Answer:
B = 2.19*10^-7 T
u = 1.92*10^-18 J/m^3
P = (4pi r^2)cεo E^2
Explanation:
In order to find the magnetic field strength of the electromagnetic wave you use the following formula:
[tex]B=\frac{E}{c}[/tex] (1)
B: magnitude of the magnetic field
E: magnitude of the electric field = 65.9V/m
c: speed of light = 3*10^8m/s
[tex]B=\frac{65.9V/m}{3*10^8m/s}=2.19*10^{-7}T[/tex]
The magnitude of the magnetic field 2.19*10^-7 T
The energy density of the electromagnetic wave is:
[tex]u=\frac{1}{2}\epsilon_oE^2[/tex] (2)
εo: dielectric permittivity = 8.85*10^-12C^2/Nm^2
[tex]u=\frac{1}{2}(8.85*10^{-12}C^2/Nm^2)(65.9V/m)^2=1.92*10^{-8}\frac{J}{m^3}[/tex]
The energy density of the electromagnetic wave is 1.92*10^-8J/m^3
The power is given by:
[tex]P=IA=c\epsilon_oE^2(4\pi r^2)[/tex]
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.
A large crate of mass m is place on the flatbed of a truck but not tied down. As the truck accelerates forward with acceleration a, the crate remains at rest relative to the truck. What force causes the crate to accelerate?
Answer:
Friction
Explanation:
There are tiny bumps and grooves on every object, which make them rough and more difficult to rub against each other. Even though the crate remains at rest at first, the frictional force causes it to stay in place and accelerate with the truck. Hope this helps!
The temperature coefficient of resistivity for the metal gold is 0.0034 (C )1, and for tungsten it is 0.0045 (C )1. The resistance of a gold wire increases by 7.0% due to an increase in temperature. For the same increase in temperature, what is the percentage increase in the resistance of a tungsten wire
Answer:
% increase in resistance of tungsten = 9.27%
Explanation:
We are given:
Co-efficient of resistivity for the metal gold; α_g = 0.0034 /°C
Co-efficient of resistivity for tungsten;α_t = 0.0045 /°C
% Resistance change of gold wire due to temperature change = 7%
Now, let R1 and R2 be the resistance before and after the temperature change respectively.
Thus;
(R2 - R1)/R1) x 100 = 7
So,
(R2 - R1) = 0.07R1
R2 = R1 + 0.07R1
R2 = 1.07R1
The equation to get the change in temperature is given as;
R2 = R1(1 + αΔt)
So, for gold,
1.07R1 = R1(1 + 0.0034*Δt)
R1 will cancel out to give;
1.07 = 1 + 0.0034Δt
(1.07 - 1)/0.0034 = Δt
Δt = 20.59°C
For this same temperature for tungsten, let Rt1 and Rt2 be the resistance before and after the temperature change respectively and we have;
Rt2 = Rt1(1 + α_t*Δt)
So, Rt2/Rt1 = 1 + 0.0045*20.59
Rt2/Rt1 = 1.0927
From earlier, we saw that;
(R2 - R1)/R1) x 100 = change in resistance
Similarly,
(Rt2 - Rt1)/Rt1) x 100 = change in resistance
Simplifying it, we have;
[(Rt2/Rt1) - 1] × 100 = %change in resistance
Plugging in the value of 1.0927 for Rt2/Rt1, we have;
(1.0927 - 1) × 100 = %change in resistance
%change in resistance = 9.27%
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 particle covers equal distance in equal intervals of time. It is said to be?
1.at rest
2.moving with constant acceleration
3.moving with constant velocity
4.moving with constant speed
4.Moving with constant speed
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
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 woman is standing in the ocean, and she notices that after a wave crest passes by, five more crests pass in a time of 50.2 s. The distance between two successive crests is 30.2 m. What is the wave's (a) period, (b) frequency, (c) wavelength, and (d) speed
Explanation:
(a) The period of a wave is the time required for one complete cycle. In this case, we have the time of five cycles. So:
[tex]T=\frac{t}{n}\\\\T=\frac{50.2s}{5}\\T=10.04s[/tex]
(b) The frequency of a wave is inversely proportional to its period:
[tex]f=\frac{1}{T}\\f=\frac{1}{10.04s}\\f=0.01Hz[/tex]
(c) The wavelength is the distance between two successive crests, so:
[tex]\lambda=30.2m[/tex]
(d) The speed of a wave is defined as:
[tex]v=f\lambda\\v=(0.1Hz)(30.2m)\\v=3.02\frac{m}{s}[/tex]
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
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 cart of mass 350 g is placed on a frictionless horizontal air track. A spring having a spring constant of 7.5 N/m is attached between the cart and the left end of the track. The cart is displaced 3.8 cm from its equilibrium position. (a) Find the period at which it oscillates. s (b) Find its maximum speed. m/s (c) Find its speed when it is located 2.0 cm from its equilibrium position.
Answer:
(a) T = 1.35 s
(b) vmax = 0.17 m/s
(c) v = 0.056 m/s
Explanation:
(a) In order to calculate the period of oscillation you use the following formula for the period in a simple harmonic motion:
[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex] (1)
m: mass of the cart = 350 g = 0.350kg
k: spring constant = 7.5 N/m
[tex]T=2\pi \sqrt{\frac{0.350kg}{7.5N/m}}=1.35s[/tex]
The period of oscillation of the car is 1.35s
(b) The maximum speed of the car is given by the following formula:
[tex]v_{max}=\omega A[/tex] (2)
w: angular frequency
A: amplitude of the motion = 3.8 cm = 0.038m
You calculate the angular frequency:
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{1.35s}=4.65\frac{rad}{s}[/tex]
Then, you use the result of w in the equation (2):
[tex]v_{max}=(4.65rad/s)(0.038m)=0.17\frac{m}{s}[/tex]
The maximum speed if 0.17m/s
(c) To find the speed when the car is at x=2.0cm you first calculate the time t by using the following formula:
[tex]x=Acos(\omega t)\\\\t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\t=\frac{1}{4.65rad/s}cos^{-1}(\frac{0.02}{0.038})=0.069s[/tex]
The speed is the value of the following function for t = 0.069s
[tex]|v|=|\omega A sin(\omega t)|\\\\|v|=(4.65rad/s)(0.038m)sin(4.65rad/s (0.069s))=0.056\frac{m}{s}[/tex]
The speed of the car is 0.056m/s
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
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
Difference between regular and irregular object.
Hope this helps....
Good luck on your assignment.....
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.
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 .
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
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]
what is the most likely elevation of point x?
A. 150 ft
B. 200 ft
C. 125 ft
D. 250 ft
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 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.
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 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.
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
A wire with mass 90.0g is stretched so that its ends are tied down at points 88.0cm apart. The wire vibrates in its fundamental mode with frequency 80.0Hz and with an amplitude of 0.600cm at the antinodes.a) What is the speed of propagation of transverse waves in the wire?b) Compute the tension in the wire.
Answer:
a) V = 140.8 m/s
b) T = 2027.52 N = 2.03 KN
Explanation:
a)
The formula for the speed of the wave is given as follows:
f₁ = V/2L
V = 2f₁L
where,
V = Speed of Wave = ?
f₁ = Fundamental Frequency = 80 Hz
L = Length of Wire = 88 cm = 0.88 m
Therefore,
V = (2)(80 Hz)(0.88 m)
V = 140.8 m/s
b)
Another formula for the speed of wave is:
V = √T/μ
V² = T/μ
T = V²μ
where,
T = Tension in String = ?
μ = Linear Mass Density of Wire = Mass of Wire/L = 0.09 kg/0.88 m
μ = 0.1 kg/m
Therefore,
T = (140.8 m/s)²(0.1 kg/m)
T = 2027.52 N = 2.03 KN
Location C is 0.021 m from a small sphere that has a charge of 5 nC uniformly distributed on its surface. Location D is 0.055 m from the sphere. What is the change in potential along a path from C to D?
Answer:
ΔV = -1321.73V
Explanation:
The change in potential along the path from C to D is given by the following expression:
[tex]\Delta V=-\int_a^bE dr[/tex] (1)
E: electric field produced by a charge at a distance of r
a: distance to the sphere at position C = 0.021m
b: distance to the sphere at position D = 0.055m
The electric field is given by:
[tex]E=k\frac{Q}{r^2}[/tex] (2)
Q: charge of the sphere = 5nC = 5*10^-9C
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
You replace the expression (2) into the equation (1) and solve the integral:
[tex]\Delta V=-kQ\int_a^b \frac{dr}{r^2}=-kQ[-\frac{1}{r}]_a^b[/tex] (3)
You replace the values of a and b:
[tex]\Delta V=(8.98*10^9Nm^2/C^2)(5*10^{-9}C)[\frac{1}{0.055m}-\frac{1}{0.021m}]\\\\\Delta V=-1321.73V[/tex]
The change in the potential along the path C-D is -1321.73V
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
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:
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 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