Answer:
Explanation:
The magnetic moment of the coil M = current x area of loop
= I x L x L
M = L² I .
Magnetic field = B
Torque = M x B
= MB sinθ where θ is angle between M and B . Direction of M is perpendicular to face of loop
Torque = MB sin90
= MB
= L² I B .
A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.24 m from the slits. The first bright fringe is located 3.30 mm from the center of the interference pattern. What is the wavelength of the laser light?
Answer:
λ = 509 nm
Explanation:
In order to calculate the wavelength of the light you use the following formula:
[tex]y=m\frac{\lambda D}{d}[/tex] (1)
where
y: distance of the mth fringe to the central peak = 3.30 mm = 3.30*10^-3 m
m: order of the bright fringe = 1
D: distance from the slits to the screen = 3.24 m
d: distance between slits = 0.500 mm = 0.500*10^-3 m
You first solve the equation (1) for λ, and then you replace the values of the other parameters:
[tex]\lambda=\frac{dy}{mD}\\\\\lambda=\frac{(0.500*10^{-3}m)(3.30*10^{-3}m)}{(1)(3.24m)}=5.09*10^7m\\\\\lambda=509*10^{-9}m=509nm[/tex]
The wavelength of the light is 509 nm
Difference between regular and irregular object.
Hope this helps....
Good luck on your assignment.....
A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm. Part A What is the gauge pressure at the bottom of the cylinder
Answer:
The gauge pressure is [tex]P_g = 2058 \ P_a[/tex]
Explanation:
From the question we are told that
The height of the water contained is [tex]h_w = 30 \ cm = 0.3 \ m[/tex]
The height of liquid in the cylinder is [tex]h_t = 40 \ cm = 0.4 \ m[/tex]
At the bottom of the cylinder the gauge pressure is mathematically represented as
[tex]P_g = P_w + P_o[/tex]
Where [tex]P_w[/tex] is the pressure of water which is mathematically represented as
[tex]P_w = \rho_w * g * h_w[/tex]
Now [tex]\rho_w[/tex] is the density of water with a constant values of [tex]\rho_w = 1000 \ kg /m^3[/tex]
substituting values
[tex]P_w = 1000 * 9.8 * 0.3[/tex]
[tex]P_w = 2940 \ Pa[/tex]
While [tex]P_o[/tex] is the pressure of oil which is mathematically represented as
[tex]P_o = \rho_o * g * (h_t -h_w )[/tex]
Where [tex]\rho _o[/tex] is the density of oil with a constant value
[tex]\rho _o = 900 \ kg / m^3[/tex]
substituting values
[tex]P_o = 900 * 9.8 * (0.4 - 0.3)[/tex]
[tex]P_o = 882 \ Pa[/tex]
Therefore
[tex]P_g = 2940 - 882[/tex]
[tex]P_g = 2058 \ P_a[/tex]
Calculate the amount of kinetic energy the car stores if it has a mass of 1200 kg and speed of 15 m/s
Answer:
KE = 135,000 j or 135 KJ
Explanation:
KE=0.5mv^2
KE=0.5*1200*15^2
KE = 135,000 joules or 135 Kilo Joules
a cannon is fired with an initial horizontal velocity of 20m/s and an initial velocity of 25m/s. After 3s in the air, the cannon hits its target. How far away(in meters) was the cannon from its target
Answer:
60 m
Explanation:
After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.
__
The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.
American eels (Anguilla rostrata) are freshwater fish with long, slender bodies that we can treat as uniform cylinders 1.0 m long and 10 cm in diameter. An eel compensates for its small jaw and teeth by holding onto prey with its mouth and then rapidly spinning its body around its long axis to tear off a piece of flesh. Eels have been recorded to spin at up to 14 revolutions per second when feeding in this way. Although this feeding method is costly in terms of energy, it allows the eel to feed on larger prey than it otherwise could. The eel has a certain amount of rotational kinetic energy when spinning at 14 spins per second. If it swam in a straight line instead, about how fast would the eel have to swim to have the same amount of kinetic energy as when it is spinning? (a) 0.5 m/s; (b) 0.7 m/s; (c) 3 m/s; (d) 5 m/s.
Answer:
(c) 3 m/s;
Explanation:
Moment of inertia of the fish eels about its long body as axis
= 1/2 m R ² where m is mass of its body and R is radius of transverse cross section of body .
= 1/2 x m x (5 x 10⁻² )²
I = 12.5 m x 10⁻⁴ kg m²
angular velocity of the eel
ω = 2 π n where n is revolution per second
=2 π n
= 2 π x 14
= 28π
Rotational kinetic energy
= 1/2 I ω²
= .5 x 12.5 m x 10⁻⁴ x(28π)²
= 4.8312m J
To match this kinetic energy let eel requires to have linear velocity of V
1 / 2 m V² = 4.8312m
V = 3.10
or 3 m /s .
A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground
Answer:
t = 2 s
Explanation:
In order to find the time taken by the stone to fall from the top of the building to the ground we can use 2nd equation of motion. 2nd equation of motion is as follows:
s = Vit + (0.5)gt²
where,
t = time = ?
Vi = Initial Velocity = 20 m/s
s = height of building = 60 m
g = 9.8 m/s²
Therefore,
60 m = (20 m/s)t + (0.5)(9.8 m/s²)t²
4.9t² + 20t - 60 = 0
solving this quadratic equation we get:
t = -6.1 s (OR) t = 2 s
Since, the time cannot be negative in magnitude.
Therefore,
t = 2 s
A stationary 6-kg shell explodes into three pieces. One 4.0 kg piece moves horizontally along the negative x-axis. The other two fragments, each 1.0 kg, move in directions that make 60o angle above and below the positive x-axis and their speeds are 60 m/s each. What is the velocity of the 4.0-kg fragment
Answer:
-15 m/s
Explanation:
The computation of the velocity of the 4.0 kg fragment is shown below:
For this question, we use the correlation of the momentum along with horizontal x axis
Given that
Weight of stationary shell = 6 kg
Other two fragments each = 1.0 kg
Angle = 60
Speed = 60 m/s
Based on the above information, the velocity = v is
[tex]1\times 60 \times cos\ 60 + 1\times 60 \times cos\ 60 - 4\ v = 0[/tex]
[tex]\frac{60}{2} + \frac{60}{2} - 4\ v = 0[/tex]
[tex]v = \frac{60}{4}[/tex]
= -15 m/s
When using a mercury barometer , the vapor pressure of mercury is usually assumed to be zero. At room temperature mercury's vapor pressure is about 0.0015 mm-Hg. At sea level, the height hhh of mercury in a barometer is about 760 mm.Required:a. If the vapor pressure of mercury is neglected, is the true atmospheric pressure greater or less than the value read from the barometer? b. What is the percent error? c. What is the percent error if you use a water barometer and ignore water's saturated vapor pressure at STP?
Answer:
Explanation:
(a)
The true atmospheric pressure will has more value than the reading in the barometer. If Parm is the atmospheric
pressure in the tube then the resulting vapour pressure is
Patm - pgh = Prapor
The final reading ion the barometer is
pgh = Palm - Proper
Hence, the true atmospheric pressure is greater.
you can find the answer in this book
physics principles with Applications, Global Edition Problem 67P: Chapter: CH 13 Problem:67p
A 2-kg block is released from rest at the top of a 20-mlong frictionless ramp that is 4 m high. At the same time, an identical block is released next to the ramp so that it drops straight down the same 4 m. What are the values for each of the following for the blocks just before they reach ground level.
Required:
a. Gravitational potential energy Block a_____ J Block b _____ J
b. Kinetic energy Block a _____ J Block b _____
c. Speed Block a _____ J Block b _____ J
d. Momentum Block a _____ J Block b _____ J
Answer:
A.) 78.4 J for both
B.) 78.4 J for both
C.) 8.85 m/s for both
D.) 17.7 kgm/s
Explanation:
Given information:
Mass m = 2 kg
Distance d = 20 m
High h = 4 m
A.) Gravitational potential energy can be calculated by using the formula
P.E = mgh
P.E = 2 × 9.8 × 4
P.E = 78.4 J
Since the two objects are identical, the gravitational potential energy of the block for both a and b will be 78.4 J
B.) According to conservative energy,
Maximum P.E = Maximum K.E.
Therefore, the kinetic energy of the two blocks will be 78.4 J
C.) Since K.E = 1/2mv^2 = mgh
V = √(2gh)
Solve for velocity V by substituting g and h into the formula
V = √(2 × 9.8 × 4)
V = √78.4
V = 8.85 m/s
The velocities of both block will be 8.85 m/s
D.) Momentum is the product of mass and velocity. That is,
Momentum = MV
Substitute for m and V into the formula
Momentum = 2 × 8.85 = 17.7 kgm/s
Both block will have the same value since the ramp Is frictionless.
The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s) of the orbit and shaded segments are the same?
Answer: not sure
Explanation:
An oscillator consists of a block of mass 0.373 kg connected to a spring. When set into oscillation with amplitude 33 cm, the oscillator repeats its motion every 0.412 s. Find the (a) period, (b) frequency, (c) angular frequency, (d) spring constant, (e) maximum speed, and (f) magnitude of the maximum force on the block from the spring.
Answer:
(a) T = 0.412s
(b) f = 2.42Hz
(c) w = 15.25 rad/s
(d) k = 86.75N/m
(e) vmax = 5.03 m/s
Explanation:
Given information:
m: mass of the block = 0.373kg
A: amplitude of oscillation = 22cm = 0.22m
T: period of oscillation = 0.412s
(a) The period is the time of one complete oscillation = 0.412s
The period is 0.412s
(b) The frequency is calculated by using the following formula:
[tex]f=\frac{1}{T}=\frac{1}{0.412s}=2.42Hz[/tex]
The frequency is 2.42 Hz
(c) The angular frequency is:
[tex]\omega=2\pi f=2\pi (2.42Hz)=15.25\frac{rad}{s}[/tex]
The angular frequency is 15.25 rad/s
(d) The spring constant is calculated by solving the following equation for k:
[tex]\omega=\sqrt{\frac{k}{m}}\\\\k=m\omega^2=(0.373kg)(15.25rad/s)^2=86.75\frac{N}{m}[/tex]
The spring constant is 86.75N/m
(e) The maximum speed is:
[tex]v_{max}=\omega A=(15.25rad/s)(0.33m)=5.03\frac{m}{s}[/tex]
(f) The maximum force applied by the spring if for the maximum elongation, that is, the amplitude:
[tex]F=kA=(86.75N/m)(0.2m)=17.35N[/tex]
The maximum force that the spring exerts on the block is 17.35N
Sea Food is a rich source of ______. *
Answer:
Sea Food is a rich source of ______. *
Explanation:
Seafood is a rich source of minerals, such as iron, zinc, iodine, magnesium, and potassium.
Answer:
Sea food is a rich source of protein
Explanation:
You should eat fish (if you are not vegan/ vegetarian ofc) at least 2 times a week and 1 has to be oily
If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. What is its vertical displacement (in m) after 10 s? (g = 10 m/s2)
Answer:
500mExplanation:
Using the equation of motion S = ut + 1/2 gt²
S = the vertical displacement (in m)
u = initial velocity of the object (in m/s)
g = acceleration due to gravity (in m/s²)
t = time taken (in secs)
Given u = 0m/s, g = 10m/s² and t = 10s, substituting this value into the equation to get the vertical displacement w have;
S = 0+1/2 (10)(10)²
S = 1000/2
S = 500m
The vertical displacement after 10seconds is 500m
A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x
Answer:
If the particle is an electron [tex]E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton, [tex]E_y = 6.08 * 10^6 N/C[/tex]
Explanation:
Initial speed at the origin, [tex]u = 3 * 10^6 m/s[/tex]
[tex]\theta = 38^0[/tex] to +ve x-axis
The particle crosses the x-axis at , x = 1.5 cm = 0.015 m
The particle can either be an electron or a proton:
Mass of an electron, [tex]m_e = 9.1 * 10^{-31} kg[/tex]
Mass of a proton, [tex]m_p = 1.67 * 10^{-27} kg[/tex]
The electric field intensity along the positive y axis [tex]E_y[/tex], can be given by the formula:
[tex]E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\[/tex]
If the particle is an electron:
[tex]E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton:
[tex]E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 6.08 * 10^6 N/C[/tex]
In the study of sound, one version of the law of tensions is:
f1= f2 √ (F1/F2)
If f1= 300, F2= 60, and f2=260, find f1 to the nearest unit.
Answer:
F1 = 80
Explanation:
f1= f2 √ (F1/F2)
Where f1 = 300, f2 = 260 and F2 = 60
Putting in the above formula
300 = 260√(F1/60)
Dividing both sides by 260
=> 1.15 = √(F1/60)
Squaring both sides
=> 1.33 = F1/60
Multiplying both sides by 60
=> F1 = 80
Giving quadrilateral a(2,-1 ) b ( 1,3) c(6,5) d(7,1) you want to prove that it is a parallelogram by showing opposite sides are congruent . what formula would you use ? show that sb is congruent to cd
Answer:
AB = CD = √17
Explanation:
The distance formula is used to find the length of a line segment between two points. Here, we want to show the distance AB is the same as the distance CD.
d = √((x1 -x1)² +(y2 -y1)²)
__
AB: d = √((1 -2)² +(3 -(-1))²) = √((-1)² +4²) = √17
CD: d = √((7-6)² +(1-5)²) = √(1² +(-4)²) = √17 . . . . same as AB
Segment AB is congruent to segment CD.
In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 88 kgkg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 NN .
How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?
Answer:
16.10 kJ
Explanation:
The thermal energy created in the slope can be found by definition of work (W):
[tex] W = E_{f} - E_{i} = K_{f} + P_{f} + Th_{f} - (K_{i} + Th_{i}) [/tex]
Where:
[tex]K_{f}[/tex] and [tex]K_{i}[/tex]: is the final and initial kinetic energy
[tex]P_{f}[/tex]: is the final potential energy
[tex]Th_{f}[/tex] and [tex]Th_{i}[/tex]: is the final and initial thermal energy
[tex]W = \frac{1}{2}mv_{f}^{2} + mgh - \frac{1}{2}mv_{i}^{2} + Th_{f} - Th_{i}[/tex]
We have that W is:
[tex] W = F*d = T*d [/tex]
Where:
F: is the force equal to the tension (T)
d: is the displacement = 120 m
And since the speed is constant, [tex]v_{i}[/tex] = [tex]v_{f}[/tex] we have:
[tex] T*d = mgh + \Delta Th [/tex]
[tex] \Delta Th = T*d - mgh = 350 N*120 m - 88 kg*9.81 m/s^{2}*30 m = 16101.6 J [/tex]
Therefore, the thermal energy created in the slope and the tube during the ascent is 16.10 kJ.
I hope it helps you!
At least how many Calories does a mountain climber need in order to climb from sea level to the top of a 5.42 km tall peak assuming the muscles of the climber can convert chemical energy to mechanical energy with an efficiency of 16.0 percent. The total mass of the climber and the equipment is 78.4 kg. (Enter your answer as a number without units.)
Answer:
Ec = 6220.56 kcal
Explanation:
In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.
You use the following formula:
[tex]W_c=Mgh[/tex] (1)
Wc: work done by the climber
g: gravitational constant = 9.8 m/s^2
M: mass of the climber = 78.4 kg
h: height reached by the climber = 5.42km = 5420 m
You replace in the equation (1):
[tex]W_c=(78.4kg)(9.8m/s^2)(5420m)=4,164,294.4\ J[/tex] (2)
Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:
[tex]0.16(E_c)=4,164,294.4J[/tex]
Ec: Calories
You solve for Ec and convert the result to Cal:
[tex]E_c=\frac{4,164,294.4}{016}=26,026,840J*\frac{1kcal}{4184J}\\\\E_c=6220.56\ kcal[/tex]
The amount of Calories needed by the climber was 6220.56 kcal
A ray in glass (n = 1.51) reaches a boundary with air at 49.2 deg. Does it reflect internally or refract into the air? Enter 0 for reflect, and 1 for refract.
Answer:
0 - Then, the ray is totally reflected
Explanation:
The ray reaches the boundary between the two mediums at 49.2°.
If the ray is totally reflected it is necessary that the crictical angle is lower that the incidet angle.
You use the following to calculate the critical angle:
[tex]\theta_c=sin^{-1}(\frac{n_2}{n_1})[/tex] (1)
n2: index of refraction of the second medium (air) = 1.00
n1: index of refraction of the first medium (glass) = 1.51
You replace the values of the parameters in the equation (1):
[tex]\theta_c=sin^{-1}(\frac{1.00}{1.51})=41.47\°[/tex]
The critical angle is 41.47°, which is lower than the incident angle 49.2°.
Then, the ray is totally reflected.
0
A bicycle rider has a speed of 20.0 m/s at a height of 60 m above sea level when he begins coasting down hill. Sea level is the zero level for measuring gravitational potential energy. Ignoring friction and air resistance, what is the rider's speed when he coasts to a height of 18 m above sea level?
Answer:
The rider's speed will be approximately 35 m/s
Explanation:
Initially the rider has kinetic and potential energy, and after going down the hill, some of the potencial energy turns into kinetic energy. So using the conservation of energy, we have that:
[tex]kinetic_1 + potencial_1 = kinetic_2 + potencial_2[/tex]
The kinetic and potencial energy are given by:
[tex]kinetic = mass * speed^2 / 2[/tex]
[tex]potencial = mass * gravity * height[/tex]
So we have that:
[tex]m*v^2/2 + mgh = m*v'^2/2 + mgh'[/tex]
[tex]20^2/2 + 9.81*60 = v'^2/2 + 9.81*18[/tex]
[tex]v'^2/2 + 176.58 = 788.6[/tex]
[tex]v'^2/2 = 612.02[/tex]
[tex]v'^2 = 1224.04[/tex]
[tex]v' = 34.99\ m/s[/tex]
So the rider's speed will be approximately 35 m/s
A flat roof is very susceptible to wind damage during a thunderstorm and/or tornado. If a flat roof has an area of 500 m2 and winds of speed 39.0 m/s blow across it, determine the magnitude of the force exerted on the roof. The density of air is 1.29 kg/m3.
Answer: The magnitude of the force exerted on the roof is 490522.5 N.
Explanation:
The given data is as follows.
Below the roof, [tex]v_{1}[/tex] = 0 m/s
At top of the roof, [tex]v_{2}[/tex] = 39 m/s
We assume that [tex]P_{1}[/tex] is the pressure at lower surface of the roof and [tex]P_{2}[/tex] be the pressure at upper surface of the roof.
Now, according to Bernoulli's theorem,
[tex]P_{1} + 0.5 \times \rho \times v^{2}_{1} = P_{2} \times 0.5 \rho \times v^{2}_{2}[/tex]
[tex]P_{1} - P_{2} = 0.5 \times \rho \times (v^{2}_{2} - v^{2}_{1})[/tex]
= [tex]0.5 \times 1.29 \times [(39)^{2} - (0)^{2}][/tex]
= [tex]0.645 \times 1521[/tex]
= 981.045 Pa
Formula for net upward force of air exerted on the roof is as follows.
F = [tex](P_{1} - P_{2})A[/tex]
= [tex]981.045 \times 500[/tex]
= 490522.5 N
Therefore, we can conclude that the magnitude of the force exerted on the roof is 490522.5 N.
A butcher grinds 5 and 3/4 lb of meat then sells it for 2 and 2/3 pounds to the customer what is the maximum amount me that the butcher can sell to the next customer
Answer:
The maximum amount of meat that the butcher can sell is [tex]3\frac{1}{12}\:lb[/tex]
Explanation:
The maximum amount can be found by taking the difference of mixed numbers.
[tex]5\frac{3}{4}-2\frac{2}{3}\\\\\mathrm{Subtract\:the\:numbers:}\:5-2=3\\\\\mathrm{Combine\:fractions:\:}\frac{3}{4}-\frac{2}{3}=\frac{1}{12}\\\\=3\frac{1}{12}\\[/tex]
Best Regards!
A wire carries a current of 4 A travelling to the left (-x direction). It is placed in a constant magnetic field of magnitude 0.05 T, pointing upward ( z direction). a. If 25 cm of the wire is in the magnetic field, what is the force on the current
Answer:
0.05 N
Explanation:
Data provided in the question
The Wire carries a current of 4A to the left direction
The constant magnetic field of magnitude = 0.05 T
Pointing upward i.e Z direction
The wire is in the magnetic field = 25 cm
Based on the above information, the force on the current is
[tex]= Current \times constant\ magnetic\ field\ of\ magnitude \times magnetic\ field[/tex]
[tex]= 4 \times 0.05 \times 0.25[/tex]
= 0.05 N
The direction will be the negative Y direction
You are trying to overhear a juicy conversation, but from your distance of 25.0 m, it sounds like only an average whisper of 25.0 dB. So you decide to move closer to give the conversation a sound level of 80.0 dB instead. How close should you come?
Answer:
r₂ = 1,586 m
Explanation:
For this problem we are going to solve it by parts, let's start by finding the sound intensity when we are 25 m
β = 10 log (I / I₀)
where Io is the sensitivity threshold 10⁻¹² W / m²
I₁ / I₀ = [tex]e^{\beta/10}[/tex]
I₁ = I₀ e^{\beta/10}
let's calculate
I₁ = 10⁻¹² e^{25/10}
I₁ = 1.20 10⁻¹¹ W / m²
the other intensity in exercise is
I₂ = 10⁻¹² e^{80/10}
I₂ = 2.98 10⁻⁹ W / m²
now we use the definition of sound intensity
I = P / A
where P is the emitted power that is a constant and A the area of the sphere where the sound is distributed
P = I A
the area a sphere is
A = 4π r²
we can write this equation for two points of the found intensities
I₁ A₁ = I₂ A₂
where index 1 corresponds to 25m and index 2 to the other distance
I₁ 4π r₁² = I₂ 4π r₂²
I₁ r₁² = I₂ r₂²
r₂ = √ (I₁ / I₂) r₁
let's calculate
r₂ = √ (1.20 10⁻¹¹ / 2.98 10⁻⁹) 25
r₂ = √ (0.40268 10⁻²) 25
r₂ = 1,586 m
What is the on ohooke benden
er ord power
What is the main difference between work, power and energy
Answer:Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.
The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ????0 . Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 2.47×1014 s−1. g
Answer:
E = 0.965eV
Explanation:
In order to calculate the minimum energy needed to eject the electrons you use the following formula:
[tex]E=h \nu[/tex] (1)
h: Planck' constant = 6.626*10^{-34}J.s
v: threshold frequency = 2.47*10^14 s^-1
You replace the values of v and h in the equation (1):
[tex]E=(6.262*10^{-34}J.s)(2.47*10^{14}s^{-1})=1.54*10^{-19}J[/tex]
In electron volts you obtain:
[tex]1.54*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=0.965eV[/tex]
The minimum energy needed is 0.965eV
The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 2006 the Earth took about 0.840 s longer to complete 365 revolutions than it did in the year 1906.
Required:
What was the average angular acceleration of the earth during this time?
Answer:
The average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²
Explanation:
We are given;
The period of 365 revolutions of Earth in 2006, T₁ = 365 days, 0.840 sec
Converting to seconds, we have;
T₁ = (365 × 24 × 60 × 60) + 0.84
T₁ = (3.1536 x 10⁷) + 0.840
T₁ = 31536000.84 s
Now, the period of 365 rotation of Earth in 2006 is; T₀ = 365 days
Converting to seconds, we have;
T₀ = 31536000 s
Hence, time period of one rotation in the year 2006 is;
Tₐ = 31536000.84/365
Tₐ = 86400.0023 s
The time period of rotation is given by the formula;
Tₐ = 2π/ωₐ
Making ωₐ the subject;
ωₐ = 2π/Tₐ
Plugging in the relevant values;
ωₐ = 2π/ 365.046306
ωₐ = 7.272205023 x 10⁻⁵ rad/s
Therefore, the time period of one rotation in the year 1906 is;
Tₓ = 31536000/365
Tₓ = 86400 s
Time period of rotation,
Tₓ = 2π /ωₓ
ωₓ = 2π / T
Plugging in the relevant values;
ωₓ = 2π/86400
ωₓ = 7.272205217 x 10⁻⁵ rad/s
The average angular acceleration is given by;
α = (ωₓ - ωₐ) / T₁
α = ((7.272205217 × 10⁻⁵) - (7.272205023 × 10⁻⁵)) / 31536000.84
α = 6.152 X 10⁻²⁰ rad/s²
Thus, the average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands
Complete question is;
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.
Answer:
Tangential speed of foot just before it lands is; v = 5.37m/s
Explanation:
Let U (potential energy) be zero on the ground.
So, initially, U = mgh
where, h = 0.98/2 = 0.49m (midpoint of the leg)
Now just before the leg hits the floor it would have kinetic energy as;
K = ½Iω²
where ω = v/r and I = ⅓mr²
So, K = ½(⅓mr²)(v/r)²
K = (1/6) × (mr²)/(v²/r²)
K = (1/6) × mv²
From principle of conservation of energy, we have;
Potential energy = Kinetic energy
Thus;
mgh = (1/6) × mv²
m will cancel out to give;
gh = (1/6)v²
Making v the subject, we have;
v = √6gh
v = √(6 × 9.81 × 0.49)
v = √28.8414
v = 5.37m/s
what is the most likely elevation of point x?
A. 150 ft
B. 200 ft
C. 125 ft
D. 250 ft