Answer:
#_galon = 8,873 gallons
Explanation:
For this exercise let's use a direct rule of proportions (rule of three). If 1 gallon is 142 MJ, how many gallons is 1260 MJ
#_galon = 1260 MJ (1 gallon / 142 MJ)
#_galon = 8,873 gallons
therefore air travel uses less fuel than car travel
An 80-kg firefighter slides down a fire pole. After 1.3 seconds of sliding, the firefighter is sliding at a velocity of 6.5 m/s, straight down the pole. Once this velocity is reached, the firefighter grips the pole so that the force of friction exerted by the firefighter's hands on the pole is equal to the force of gravity. At this point what is the downward acceleration of the firefighter
Answer:
a= 0
Explanation:
In the vertical direction, if the friction force (directed upward) is equal to the force of gravity (downward) this means that no net force is acting on the firefighter.According to Newton's 2nd Law, if no net force is present, the acceleration in this direction is just zero, as follows:[tex]F_{net} = m*a = 0 (1)[/tex]
⇒ a = 0
The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons).
(a) What is the force between the two alpha particles when they are 6.60 ✕ 10−15 m apart? N
(b) What is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u. m/s2
Answer:
a) F = 21.16 N, b) a = 3.17 10²⁸ m / s
Explanation:
a) The outside between the alpha particles is the electric force, given by Coulomb's law
F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]
in that case the two charges are of equal magnitude
q₁ = q₂ = 2q
let's calculate
F = [tex]9 \ 10^9 \ \frac{ (2 \ 1.6 \ 10^{-19} )^2 }{ (6.60 \ 10^{-15} )^2 }[/tex]
F = 21.16 N
this force is repulsive because the charges are of the same sign
b) what is the initial acceleration
F = ma
a = F / m
a = [tex]\frac{21.16}{4.0026 \ 1.67 \ 10^{-27} }[/tex]21.16 / 4.0025 1.67 10-27
a = 3.17 10²⁸ m / s
this acceleration is in the direction of moving away the alpha particles
Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water
Answer:
See explanation below
Explanation:
This is a typical exercise of free falling. In this case a rock thrown straight down from the bridge, and we are asked to determine the final velocity of the rock and it's displacement at those given times.
First, just for this problem, as the rock is going straight down, we'll say that the downward direction is positive, therefore, the following expressions to calculate velocity and speed will be:
V = V₀ + gt (1)
X = V₀t + gt²/2 (2)
In this case, g = 9.8 m/s²
Now, let's see the displacement and velocity for each given time:
a) For t = 0.5 s
V = 14 + (9.8)*0.5
V = 18.9 m/s
X = (14*0.5) + (9.8)(0.5)²/2
X = 7 + 1.225
X = 8.225 m
b) For t = 1.00 s
V = 14 + (9.8)*1
V = 23.8 m/s
X = (14*1) + (9.8)(1)²/2
X = 14 + 4.9
X = 18.9 m
c) For t = 1.5 s
V = 14 + (9.8)*1.5
V = 28.7 m/s
X = (14*1.5) + (9.8)(1.5)²/2
X = 21 + 11.025
X = 32.025 m
d) For t = 2 s
V = 14 + (9.8)*2
V = 33.6 m/s
X = (14*2) + (9.8)(2)²/2
X = 28 + 19.6
X = 47.6 m
e) For t = 2.50 s
V = 14 + (9.8)*2.5
V = 38.5 m/s
X = (14*2.5) + (9.8)(2.5)²/2
X = 35 + 30.625
X = 65.625 m
Hope this helps
A car is traveling along a straight road at a velocity of +30.0 m/s when its engine cuts out. For the next 1.79 seconds, the car slows down, and its average acceleration is . For the next 4.03 seconds, the car slows down further, and its average acceleration is . The velocity of the car at the end of the 5.82-second period is +18.4 m/s. The ratio of the average acceleration values is = 1.53. Find the velocity of the car at the end of the initial 1.79-second interval.
Answer:
first value+2nd +3rd
Explanation:
thug life and there
What is the main cause of tides on Earth? *
1. Sun's gravity
2. Moon's gravity
Electric power is to be generated by installing a hydraulic turbine-generator at a site 120 m below the free surface of a large water reservoir that can supply water at a rate of 2300 kg/s steadily. Determine the power generation potential.
Answer:
the power generation potential is 2.705 x 10⁶ J/s.
Explanation:
Given;
height below the free surface of a large water reservoir, h = 120 m
mass flow rate of the water, m' = 2300 kg/s
The power generation potential is calculated as;
[tex]Power = \frac{Energy}{time} = \frac{F\times h}{t} = \frac{(mg) \times h}{t} = \frac{m}{t}\times gh = m' \times gh\\\\Power = m' \times gh\\\\Power = 2300 \ kg/s \ \times \ 9.8 \ m/s^2 \ \times \ 120 \ m\\\\Power = 2.705 \times 10^6 \ J/s[/tex]
Therefore, the power generation potential is 2.705 x 10⁶ J/s.
Which statement applies only to magnetic force instead of both electric and magnetic forces? O A. It acts between a north pole and a south pole. O B. It can push objects apart. O C. It can pull objects together. D. It acts between objects that do not touch.
Answer:
the answer would be A. electricity don't specify the direction of any cardinal points the flow of charges moves.
Answer:
A
Explanation:
I did the test on ap3x
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 520 lines/mm , and the light is observed on a screen 1.4 m behind the grating.
Required:
What is the distance between the first-order red and blue fringes?
Answer:
0.143 m
Explanation:
Since
d = 1/N = 1/520 = 1.92 * 10^-3 mm
For red light;
θ = sin^-1 (1 * λred/d) = sin^-1 (1 * 656 * 10^-9/1.92 * 10^-6) = 19.98°
L = 1.4 * (tan 19.98) = 0.509 m
For blue light;
θ = sin^-1 (1 * λblue/d) = sin^-1 (1 * 486 * 10^-9/1.92 * 10^-6) = 14.66°
L = 1.4 * (tan 14.66°) = 0.366 m
Distance between the first-order red and blue fringes= 0.509 m - 0.366 m = 0.143 m
when air mass is caught between two cold fronts the result is a _______ front.
Answer choices
A.occluded
B.warm
C.cold
D.stationary
HELP ASAP PLEASE
Which 2 factors must be present for chemical vapor deposition to be successful:
A.) The size of the diamond is larger than the most.
B.) The conditions during cooling are controlled.
C.) The heat in a vacuum forms a gas of single atoms.
D.) The heat in a vacuum is decreased to freezing.
E.) The pressure of a reaction vessel is negative.
A ball is projected at an angle of 53º. If the initial velocity is 48 meters/second, what is the vertical component of the velocity with which it was
launched?
OA. 31 meters/second
OB. 38 meters/second
OC
44 meters/second
OD
55 meters/second
Answer: B
Explanation:
The vertical component of a vector such as velocity is the magnitude of the vector multiplied by the sine of the angle.
[tex]V_y=48*sin(53)=38.3m/s[/tex]
Along the remote Racetrack Playa in Death valley, California, stones sometimes gouge out prominent trails in the desert floor, as if they had been migrating. For years curiosity mounted about why the stones moved. One explanation was that strong winds during the occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about 0.80. What horizontal force is needed on a 25 kg stone (a typical mass) to maintain the stone's motion once a gust has started it moving
Answer:
F = 196 N
Explanation:
For this exercise we will use Newton's second law, we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically
Y axis
N- W = 0
N = mg
X axis
F -fr = ma
In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.
F- fr = 0
F = fr
the friction force has the equation
fr = μ N
fr = μ mg
we substitute
F = μ mg
let's calculate
F = 0.80 9.8 25
F = 196 N
Explain what is happening in this picture
Answer:
in this video waves are coming up for the BOTTOM to the top of the sandbar
Determine the gravitational potential energy, in kJ, of 1 m3 of liquid water at an elevation of 30 m above the surface of Earth. The acceleration of gravity is constant at 9.7 m/s2 and the density of the water is uniform at 1000 kg/m3. Determine the change in gravitational potential energy if the elevation decreases by 10 m.
Answer:
Explanation:
Gravitational potential energy = mgh where m is mass , g is acceleration due to gravity and h is height from the ground .
In the first case mass m = volume x density
= 1 x 1000 = 1000 kg
height h = 30 m
potential energy = 1000 x 30 x 9.8 = 294000 J = 294 kJ .
When height decreases by 10 m , potential decreases as follows .
Decrease in potential energy
= mass x gravitational energy x decrease in height
= 1000 x 9.8 x 10
= 98000 J
= 98 kJ .
what is the difference between heat and temperature
The cart is given an initial push up the ramp. After this push, as the car moves up the ramp, the direction of the acceleration of the cart is ________ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _______.
Answer:
Explanation:
The only force acting on the cart is a component of its weight parallel to ramp downwards . No other force acts parallel to the ramp .
Even when the cart is moving up after the initial push , its weight is acting downwards so acceleration is acting downwards .
When the cart is stationary at the top position , its weight is acting downwards so acceleration is downwards at that moment also . When the cart is going downwards , still its weight is acting down so acceleration is acting downwards .
After this push, as the car moves up the ramp, the direction of the acceleration of the cart is _down _______ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is _down ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _downwards ______.
A circuit has 12 Amps and 220 Volts. What is the Resistance of the circuit?
Answer:
:To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)
To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)
To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)
To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)
How would you compare the acceleration between the unbalanced net force of 100 N and of 50 N
Answer:
The acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.
Explanation:
Given;
first net force, F₁ = 100 N
second net force, F₂ = 50 N
If we consider equal mass for the two net forces, and apply Newton's second law of motion, the acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.
Let a₁ be the acceleration produced by the first net force
then, a₂ be the acceleration produced by the second net force
Thus, a₁ = 2a₂
What Is a Sound Wave? Learning Goal: To understand the nature of a sound wave, including its properties: frequency wavelength, loudness, pitch, and timbre. Sound is a phenomenon that we experience constantly in our everyday life. Therefore, it is important to understand the physical nature of a sound wave and its properties to correct common misconceptions about sound propagation Most generally, a sound wave is a longitudinal wave that propagates in a medium (ie, air) The particles in the medium oscillate back and forth along the direction of motion of the wave. This displacement of the particles generates a sequence of compressions and rarefactions of the medium Thus, a sound wave can also be described in terms of pressure variations that travel through the medium. The pressure fluctuates at the same frequency with which the particles positions oscillate When the human ear perceives sound. It recognizes a series of pressure fluctuations rather than displacements of individual air particles. Part 1 Figure 1 of 2 > Fi MA length Part A Based on the information presented in the introduction of this problem, what is a sound wave? Propagation of sound particles that are offerent from the particles that comprise the medium Propagation of energy that does not require a medium Propagation of pressure fluctuations in a medium Propagation of energy that passes through empty spaces between the partides that com Submit Request Anst Part B Complete previous parts) Part hall to the other? Does air play a role in the propagation View Available Hints) SUITE Part D The graphs shown in (Figure 1) represent pressure variation versus time recorded by Enter the letters of all the correct answers in alphabetical order.
Answer:
A) Propagation of pressure fluctuations in a medium
B) air is the medium in which the wave is transported,
Explanation:
Part A.
A sound wave is a longitudinal oscillation of the molecules that forms in a material medium, they can be solid, liquid or gases, therefore the wave propagates in the same direction as the oscillation of the particles.
The most correct answer is:
* Propagation of pressure fluctuations in a medium
Part b
air is the medium in which the wave is transported, otherwise it cannot propagate
Nikki was walking around a department store shopping one day, and did not realize that the shirt she was wearing looked just like the shirts worn by employees. When a stranger asked, "do you work here," she thought it was funny. The other customers' assumption that Nikki was a store employee demonstrates the Gestalt principle of _______.
Answer: Similarity
By.
A point charge of -11 [Coulombs] is placed inside a spherical conducting shell with net charge of 5 [Coulombs]. Calculate the net charge on the outer surface of the conducting shell. Enter your answer without units (example 100 for 100 [Coulombs] or -100 for -100 [Coulombs] ).
Answer:
20 C
Explanation:
To do this, is pretty easy, we just need to do a little reasoning of what is happening.
When any charge called q is placed inside this metallic shell which is spheric, all the opposite and even equal charges are induced on the inner and outer surface of the shell. Hence, we can say that if in the inner shell we have +q, in the outer will be -q.
Now, here we have the shell with 5 C, and when the charge of -11 C is placed inside the shell we can have the following changes on the inner surface and the outer surface:
Inner surface: +11 C
Outer surface: 9 + 11 = 20 C
Net charge on the outer surface: 20 CHope this helps
A person is holding a bucket by applying a force of 10N. He moves a horizontal distance of 5m and then climbs up a vertical distance of 10m. Find the total work done by him?
Answer:
dgfggddhdbxbxjxddhsnsxnc
Which three statements are true of all matter?
A.
It is filled with air.
B.
It takes up space.
C.
It contains aluminum.
D.
It has mass.
E.
It is made up of atoms
Answer:
B, D and E, not all matter can be filled with air
An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field towards the other end. If they move fast enough when they strike the positive electrode at the other end, they will give up their energy as X-Rays
(a) Through what potential difference should electrons be accelerated so that their speed is 1% of the speed of light?
(b) What potential difference would be needed to give the protons same kinetic energy as electrons?
(c) What speed would this potential difference give to the protons, both in m/s and as a % of the speed of light.
Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV = [tex]\frac{1}{2}[/tex] mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV = [tex]\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }[/tex]
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
[tex]K_e = K_p[/tex]
K_p = ½ m v_e²
K_p = [tex]\frac{1}{2}[/tex] 9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² = [tex]\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }[/tex]
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c = [tex]\frac{7 \ 10^4 }{ 3 \ 10^8}[/tex]
v/c= 2.33 10⁻⁴
True or false it is impossible to determine weather you are moving unless you can touch another object
Answer: false
Explanation:
Answer:
false
Explanation:
According to Newton's First Law of motion, an object remains in the same state of motion unless a resultant force acts on it.
A bartender slides a beer mug at 1.3 m/s towards a customer at the end of a frictionless bar that is 1.3 m tall. The customer makes a grab for the mug and misses, and the mug sails off the end of the bar. (a) How far away from the end of the bar does the mug hit the floor
Answer:
x = 0.67 m
Explanation:
For this problem, let's use the projectile launch equations, as the jug goes through the bar, it comes out with horizontal speed vx = 1.3 m / s, which does not decrease as there is no friction.
Let's find the time or it takes to get to the floor
y = y₀ + v_{oy} - ½ g t²
in this case I go = 0 and when I get to the floor y = 0
0 = y₀ + 0 - ½ g t²
t² = 2y₀ / g
t² = 2 1.3 / 9.8 = 0.2653
t = 0.515 s
now let's find the distance traveled in this time
x = vx t
x = 1.3 0.515
x = 0.6696 m
x = 0.67 m
PHYSICS QUESTION PLS HELP
The coaster starts at rest, so the kinetic energy (KE) at point A is 0. It is situated 33 m above ground, so its potential energy (PE) at A is
mgh = (3000 kg) (9.80 m/s²) (33 m) = 970,200 J
The total energy is the same, 970,200 J.
Assuming no energy is lost to friction or sound etc, energy is conserved throughout the coaster's motion, so the total energy should be the same at each point.
At point B, the coaster has dropped to a height of 10 m, so it has PE
mgh = (3000 kg) (9.80 m/s²) (10 m) = 294,000 J
which means it must have KE
970,200 J = KE + 294,000 J → KE = 676,200 J
which gives the coast a speed v at point B of
1/2 mv ² = 1/2 (3000 kg) v ² = 676,200 J → v ≈ 21.2 m/s
At point C, the coaster has a speed of 16.0 m/s, so it has KE
1/2 mv ² = 1/2 (3000 kg) (16.0 m/s)² = 384,000 J
and hence PE
970,200 J = 384,000 J + PE → PE = 586,200 J
This lets us determine the height h at C:
mgh = (3000 kg) (9.80 m/s²) h = 586,200 J → h ≈ 19.939 m
which means the loop has diameter h - 10 m ≈ 9.94 m.
At point D, the coaster is 15 m above the ground so its PE at D is
mgh = (3000 kg) (9.80 m/s²) (15 m) = 441,000 J
and so its KE is
970,200 J = KE + 441,000 J → KE = 529,200 J
and hence has speed v at D
1/2 mv ² = 1/2 (3000 kg) v ² = 529,200 J → v ≈ 18.9 m/s
A Car is moving at a speed of 20 m/s. How Much Distance it will cover in 1 min? Express the answer in km.
Answer:
d=20m/sx60s=1200m=1200/1000Km=1.2km
Explanation:
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta leading away from the heart. Since blood within the heart is essentially stationary, this pressure difference can be inferred from a measurement of the speed of blood flow in the aorta. Take the speed of sound in stationary blood to be c.
a. Sound sent by a transmitter placed directly inline with the aorta will be reflected back to a receiver and show a frequency shift with each heartbeat. If the maximum speed of blood in the aorta is v, what frequency will the receiver detect? Note that you cannot simply use the textbook Doppler Shift formula because the detector is the same device as the source, receiving sound after reflection.
b. Show that in the limit of low blood velocity (v <
f= 2fo v/c
Answer:
a) f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex] , b) Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo[tex]\frac{c+v}{c}[/tex]
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’ [tex]\frac{c}{ c-v}[/tex]
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀ [tex]\frac{c}{c-v}[/tex]
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex]
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
[tex]( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}[/tex]
f ’’ = fo [tex]( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )[/tex]
f ’’ = fo [tex]( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )[/tex]
leave the linear term
f ’’ = f₀ + f₀ 2[tex]\frac{v}{c}[/tex]
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
keli learned that an air mass is a very large body of air with similar temperature humidity and pressure and the air mass are constantly in motion she knows that you're messing depending on the temperature and moisture content tent of region where they form she looked up more information about what makes them move what are the major causes for moving & Masten North America choose two that apply.
Answer choices
A. changing humidity
B. low temperature
C. jet storm
D. prevailing westerlies
Air masses from the tropics and the equator are warm as they form over lower latitudes. The major causes for moving air masses North America exists jet storm.
What is meant by air mass?An air mass is a volume of air that in meteorology is identified by its temperature and humidity. Many hundreds or thousands of square miles are covered by air masses, which adjust to the properties of the land underneath them. Latitude and their continental or maritime source regions are used to categories them.
Warmer air masses are referred to as tropical, whilst colder air masses are referred to as polar or arctic. Superior and maritime air masses are moist, whereas continental and superior air masses are dry. Air masses with various densities are divided by weather fronts. Once an air mass has left its original location, nearby plants and bodies of water can quickly change the way it behaves. Classification systems address both the properties and modification of an air mass.
Air masses from the tropics and the equator are warm as they form over lower latitudes. They move poleward along the southern edge of the subtropical ridge and are drier and hotter than those that originate over seas. Trade air masses are another name for tropical maritime air masses. The Caribbean Sea, southern Gulf of Mexico, and tropical Atlantic Oceans, east of Florida via the Bahamas, are the origins of maritime tropical air masses that have an impact on the United States.
Monsoon air masses are moist and unstable. Rarely do dry superior air masses touch the ground. A trade wind inversion, which is a warmer and drier layer over the more moderately moist air mass below, is typically created over maritime tropical air masses when they are located above them.
Therefore, the correct answer is option C. jet storm.
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