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 acceleration of a satellite moving in a circular orbit around the earth of radius 2r
Explanation:
You do the radius times the circumference of the earth
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]
a device that spreads light into different wavelengths is a what?
maybe a spectrograph ?
if the forces on an object are balanced the resultant force is equal to zero true false
Answer:
If the forces are balanced, the resultant force is zero. If the forces on an object are unbalanced, this is what happens: a stationary object starts to move in the direction of the resultant force. a moving object changes speed and/or direction in the direction of the resultant force.
Explanation:
How old do you need to be in order to qualify to be a U.S. Senator
Answer: 30 Years Old
Explanation: The constitution has around three qualifications for service in the U.S. Senate, Your age must be at least 30 years.
For this assignment, you should mathematically solve and record a video testing your solution for the following prompt: Two rolls of toilet paper, of equal mass and radius, are dropped from different heights so that they hit the ground at the same time. One roll of toilet paper is dropped normally while the other is dropped while a person holds onto a sheet of toilet paper such that the roll unravels as it descends. Determine the ratio of heights h1/h2, where h1 represents the height of the toilet paper dropped normally and h2 represents the height of the toilet paper that unravels, so that both rolls hit the ground at the same time.
Answer:
h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]
Explanation:
Using two rolls of tissue paper : One roll dropped normally while the other drops as some holds onto a sheet of the toilet paper ( I.e. the tissue paper drops rotating about its axis )
Determine the ratio of heights h1/h2
mass of tissues = same
radius of tissues = same
h1 = height of tissue 1
h2 = height of tissue 2
For the first tissue ( Tissue that dropped manually )
potential energy = kinetic energy
mgh = 1/2 mv^2
therefore the final velocity ( v^2 ) = 2gH ----- ( 1 )
second tissue ( Tissue that dropped while rotating )
gh = [tex]\frac{v^2}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] ) ------ ( 2 )
To determine the ratio of heights we will equate equations 1 and 2
hence :
gh = [tex]\frac{2gH}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] )
∴ h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]
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
Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun's electric field were 2.95 × 104 N/C? (b) What speed would the proton attain if the field accelerated the proton through a distance of 1.26 cm?
Answer:
(A) the acceleration experienced by the proton 2.821 x 10¹² m/s²
(B) the speed of the proton is 2.67 x 10⁵ m/s
Explanation:
Given;
electric field experienced by the proton, E = 2.95 x 10⁴ N/C
charge of proton, Q = 1.6 x 10⁻¹⁹ C
mass of proton, m = 1.673 x 10⁻²⁷ kg
distance moved by the proton, d = 1.26 cm = 0.0126 m
(a)
The force experienced by the proton is calculated as;
F = ma = EQ
where;
a is the acceleration experienced by the proton
[tex]a = \frac{EQ}{m} \\\\a = \frac{2.95\times 10^4 \ \times \ 1.6\times 10^{-19}}{1.673 \times 10^{-27}} \\\\a = 2.821 \times 10^{12} \ m/s^2[/tex]
(b) the speed of the proton is calculated;
v² = u² + 2ad
v² = 0 + (2 x 2.821 x 10¹² x 0.0126)
v² = 7.109 x 10¹⁰
v = √7.109 x 10¹⁰
v = 2.67 x 10⁵ m/s
Explain what is happening in this picture
Answer:
in this video waves are coming up for the BOTTOM to the top of the sandbar
A spring has a spring constant of 25 Newtons per meter. The minimum force required to
stretch the spring 0.20 meter from its equilibrium position is approximately
Answer:
6.3N
Explanation:
Guessed it right on castle learning
Answer:
6.3 N
Explanation:
F=kx
F=(25N/m)(0.25m)
6.3 N
I need help will mark brainliest
Answer: ITS 1 TRUST ME MAN BYE K
Explanation: OK BYE TRUST YEAH
What must the charge (sign and magnitude) of a particle of mass 1.43 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.
Answer:
the sign and magnitude of the charge is - 2 x 10⁻⁵ C.
Explanation:
Given;
mass of the particle, m = 1.43 g = 0.00143 kg
electric field experienced by the particle, E = 700 N/C
The force experienced by the particle is calculated as;
F = mg = EQ
Where;
Q is the magnitude of the charge
[tex]Q = \frac{mg}{E} \\\\Q = \frac{0.00143 \times 9.8}{700} \\\\Q = 2\times 10^{-5} \ C[/tex]
The force must be upward in opposite direction to the electric field. Since the force and the electric field are in opposite direction, the charge must be negative.
Therefore, the sign and magnitude of the charge is - 2 x 10⁻⁵ C.
The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed.
The charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.
What is electric charge?The electric force experienced by the body when placed it into the electromagnetic field is called electric charge.
Given information-
The mass of the particle is 1.43 g or 0.00143 kg.
The magnitude of the downward-directed electric field is 700 N/C.
The magnitude of the free-fall acceleration is 9.80 meter per second squared.
The electric field is defined as the electric force per unit charge. It can be given as,
[tex]E=\dfrac{F}{q}[/tex]
Rewrite the equation for the charge,
[tex]q=\dfrac{F}{E}[/tex]
Force experienced by the particle is equal to the product of mass and free fall acceleration (gravity). Thus,
[tex]q=\dfrac{0.00143\times9.8}{700}\\q=2\times10^{-5}[/tex]
Thus the magnitude of the charge is [tex]2\times10^{-5}[/tex] C.
The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed. For the opposite direction the sign of the charge should be negative.
Thus the charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.
Learn more about the electric charge here;
https://brainly.com/question/14372859
Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 piston with a mass of 10 kg that can slide up and down vertically without friction. Both containers are placed on identical heaters and heated for equal amounts of time.
Required:
a. Will the final temperature of the gas in A be greater, less than, or equal to the temperature in B?
b. Show both processes on a single PV diagram.
c. What are the initial pressures in containers A and B?
d. Suppose the heaters have 25 W of power and are turned on for 15s. What is the final volume of container B?
Answer:
1) Final Temperature of the gas in A will be GREATER than the temperature in B
2) Diagram of both processes on a single PV has been uploaded below
3) The Initial pressures in containers A and B is 3039.87 J/liters
4) the final volume of container B is 923.36 cm³
Explanation:
Given that;
Temperature = 20°C = 293 K
mass of piston = 10 kg
Area = 100cm³
Volume V = 800 cm³ = 0.8 L
ideal gas constant R = 8.3 J/K·mol
1)
Final Temperature of the gas in A will b GREATER than the temperature in B
2)
Diagram of both processes on a single PV has been uploaded below,
3)
Initial pressures in containers A and B
PV = nRT
P = RT/V
we substitute
P = (8.3 × 293) / 0.8
P = 2431.9 / 0.8
P = 3039.87 J/liters
Therefore, The Initial pressures in containers A and B is 3039.87 J/liters
4)
Given that;
power = 25 W
time t = 15s
the final volume of container B = ?
we know that;
work done = power × time
work done = 25 × 15 = 375
Also work done = P( V₂ - V₁ )
so we substitute
375 = 3039.87 ( V₂ - 0.8 )
( V₂ - 0.8 ) = 375 / 3039.87
V₂ - 0.8 = 0.12336
V₂ = 0.12336 + 0.8
V₂ = 0.92336 Litres
V₂ = 923.36 cm³
Therefore, the final volume of container B is 923.36 cm³
a body accelerates uniformly from rest at 2m/s^2 for 5 seconds. Calculate its averege velocity in this time
HERE IS YOUR ANSWER!
calculate the force needed to push the ball up a 4 m ramp if the work is equal to 16 joules.
What are the two rules that light follows.
ok so i dont know srry5
A roller coaster moving along its track rolls into a circular loop of radius r. In the loop, it is only affected by its initial velocity, gravity, and the shape of the track. Let v denote the instantaneous speed and a denote the magnitude of the instantaneous acceleration of the roller coaster in the loop. Which of the following is true in the loop?
a. The roller coaster is not in uniform circular motion, but we still have a=v^2/r everywhere on the loop
b. The roller coaster is not in uniform circular motion, but the tangential acceleration is so small that we can approximate a by v^2/r everywhere on the loop
c. The roller coaster is in uniform circular motion
d. The roller coaster is not in uniform circular motion, and a=v^2/r is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes
Answer:
c. The roller coaster is in uniform circular motion
Explanation:
Since the loop is circular with radius r, and its instantaneous speed, v is always constant, and also, its centripetal acceleration, a' = v²/r.
Since the angular speed, ω = v/r does not change, the magnitude of its tangential acceleration is zero although there is a change in its direction because the direction of its initial velocity changes. That is a" = rα and α = Δω/Δt since Δω = 0, α = 0 and a" = r(0) = 0
So, there is no tangential acceleration. Since there is no tangential acceleration, our instantaneous acceleration which is the vector sum of our centripetal acceleration and tangential acceleration is a = √(a'² + a"²) = √(a'² + 0²) = √a'² = a' = v²/r
So, a is always v²/r.
Since the instantaneous acceleration is always (a = v²/r) constant, the motion is uniform. So, it is uniform circular motion.
The roller coaster is not in uniform circular motion, and [tex]a = \frac{v^2}{r}[/tex] is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes.
The given parameters;
radius of the circular path, = r instantaneous speed = v instantaneous acceleration = aThe motion tension on the loop at the lowest point in the circular motion is given as;
[tex]T = mg + \frac{mv^2}{r}[/tex]
The motion tension on the loop at the highest point in the circular motion is given as;
[tex]T = \frac{mv^2}{r} - mg[/tex]
This shows that circular motion is affected by;
acceleration due to gravity, gradius of the circular path, rspeed of the motion, vmass of the object, mThus, we can that the roller coaster is not in uniform circular motion, and [tex]a = \frac{v^2}{r}[/tex] is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes.
Learn more here:https://brainly.com/question/14672628
The Earth’s orbit _____.
is an ellipse
goes around the moon
is a circle
causes day and night
please help!!!
When a switch is turned from the off to the on position, it is changing the circuit in which of the following ways? O An open circuit is being changed into a closed circuit. A closed circuit is being changed into an open circuit. O A parallel circuit is being changed into a series circuit. A series circuit is being changed into a parallel circuit.
Answer:
i Believe the correct answer is "An open circuit being changed into a closed circuit"
Explanation:
A 20 kg box has an initial velocity of 2 m/s starting at the bottom of a 30-degree inclined plane. A person pushes on the box directly up the frictionless inclined plane so that it travels up the inclined plane at a constant velocity of 2 m/s. Calculate the how much is done by the person after 5 seconds have past.
Answer:
Explanation:
The box is moving with constant velocity so acceleration of box is zero . That means net force on the box is zero .
The weight component acting on box parallel to incline plane
= mg sin 30⁰ = 20 x 9.8 x sin 30 = 98 N
This force is acting down the plane , hence to make the net force zero acting on box , force exerted by person will also be 98 N up the incline .
Force exerted by person = 98 N
distance travelled in 5 s
= velocity x time
= 2 x 5 = 10 m
Work done by person
= 98 x 10
= 980 J .
Two motorcycles are traveling due east with different velocities. However, 5.68 seconds later, they have the same velocity. During this 5.68-second interval, motorcycle A has an average acceleration of 3.87 m/s2 due east, while motorcycle B has an average acceleration of 18.2 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 5.68-second interval, and (b) which motorcycle was moving faster?
Answer:
The answer is below
Explanation:
Let a be the initial velocity of motorcycle A and b be the initial velocity of motorcycle B.
After 5.68 seconds, both motorcycle had the same velocity (v), therefore for motorcycle A:
(a - v) / 5.68 = 3.87
a - v = 21.9816
v = a - 21.9816
For motorcycle B:
(b - v) / 5.68 = 18.2
b - v = 103.376
v = b - 103.376
Therefore:
a - 21.9816 = b - 103.376
b - a = -21.9816 + 103.376
b - a = 81.3944
a) The difference between their speeds at the beginning was 81.3944 m/s
b) Since b - a = 81.3944. This means that the initial velocity of motorcycle B is greater than that of motorcycle A by 81.3944 m/s.
Therefore motorcycle B was moving faster
Balance the following equation:
H3B03 →_B203 +_H20
a. 1, 3,2
b. 2,4,6
C. 4, 2, 6
d. 6, 4,2
It's c I think ( 4 , 2 , 6 , )
What is the shortest time that a jet pilot starting from rest can take to reach Mach-3.60 (3.60 times the speed of sound) without graying out? (Use 331 m/s for the speed of sound in cold air.)
Answer:
30.4 s
Explanation:
A pilot , with plane accelerated at 4 g starts greying out . In the problem , the acceleration of jet is 4 g
a = 4 x 9.8 = 39.2 m /s²
initial velocity u = 0
Final velocity = 3.60 times speed of sound
= 3.6 x 331 = 1191.6 m /s
v = u + at
Putting the values
1191.6 = 0 + 39.2 t
t = 30.4 s .
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
Which statement BEST explains why a bouncing basketball will not remain in motion forever?
Group of answer choices
The energy is transferred to sound and heat energy.
The energy is used up and destroyed.
The energy is transferred to light and potential energy.
The energy is transferred to chemical and heat energy.
Answer:
The energy is transferred to chemical and heat energy.
Explanation:
If you define "bouncing" as leaving the ground for any amount of time, the ball stops bouncing when the elastic energy stored in the compression phase of the bounce is not enough to overcome the weight of the ball. This is the proof of the answer i Hope this helps :)
77. A drag racing vehicle travels from 0 to 100 mph in 5 seconds north. What is the acceleration?
a).004
s2
b).0056 m/s2 c).0079"
d).01 m/s2
M
m
Answer:
a
Explanation:
i just took the test
Starting from the front door of your ranch house, you walk 55.0 m due east to your windmill, turn around, and then slowly walk 35.0 m west to a bench, where you sit and watch the sunrise. It takes you 30.0 s to walk from your house to the windmill and then 36.0 s to walk from the windmill to the bench.
Required:
a. For the entire trip from your front door to the bench, what is your average velocity?
b. For the entire trip from your front door to the bench, what is your average speed?
Answer:
Explanation:
Average velocity = Total displacement / total time
Average speed = total distance covered / total time
a )
For the entire trip from your front door to the bench
Total displacement = 55 - 35 = 20 m [ first displacement is positive and second displacement is negative , because second displacement is in opposite direction ]
Total displacement = 20 m
Total time = 30 + 36 = 66 s
Average velocity = 20 / 66
= .303 m / s
b )
For the entire trip from your front door to the bench
Total distance covered = 55 + 35 = 90 m
Total time = 30 + 36 = 66 s
Average speed = 90 / 66
= 1.36 m / s
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:
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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 ball 12 m in 4 seconds and then 2.5 seconds later it rolls 8 m in 2 seconds what is its acceleration
Answer:
If it accelerates at 20 m/s2 for a period of 22 seconds, what is its final velocity? ... How fast is the ball falling after 5 seconds? v = v0 + gt v = 0 + 10(5) v = 50 m/s. 4. ... + ½ 2.5(15)2 x = 281 m. 5. What is the total displacement of the car in question 2? ... 8. A base jumper falls until he reaches a speed of 200 m/s
Explanation: