenon that takes place in every place. It is a phenomenon in which two waves superpose to create the resultant wave of the higher, lower, or same amplitude.
There are two kinds of Interference which are constructive interference and destructive interference. In constructive interference, the crest of one wave falls on the crest of another wave, and the amplitude increases. In destructive interference, the crest of one wave falls on the trough of another wave, and the amplitude decreases.
For the minima, the path difference is given by (2n -1)/λ.
Here the value of n is given, n = 2
Path difference, = (2×2 - 1)/λ = (3/2)λ
Therefore, the path difference in the distance is (3/2)λ.
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A transverse wave produced near one end of an extremely long vibrating string is described by the equation below. (Ignore wave reflections from the other end of the string, which is extremely far ahead.) The linear density of this string is 0.0073 kg/m. What must be the tension applied to this string?
Answer:
The tension that must be applied to this string = 477 N
Explanation:
y(x,t)= 0.0321m sin (2.05x-524t + pi/4)
Comparing this to the general wave equation
y(x,t) = A sin (kx - wt + Φ)
where
A = amplitude of the wave = 0.0321 m
k = 2.05 /m
w = 524 /s
Φ = phase angle = pi/4
Velocity of a wave is given by
v = (w/k) = (524/2.05) = 255.61 m/s
Tension in the string is then related to the velocity of wave produced and the linear density through
v = √(T/μ)
v = velocity = 255.61 m/s
T = ?
μ = 0.0073 kg/m
v² = (T/μ)
T = v²μ = (255.61² × 0.0073) = 476.96 N
T = 477 N
Hope this Helps!!!
Alligators and other reptiles don't use enough metabolic energy to keep their body temperatures constant. They cool off at night and must warm up in the sun in the morning. Suppose a 300 kg kg alligator with an early-morning body temperature of 25 ∘ C ∘C is absorbing radiation from the sun at a rate of 1200 W How long will the alligator need to warm up to a more favorable 30 C?
Answer:
Explanation:
Amount of heat required can be found from the following relation
Q = mcΔT
m is mass of the body , c is specific heat and ΔTis rise in temperature .
Here m = 300 kg
c = 3350 J /kg k
ΔT = 30 - 25
= 5 °C
Putting the values in the expression above
Q = 300 x 3350 x 5
= 5025000 J
Rate at which energy is absorbed = 1200 J /s
Time required
= 5025000 / 1200
= 4187.5 S
= 69.8 minute
= 1 hour 9.8 mimutes.
A 500 pF capacitor is charged up so that it has 10μC of charge on its plates. The capacitor is then quicklyconnected to a 10 H inductor. Calculate themaximum energystored in the magnetic field of the inductoras the circuit oscillates. What is the current through the inductor when the maximum energy stored in theinductor is reached?
Answer:
Explanation:
Given that,
A capacitor of capacitance
C = 500pF
Charge on capacitor is
Q = 10μC
Capacitor is then connected to an inductor of inductance 10H
L = 10H
Since we want to calculate the maximum energy stored by the inductor, then, we will assume all the energy from the capacitor is transfer to the inductor
So energy stored in capacitor can be determined by using
U = ½CV²
Then, Q = CV
Therefore V = Q/C
U = ½ C • (Q/C)² = ½ C × Q²/C²
U = ½Q² / C
Then,
U = ½ × (10 × 10^-6)² / (500 × 10^-9)
U = 1 × 10^-4 J
U = 0.1 mJ
So the energy stored in this capacitor is transfers to the inductor.
So, energy stored in the inductor is 0.1mJ
B. Current through the inductor
Energy in the inductor is given as
U = ½Li²
1 × 10^-4 = ½ × 10 × i²
1 × 10^-4 = 5× i²
i² = 1 × 10^-4 / 5
i² = 2 × 10^-5
I = √(2×10^-5)
I = 4.47 × 10^-3 Amps
Then,
I = 4.47 mA
A sanding disk with rotational inertia 2.0 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnitude 11 N·m about the central axis of the disk. About that axis and with torque applied for 19 ms, what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?
Answer
Given,
Rotational inertia = 2.0 x 10-3 kg·m²
Torque = 11 N.m
time, t = 19 ms
a) Angular momentum
[tex]\tau = \dfrac{\Delta L}{\Delta t}[/tex]
L is angular momentum
[tex]\Delta L = \tau \Delta t[/tex]
[tex]\Delta L = 11\times 19 \times 10^{-3}[/tex]
[tex]\Delta L = 0.209\ Kg m^2/s[/tex]
b) Angular velocity
We know.
[tex]L = I \omega[/tex]
[tex]\omega = \dfrac{L}{I}[/tex]
[tex]\omega = \dfrac{0.209}{2\times 10^{-3}}[/tex]
[tex]\omega = 104.5\ rad/s[/tex]
A student sits on a rotating stool holding two 3.0-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg⋅m^2 and is assumed to be constant. The student then pulls in the objects horizontally to 0.30 m from the rotation axis. Find the kinetic energy of the student before and after the objects are pulled in.
Answer:
2.86J
Explanation:
M = 3.0kg
R₁ = 1.0m
R₂ = 0.3m
I₁ = I(mass) + I(student + stool)
I₁ = 2mr₁² + I(student + stool)
I₁ = 2*(3*1²) + 3.0
I₁ = 9.0kgm²
the initial moment of inertia of the system = 9.0kgm²
I₂ = 2mr₂² + I(student + stool)
I₂ = 2*(3 * 0.3²) + 3.0
I₂ = 0.54 + 3.0
I₂ = 3.54kgm²
the final moment of inertia of the system is 3.54kgm²
From conservation of angular momentum
I₁ω₁ = I₂ω₂
ω₂ = (I₁ * ω₁) / I₂
ω₂ = (0.75 * 9) / 3.54
ω₂ = 1.09rad/s
kinetic energy of rotation (k.e) =½ Iω²
K.E = (K.E)₂ - (K.E)₁
k.e = [½ * 3.54 * (1.90)²] - [½ * 9.0 * 0.75²]
K.E = 6.3897 - 2.53125
K.E = 2.85845
K.E = 2.86J
Emilie Sannom drives a rocket powered car to an initial speed of 228 mph(102 m/s) before attempting to make a jump of deadlizard canyon on a ramp angled at 42 degrees. Assume that the rocket has burned out when she leaves the ramp. If the canyon is 420 m across and both sides are level, will she be able to make the jump? If so, how far from the edge will she land?
Answer:
Explanation:
Range= U²sin2theta/g
= 102²* sin (2*42)/9.8
= 1054.74m
Yes she will be abable to jump
The distance from the edge will be
(1054.74- 420)m
= 634.74m
5/137 Under the action of its stern and starboard bow thrusters, the cruise ship has the velocity vB = 1 m/s of its mass center B and angular velocity ω = 1 deg/s about a vertical axis. The velocity of B is constant, but the angular rate ω is decreasing at 0.5 deg/s2. Person A is stationary on the dock. What velocity and acceleration of A are observed by a passenger fixed to and rotating with the ship? Treat the problem as two‐dimensional.
The image is missing, so i have attached it;
Answer:
A) V_rel = [-(2.711)i - (0.2588)j] m/s
B) a_rel = (0.8637i + 0.0642j) m/s²
Explanation:
We are given;
the cruise ship velocity; V_b = 1 m/s
Angular velocity; ω = 1 deg/s = 1° × π/180 rad = 0.01745 rad/s
Angular acceleration;α = -0.5 deg/s² = 0.5 x π/180 rad = -0.008727 rad/s²
Now, let's write Velocity (V_a) at A in terms of the velocity at B(V_b) with r_ba being the position vector from B to A and relative velocity (V_rel)
Thus,
V_a = V_b + (ω•r_ba) + V_rel
Now, V_a = 0. Thus;
0 = V_b + (ω•r_ba) + V_rel
V_rel = -V_b - (ω•r_ba)
From the image and plugging in relevant values, we have;
V_rel = -1[(cos15)i + (sin15)j] - (0.01745k * -100j)
V_rel = - (cos15)i - (sin15)j - 1.745i
Note that; k x j = - i
V_rel = [-(2.711)i - (0.2588)j] m/s
B) Let's write the acceleration at A with respect to B in terms of a_b.
Thus,
a_a = a_b + (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel
a_a and a_b = 0.
Thus;
0 = (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel
a_rel = - (α*r_ba) - (ω(ω•r_ba)) - (2ω*v_rel)
Plugging in the relevant values with their respective position vectors, we have;
a_rel = - (-0.008727k * -100j) - (0.01745k(0.01745k * -100j)) - (2*0.01745k * [-(2.711)i - (0.2588)j])
a_rel = 0.8727i - (0.01745² x 100)j + 0.0946j - 0.009i
Note that; k x j = - i and k x i = j
Thus,simplifying further ;
a_rel = 0.8637i + 0.0642j m/s²
Final answer:
From the ship's frame of reference, person A has a velocity of -1 m/s (opposite to the ship's motion). The acceleration observed would be the negative ship's angular deceleration converted to linear acceleration, needing more data for calculation.
Explanation:
The problem presents a rotational kinematics and relative motion scenario common in physics. Person A is stationary on the dock while the ship moves with velocity vB = 1 m/s and has an angular velocity of ω = 1 deg/s, decreasing at 0.5 deg/s2. From the ship's reference frame, the velocity of person A will appear as the opposite of the ship's velocity, which is -1 m/s. However, since person A is stationary, A's acceleration as observed from the ship would be the opposite of the ship's angular deceleration translated into linear acceleration at the radius person A is perceived to be from the ship's center of rotation, which requires further information to calculate.
A 113 ‑turn circular coil of radius 2.71 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is connected to a 10.3 Ω resistor to create a closed circuit. During a time interval of 0.153 s, the magnetic field strength decreases uniformly from 0.441 T to zero. Find the energy, in millijoules, that is dissipated in the resistor during this time interval.
Answer:
357 mJ
Explanation:
Parameters given:
Number of turns, N = 113
Radius, r = 2.71 cm = 0.0271 m
Resistance connected to coil, R = 10.3 ohms
Time interval, dt = 0.153 s
Change in magnetic field, dB = 0 - 0.441 T = - 0.441 T
First, we need to find the induced EMF in the coil. It is given as:
V = - (N * A * dB)/dt
A = area of coil = pi * r² = 3.142 * 0.0271² = 0.0023 m²
Therefore:
V = - (113 * 0.0023 * - 0.441) / 0.153
V = 0.75 V
Electrical energy dissipated in the resistor is given as:
E = V² / (R * t)
Where t is the time (0.153 secs)
E = (0.75²) / (10.3 * 0.153)
E = 0.357 J = 357 mJ
The energy dissipated in the resistor is 357 mJ
A star ending its life with a mass of four to eight times the Sun's mass is expected to collapse and then undergo a supernova event. In the remnant that is not carried away by the supernova explosion, protons and electrons combine to form a neutron star with approximately twice the mass of the Sun. Such a star can be thought of as a gigantic atomic nucleus. Assume r-aA1/3. If a star of mass 3.88 x 1030 kg is composed entirely of neutrons (mn 1.67 x 1027 kg), what would its radius be?
Answer:
r = 16 Km
Explanation:
given
m_n= 1.67 x 10^-27 Kg
M_star = 3.88 x 10^30 Kg
A= M_star/m_n
A= 3.88*10^30/1.67 x 10^-27
A=2.28 *10^57 neutrons A = The number of neutrons
we use the number of neutrons as a mass number because the star has only neutrons. = 1.2 x 10-15 m
r = r_o*A^1/3
r = 1.2*10^-15*2.28 *10^57^1/3
r = 16 Km
(hrw8c10p78) Two uniform solid spheres have the same mass, 1.65 kg, but one has a radius of 0.206 m while the other has a radius of 0.804 m. For each of the spheres, find the torque required to bring the sphere from rest to an angular velocity of 367 rad/s in 14.5 s. Each sphere rotates about an axis through its center. Torque on sphere with the smaller radius.
Complete question is:
Two uniform solid spheres have the same mass of 1.65 kg, but one has a radius of 0.206 m and the other has a radius of 0.804 m. Each can rotate about an axis through its center. (a) What is the magnitude τ of the torque required to bring the smaller sphere from rest to an angular speed of 367 rad/s in 14.5 s? (b) What is the magnitude F of the force that must be applied tangentially at the sphere’s equator to give that torque? What are the corresponding values of (c) τ and (d) F for the larger sphere?
Answer:
A) τ = 0.709 N.m
B) F = 3.44 N
C) τ = 10.8 N.m
D) F = 13.43N
Explanation:
We are given;
Mass if each sphere = 1.65kg
Radius of the first sphere; r1 = 0.206m
Radius of second sphere; r2 = 0.804m
A) initial angular speed of smaller sphere; ω_i = 0 rad/s
Final angular speed of smaller sphere; ω_f = 367 rad/s
Time;t = 14.5 s
The constant angular acceleration is calculated from;
ω_f = ω_i + αt
367 = 0 + α(14.5)
Thus,
α = 367/14.5 = 25.31 rad/s²
The torque is given by the formula;
τ = Iα
Where τ is torque ; I is moment of inertia given as (2/5)Mr²
α is angular acceleration
Thus;
τ = (2/5)(1.65)(0.206)² x 25.31
τ = 0.709 N.m
B) The magnitude of the force that must be applied to give the torque τ is gotten from the formula;
τ = F•r•sin90°
0.709 = F x 0.206 x 1
F = 0.709/0.206
F = 3.44 N
C) Now for the larger sphere, we'll repeat the same procedure in a above. Thus;
The torque is given by the formula;
τ = Iα
Where τ is torque ; I is moment of inertia given as (2/5)Mr²
α is angular acceleration
Thus;
τ = (2/5)(1.65)(0.804)² x 25.31
τ = 10.8 N.m
D) Now for the larger sphere, we'll repeat the same procedure in b above. Thus, τ is gotten from the formula;
τ = F•r•sin90°
10.8 = F x 0.804 x 1
F = 10.8/0.804
F = 13.43N
An astronaut working with many tools some distance away from a spacecraft is stranded when the "maneuvering unit" malfunctions. How can the astronaut return to the spacecraft by sacrificing some of the tools? (Note: the maneuvering unit is connected to the astronaut's spacesuit and is not removable)
Answer:
He can return to the spacecraft by sacrificing some of the tools employing the principle of conservation of momentum.
Explanation:
By carefully evaluating his direction back to the ship, the astronaut can throw some of his tools in the opposite direction to that. On throwing those tools of a certain mass, they travel at a certain velocity giving him velocity in the form of recoil in the opposite direction of the velocity of the tools. This is same as a gun and bullet recoil momentum conservation. It is also the principle on which the operational principles of their maneuvering unit is designed.
An astronaut stranded far from their spacecraft can return by using the principle of Newton's third law of motion. By throwing tools in the opposite direction of the spacecraft, the astronaut can create a reaction force that propels them back to the spacecraft.
Explanation:In this scenario, the astronaut working with many tools some distance away from their spacecraft can use the concept of Newton's third law of motion, also known as action-reaction principle, to return to the spacecraft. According to this law, for every action, there is an equal and opposite reaction. So, the astronaut can return to the spacecraft by sacrificing some of the tools that they are carrying, and throwing the tools in the opposite direction of the spacecraft. When the tools are thrown away, the reaction force will propel the astronaut towards the spacecraft. This is a effective method of maneuvering in space because in the vacuum of space, even small forces can have a significant impact due to the absence of air resistance or friction. This way, the astronaut can cleverly convert a threat into an opportunity and solve their problem with minimal risks.
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Which of the following decreases the dissolving rate of a solid in water?
Answer:
Using larger pieces of solids
Using larger pieces of solid the dissolving rate of a solid in water can be decreased. Therefore, the correct option is option C.
What is solution?A solution is indeed a specific kind of homogenous combination made up of two or even more components that is used in chemistry. A solute is a material that has been dissolved in a solvent, which is the other substance in the combination.
The solvent particles will pull the solute particles apart surround them if the attractive forces here between solute and solvent particles are stronger than the attractive forces keeping the solute particles together. Using larger pieces of solid the dissolving rate of a solid in water can be decreased.
Therefore, the correct option is option C.
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Your question is incomplete but most probably your full question was,
Which of the following decreases the dissolving rate of a solid in water?
A. raising the temperature
B. stirring constantly
C. using larger pieces of solid
D. crushing the solid
A partially evacuated vertical cylindrical container is covered by a circular lid that makes an airtight seal. The pressure in the room is 1.01 x 105 Pa and the pressure inside the container is 0.41 x 105 Pa. What other two quantities would you need to know in order to calculate the minimum upward applied force required to lift the lid? Select two answers.
The pressure difference (∆P) would be the atmospheric pressure in the room (1.01 x 10⁵ Pa) minus the pressure inside the container (0.41 x 10⁵ Pa).
To calculate the minimum upward applied force required to lift the lid of a partially evacuated vertical cylindrical container, you would need to know two additional quantities: the radius of the lid, and consequently the surface area of the lid.
The force required to lift the lid can be calculated using the equation F = P x A, where F is the force, P is the difference in pressure (external minus internal), and A is the surface area of the lid. Recall that the area of the circular lid would be calculated using the formula A = πr², where r is the radius of the circular lid.
In this case, the pressure difference (∆P) would be the atmospheric pressure in the room (1.01 x 10⁵ Pa) minus the pressure inside the container (0.41 x 10⁵ Pa).
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An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 60.0 cm long is at a location 4.00 mi from the transmitter. Compute the amplitude of the emf that is induced by this signal between the ends of the receiving antenna.
To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.
The intensity of the wave at the receiver is
[tex]I = \frac{P_{avg}}{A}[/tex]
[tex]I = \frac{P_{avg}}{4\pi r^2}[/tex]
[tex]I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m[/tex]
[tex]I = 6.529*10^{-6}W/m^2[/tex]
The amplitude of electric field at the receiver is
[tex]I = \frac{E_{max}^2}{2\mu_0 c}[/tex]
[tex]E_{max}= \sqrt{2I\mu_0 c}[/tex]
The amplitude of induced emf by this signal between the ends of the receiving antenna is
[tex]\epsilon_{max} = E_{max} d[/tex]
[tex]\epsilon_{max} = \sqrt{2I \mu_0 cd}[/tex]
Here,
I = Current
[tex]\mu_0[/tex] = Permeability at free space
c = Light speed
d = Distance
Replacing,
[tex]\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}[/tex]
[tex]\epsilon_{max} = 0.05434V[/tex]
Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V
Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A. Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A. The angular velocity of A is twice that of B. The angular velocity of A is half that of B. The angular velocity of A equals that of B.
Answer:
The angular velocity of A equals that of B.
Explanation:
Since both coins are on the same turn table that is rotating with a given angular velocity about a particular axis, then the angular velocity of coin A will be equal to the angular velocity of coin B about that same axis.
The metal zinc has free electron concentration per unit volume, nv, of 1.31029/m3 and an electron mobility me of 8104 m2/V.s. The charge carried by an electron, e, is 1.61019 coulomb. Based on this information, what is the electrical conductivity of zinc? Handbooks list the measured resistivity of zinc as 5.9 mU.cm. Is this consistent with your calculation? (Watch the units.)
Answer:
1.7*10^{12}Ucm^-1
Explanation:
The answer to this question is obtained by using the following formula:
[tex]\sigma=nq\mu_e[/tex]
sigma: conductivity
q: charge of the electron = 1.61019*10^{-19}C
mu_e: electron mobility = 8104 m^2/Vs
n: free electron concentration = 1.31029/m^3
By replacing you get:
[tex]\sigma=(1.31*10^{29}/m^3)(1.61*10^{-19}C)(8104 m2/Vs)=1.709*10^{14}\Omega^{-1}m^{-1} \\\\1.709*10^{14}\Omega^{-1} (1*10^{2}cm})^{-1}=1.7*10^{12}\Omega^{-1}cm^{-1}[/tex]
the result obained is not 5.9mU.cm. This is because temperature effects has not taken into account.
An 800 N box is pushed across a level floor for a distance of 5.0 m with a force of 400 N. How much work was done on this box.
Answer: 2000 J.
Explanation: Since work is force*displacement, we just have to multiply the force by the distance: w = f*d = 400 N*5.0 m = 2000 J.
Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the right as positive and forces pointing to the left as negative. Rank in increasing order by placing the most negative on the left and the most positive on the right. To rank items as equivalent, overlap them.
The electric force between charges can be determined using Coulomb's law. Positive-positive and negative-negative combinations have repulsive forces, while positive-negative combinations have attractive forces. Neutral charges do not exert any force.
The electric force between two charges can be determined using Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Using this information, we can rank the six combinations of electric charges based on the electric force acting on q1:
Combination 1: Positive charge (+) and positive charge (+) - repulsive forceCombination 2: Negative charge (-) and negative charge (-) - repulsive forceCombination 3: Positive charge (+) and negative charge (-) - attractive forceCombination 4: Negative charge (-) and positive charge (+) - attractive forceCombination 5: Positive charge (+) and neutral charge (0) - no forceCombination 6: Negative charge (-) and neutral charge (0) - no forceLearn more about Electric force here:
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Explain what causes spring and neap tides, and identify
the phases of the moon when these tides occur. Be sure
to mention what happens to Earth's crust, ocean water,
and the positions of the Sun, Earth, and the moon.
this is the sample Answer: Spring tides occur when the moon is full or new. Earth, the moon, and the Sun are in a line. The moon’s gravity and the Sun’s gravity pull Earth’s crust and ocean water. This causes tides to be higher than normal. At neap tide, the moon and the Sun are at right angles to each other. This happens during the first and third quarters of the lunar cycle. At neap tide, the Sun’s gravity and the moon’s gravity are balanced. High tides are lower; low tides are higher.
Explanation:
just did the assament
Answer:
Spring tides occur when the moon is full or new. Earth, the moon, and the Sun are in a line. The moon’s gravity and the Sun’s gravity pull Earth’s crust and ocean water. This causes tides to be higher than normal. At neap tide, the moon and the Sun are at right angles to each other. This happens during the first and third quarters of the lunar cycle. At neap tide, the Sun’s gravity and the moon’s gravity are balanced. High tides are lower; low tides are higher.
Explanation:
Just did the quiz.
Consider 500 g of silver (atomic mass 107 g/mol). Assume that the temperature is high enough that equipartition applies. We heat the silver by immersing it in water; by measuring the temperature change of the water we compute that 300 J of heat energy entered the silver.
What was the change in temperature of the silver?
A. 2.57 K
B. 0.6 K
C. 4.99 K
D. 169 K
E. 0.0466 K
Answer:
A. 2.57 K
Explanation:
From specific heat capacity,
Q = cmΔT........................ Equation 1
Where Q = Amount of heat that entered into silver, m = mass of silver, c = specific heat capacity of silver, ΔT = change in temperature of the silver.
make ΔT the subject of the equation
ΔT = Q/cm................... Equation 2
Given: Q = 300 J, m = 500 g = 0.5 kg
Constant: c = 233 J/kg.K
Substitute into equation 2
ΔT = 300/(0.5×233)
ΔT = 300/116.5
ΔT = 2.57 K
Hence the right option is A. 2.57 K
The change in the temperature of the silver should be considered as the 2.57 K.
Change in temperature:
From specific heat capacity,
Q = cmΔT........................ Equation 1
Here
Q = Amount of heat that entered into silver,
m = mass of silver,
c = specific heat capacity of silver,
ΔT = change in temperature of the silver.
Now
ΔT = Q/cm................... Equation 2
Since
Q = 300 J, m = 500 g = 0.5 kg
Constant: c = 233 J/kg.K
So,
ΔT = 300/(0.5×233)
ΔT = 300/116.5
ΔT = 2.57 K
hence, The change in the temperature of the silver should be considered as the 2.57 K.
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The sun, like all stars, releases energy through nuclear fusion. In this problem, you will find the total number of fusion reaction events that occur inside the sun every second. You will be considering the proton-proton chain, in which four hydrogen nuclei are converted into a helium nucleus and two positrons. The net reaction for the proton-proton chain is
411H? 42He+2?+ .
To find the energy released by this reaction, you will need the following mass data:
mass of 11H=1.007825u
and
mass of 42He=4.002603u .
Using the masses of the neutral atoms in your calculation accounts for the energy released by the annihilation of the positrons with electrons, so you can work this problem without reference to the positrons or their rest mass.
a. What is the total energy Q released in a single fusion reaction event for the equation given in the problem introduction?
Answer:
Explanation:
DetaM=4 x 1.02875 - 4.002603
DetaM= 0.028697u
Using E= mc²
= 0.028697 x 1.49x*10^-10
= 4.2x10^-12J
A truck with a mass of 1400 kg and moving with a speed of 14.0 m/s rear-ends a 641-kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision.
After collision, truck remains at 14.0 m/s, car stops (0 m/s) due to elastic collision.
To find the velocities of both vehicles after the collision, we can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.
The momentum [tex]\( p \) of an object is given by its mass \( m \) multiplied by its velocity \( v \)[/tex]:
[tex]\[ p = mv \][/tex]
Before the collision, the total momentum is the sum of the momentum of the truck and the momentum of the car. After the collision, the total momentum remains the same.
Let [tex]\( v_t \) be the velocity of the truck after the collision, and \( v_c \)[/tex] be the velocity of the car after the collision.
Before the collision:
[tex]\[ \text{Total momentum} = \text{momentum of truck} + \text{momentum of car} \][/tex]
[tex]\[ (1400 \, \text{kg}) \times (14.0 \, \text{m/s}) + (641 \, \text{kg}) \times (0 \, \text{m/s}) = (1400 \, \text{kg}) \times (v_t) + (641 \, \text{kg}) \times (v_c) \][/tex]
[tex]\[ 1400 \times 14.0 = 1400v_t + 641v_c \][/tex]
[tex]\[ 19600 = 1400v_t + 641v_c \][/tex]
Since the collision is approximately elastic, we also have the condition that the relative velocity of approach equals the relative velocity of separation.
[tex]\[ v_t - v_c = 14.0 \, \text{m/s} \][/tex]
We now have a system of two equations:
[tex]\[ 19600 = 1400v_t + 641v_c \][/tex]
[tex]\[ v_t - v_c = 14.0 \][/tex]
We can solve this system of equations to find the values of [tex]\( v_t \)[/tex] and [tex]\( v_c \)[/tex].
[tex]\[ v_t = \frac{19600 - 641v_c}{1400} \][/tex]
Substitute this expression for [tex]\( v_t \)[/tex] into the second equation:
[tex]\[ \frac{19600 - 641v_c}{1400} - v_c = 14.0 \][/tex]
Multiply both sides by 1400 to eliminate the denominator:
[tex]\[ 19600 - 641v_c - 1400v_c = 14.0 \times 1400 \][/tex]
[tex]\[ 19600 - 2041v_c = 19600 \][/tex]
Now, isolate [tex]\( v_c \)[/tex] :
[tex]\[ -2041v_c = 0 \][/tex]
[tex]\[ v_c = 0 \][/tex]
Substitute [tex]\( v_c = 0 \)[/tex] into [tex]\( v_t = \frac{19600 - 641v_c}{1400} \)[/tex] :
[tex]\[ v_t = \frac{19600 - 641 \times 0}{1400} \][/tex]
[tex]\[ v_t = \frac{19600}{1400} \][/tex]
[tex]\[ v_t = 14.0 \, \text{m/s} \][/tex]
So, after the collision, the truck's speed remains 14.0 m/s, and the car's speed becomes 0 m/s.
The speed of the truck and car after the collision is approximately 9.47 m/s
Use the principle of conservation of momentum to solve this problem. The total momentum before the collision is equal to the total momentum after the collision. We can express this as:
[tex]$$ m_{\text{truck}} \cdot v_{\text{truck, initial}} + m_{\text{car}} \cdot v_{\text{car, initial}} = (m_{\text{truck}} + m_{\text{car}}) \cdot v_{\text{final}} $$[/tex]
Given the following information:
- Mass of the truck, [tex]\(m_{\text{truck}}\)[/tex]: 1400 kg
- Initial velocity of the truck, [tex]\(v_{\text{truck, initial}}\)[/tex]: 14.0 m/s
- Mass of the car, [tex]\(m_{\text{car}}\):[/tex] 641 kg
- Initial velocity of the car, [tex]\(v_{\text{car, initial}}\)[/tex]: 0 m/s (since the car is stopped)
We can solve for the final velocity, [tex]\(v_{\text{final}}\):[/tex]
[tex]$$ 1400 \cdot 14.0 + 641 \cdot 0 = (1400 + 641) \cdot v_{\text{final}} $$[/tex]
Solving for [tex]\(v_{\text{final}}\):[/tex]
[tex]$$ v_{\text{final}} = \frac{1400 \cdot 14.0}{1400 + 641} $$[/tex]
Calculating the value:
[tex]$$ v_{\text{final}} \approx 9.47 \, \text{m/s} $$[/tex]
Therefore, the speed of the truck and car after the collision is approximately 9.47 m/s.
Light striking a metal surface causes electrons to be emitted from the metal via the photoelectric effect.The intensity of the incident light and the temperature of the metal are held constant. Assuming that the initial light incident on the metal surface causes electrons to be ejected from the metal, what happens if the frequency of the incident light is increased? Check all that apply.A. The work function of the metal increases.B. The number of electrons emitted from the metal per second increases.C. The maximum speed of the emitted electrons increases.D. The stopping potential increases.
Answer:
C. The maximum speed of emitted electrons increase
D. The stopping potential increases
Explanation:
Albert Einstein provided a successful explanation of the photoelectric effect on the basis of quantum theory. He proposed that,
“An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time interval that it has almost no chance to absorb a second photon. An increase in intensity of light source simply increase the number of photons and thus, the number of electrons, but the energy of electron remains same. However, increase in frequency of light increases the energy of photons and hence, the energy of electrons too.”
Therefore, due to this increase in energy of photons, the kinetic energy of emitted electrons also increase. And the increase in kinetic energy follows with the increase in velocity.
Now, the stopping potential is directly proportional to Kinetic Energy of the electrons. Thus, the increase in Kinetic Energy, results in an increase in stopping potential.
Therefore, two options apply here:
C. The maximum speed of emitted electrons increases
D. The stopping potential increases
If the frequency of the incident light in the photoelectric effect is increased, C. The maximum speed of the emitted electrons increases and D. The stopping potential increases; however, the work function does not change and the number of electrons emitted per second remains the same at a constant light intensity.
For the question about what happens if the frequency of the incident light in the photoelectric effect is increased, the correct options are:
The maximum speed of the emitted electrons increases.The stopping potential increases.The work function of the metal (Option A) is a material property and does not change with the frequency of the incident light. Therefore, this option is incorrect. The number of electrons emitted from the metal per second (Option B) depends on the intensity of the light, not its frequency, so increasing the frequency at constant intensity does not increase the emission rate. Thus, this option is also incorrect. When the frequency of the incident light is increased, the kinetic energy of the emitted electrons increases (as per Einstein's equation for the photoelectric effect), which correlates with Option C. The stopping potential, which is the electric potential needed to stop the fastest photoelectrons, also increases with the electron's kinetic energy, confirming Option D as correct.
Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms are not quantized. The energies of atoms are quantized. Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted.
Answer:
This is because The energies of atoms are quantized.
Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed
Your group has been asked to examine collisions in which the pucks stick together after the collision. To get the pucks to stick together, your group has wrapped the circumference of each puck with velcro (the regulation puck has the velcro hooks and the practice puck has the velcro loops) with the hooks or loops outward in each case. To get consistent puck speeds before the collision, your group has constructed a pair of rubber-band launchers, in which a very strong rubber band is stretched tightly between two posts and then used like a slingshot. A few trials are sufficient to determine that these launchers, if used carefully, can launch the pucks with predictable speeds and with essentially no spin?
Answer:
If necessary, readjust the f-stop of the CCD camera until only the LED’s from the pucks
are visible. You may find that a vertical stripe appears associated with the LED. This
is called streaking or ‘blooming’ and is a limitation of CCD technology in the presence of
localized bright spots. Some amount of streaking is acceptable, and can be compensated for
later in the computer analysis.
Once the exposure level of the camera has been set you should capture a ‘dark frame’.
For this, first remove all pucks, hands, etc., from the air table and click on the box to the
right of Background Frame. The dark frame can later be subtracted from your images of
collisions to suppress any constant background such as the edge of the table.
According to the Heisenberg uncertainty principle, quantum mechanics differs from classical mechanics in that: Select the correct answer below: Quantum mechanics involves particles that do not move. It is impossible to calculate with accuracy both the position and momentum of particles in classical mechanics. The measurement of an observable quantity in the quantum domain inherently changes the value of that quantity. All of the above
Answer:
Statement 3 is correct.
Heisenberg's uncertainty principle explains that the measurement of an observable quantity in the quantum domain inherently changes the value of that quantity
Explanation:
Classical mechanics is the study of motion of big, relatable bodies that we come in contact with in our day to day lives.
Quantum mechanics refers to this same study, but for particles on a subatomic level.
Obviously, Classical mechanics' theories and principles were first discovered and they worked for their intended uses (still work!). But when studies on particles on a sub-atomic level intensified, it became impractical to apply those theories and principles to these sub-atomic particles that displayed wave-particle duality nature properly.
Heisenberg's Uncertainty principle came in a time that explanations and justifications were needed to adapt these theories to sub-atomic particles.
The principle explains properly that it is impossible to measure the position and velocity (momentum) of a sub-atomic particle in exact terms and at the same time.
Mathematically, it is presented as
Δx.Δp ≥ ℏ
Where ℏ= adjusted Planck's constant.
ℏ= (h/2π)
And Δx and Δp are the uncertainties in measuring the position and momentum of sub-atomic particles.
The major reason for this is the wave-particle duality of sub-atomic particles. They exist as waves and particles at the same time that a complete knowledge of their position mean that a complete ignorance of their velocity and vice versa.
Taking the statements one at a time
Statement 1
Quantum Mechanics studies sub-atomic particles which are mostly always in motion. So, this is false.
Statement 2
It is impossible to calculate with accuracy both the position and momentum of particles in quantum mechanics not classical mechanics. As stated above, the reason for the uncertainty is the wave-particle duality of sub-atomic particles which the particle in classical mechanics do not exhibit obviously enough.
Statement 3
Any attempt to measure precisely the velocity of a subatomic particle, will knock it about in an unpredictable way, so that a simultaneous measurement of its position has no validity.
An essential feature of quantum mechanics is that it is generally impossible, even in principle, to measure a system without disturbing it. This is basically the uncertainty principle rephrased. This is the only true statement.
Hope this Helps!!!
In a Young's double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separation between adjacent bright fringes on a screen 5 m from the slits is:______a. 0.10 cm b. 0.25 cm c. 0.50 cm d. 1.0 cm e. none of the above
Answer:
b. 0.25cm
Explanation:
You can solve this question by using the formula for the position of the fringes:
[tex]y=\frac{m\lambda D}{d}[/tex]
m: order of the fringes
lambda: wavelength 500nm
D: distance to the screen 5 m
d: separation of the slits 1mm=1*10^{-3}m
With the formula you can calculate the separation of two adjacent slits:
[tex]\Delta y=\frac{(m+1)(\lambda D)}{d}-\frac{m\lambda D }{d}=\frac{\lambda D}{d}\\\\\Delta y=\frac{(500*10^{-9}nm)(5m)}{1*10^{-3}m}=2.5*10^{-3}m=0.25cm[/tex]
hence, the aswer is 0.25cm