Relativistic kinetic energy is KErel = (γ − 1)mc2. γ=1+1v22c2 γ = 1 + 1 v 2 2 c 2 . A binomial expansion is a way of expressing an algebraic quantity as a sum of an infinite series of terms.
Hope This Helps Youwhat is the size of the electron charge?
Answer:
It’s like medium uk
Explanation:
The size of electron charge is estimated to be 1.6 × 10-¹⁹
How do you find a wavelength?
In which stage of a star's life cycle, does gravity and fusion become balanced, and an adult star forms?
A. Main-sequence
B. Red supergiant
C. Protostar
D. White dwarf
Answer:
im guessing red giant-
A tennis ball hits with an initial velocity of 25 m/s [10 o above the
horizontal] starting from a height of 2.1 m, lands on the ground. There
is a net 15 m away and 0.9 m high. Will the ball clear the net at its
peak, and by how much?
First, calculate the components of its initial velocity v0:
v0x = v0cos10 = 24.6 m/s
v0y = v0sin10 = 4.34 m/s
The ball reaches its peak when vy = 0. Let's calculate the time it takes for vy to become zero:
vy = v0y - gt ---> t = v0y/g = 0.44 s
The horizontal distance it travels in this time is
x = v0xt = (24.6 m/s)(0.44 s)
= 10.8 m
Note that the net is 15 m away. After traveling a horizontal distance of 10.8 m, the height of the ball is
y = -(1/2)gt^2 + v0yt + 2.1
= -(4.9 m/s^2)(0.44 s)^2 + (4.34 m/s)(0.44 s) + 2.1 m
= -0.95 m + 1.9 m + 2.1 m
= 3.05 m
Note that this is the height of the ball at its peak. While the ball is well above the net at its peak, it is well short of its required horizontal distance to clear it. Instead, let's find the time it takes for the tennis ball to travel a horizontal distance of 15 m first:
x = v0xt ----> t = x/v0x = (15 m)/(24.6 m/s) = 0.61 s
Then calculate the height y when t = 0.61 s. If y > 0.9 m (height of the net), then the ball will clear the net.
y = -4.9t^2 + v0yt + 2.1
= -4.9(0.61 s)^2 + (4.34 m/s)(0.61 s) + 2.1 m
= -1.82 m + 2.65 m + 2.1 m
= 2.93 m
Yes, the ball will clear the net by 2.03 m.
Quanto tempo deve ficar ligado um ferro eletrico de 1000 w para que tenha o mesmo consumo de energia que um chuveiro de 4400 w que fica ligado 10 minutos
Answer:
Thus, the time for the first lamp is 44 minutes.
Explanation:
Power of first lamp, P' = 1000 W
Power of second lamp, P'' = 4400 W
time for second lamp, t'' = 10 minutes
Let the time for first lamp is t'.
As the energy is same, so,
P' x t' = P'' x t''
1000 x t' = 4400 x 10
t' = 44 minutes
what is velocity Write its formula
Answer:
Explanation:
A student was asked to explain how a prism separates white light into its separate colors. The student explanation is below but contains a few mistakes that have been underlined. Re-write the student explanation and correct any errors that you find.
When a beam of white light enters a glass prism, the light waves increase in speed which causes them to bend and reflect inside the glass. White light is actually made up of a combination of different frequencies of light that bend at the same angle. When the light leaves the prism, it bends again, and it is separated into each different color and frequency forming a spectrum that looks like a rainbow.
Answer:
pp i hate khan academy pls help me
The Keck Observatory is home to the largest Earth-based twin telescopes. The telescopes are located away from city lights, above the clouds where the atmosphere is clear, calm, and dry. In 1999, additional steps were taken to help the telescopes overcome any distortions caused by changes in the atmosphere by adding a laser sighting system. Now, the Keck telescopes can probe distant galaxies and capture images with more detail than even the Hubble Space Telescope. Scientists can now use the Keck telescopes to investigate many questions. What is one question the Keck telescopes would NOT help scientists answer
Answer:
Option D
Explanation:
The options for the question are
a) How do galaxies rotate?
b) What is the weather on Neptune?
c) What is the core of Saturn made of?
d) What other solar systems have planets?
Solution
The Hubble space telescope was designed and integrated into the extraterrestrial system in order to capture information about the surrounding universe. If the Keck Observatory has a better observation capacity than the Hubble space telescope then the scientist would be interested to know the surrounding planets in the solar system.
Hence, option D is correct
A football player kicks a football in a field goal attempt. When the football reaches its maximum height, what is the relationship between the direction of the velocity and acceleration vectors
Answer:
The correct answer is - At the maximum height, the velocity and acceleration vectors are perpendicular to each other.
Explanation:
When the football reaches maximum height, then the vertical component of velocity will be zero and therefore the only component of velocity left will be the horizontal component, and acceleration of the object will be downward, due to gravity.
So at the maximum height, there is horizontal velocity only which means velocity is horizontal and acceleration is vertical thus, the velocity and acceleration are perpendicular to each other.
Which two changes would decrease the gravitational force between two
objects?
A. Increase the distance between the objects.
B. Increase the mass of one of the objects.
C. Decrease the distance between the objects.
D. Decrease the mass of one of the objects.
E. Increase the mass of both objects.
Answer:
A. Increase the distance between the objects.
D. Decrease the mass of one of the objects.
Explanation:
The smaller the mass of an object the less gravity it has and the farther away two objects are the less gravitational force
A table is moved using 60 N of force.
How far is the table moved if 900 J of work is done on the table?
960 m
840 m
15 m
0.06 m
Answer:
15m
Explanation:
W=f×s
[tex]s = \frac{w}{f} \\ s = \frac{900j}{60n} \\ s = 15m [/tex]
Given:
Force,
F = 60 NWork done,
W = 900 JWe know,
→ [tex]W = F\times s[/tex]
or,
→ [tex]s = \frac{W}{F}[/tex]
By putting the values, we get
[tex]= \frac{900}{60}[/tex]
[tex]= 15 \ m[/tex]
Thus the above response i.e., "Option c" is right.
Learn more about work done here:
https://brainly.com/question/8625856
which of the two bodies a and b in the following graph is moving with higher speed and why
.........................................
pls say it fst its very urgent
name few biodegradable materials that get decomposed in a week
Answer:
Cardboard, Paper towels, Food waste, and wooden based items
Answer:
How fast do things biodegrade?
Vegetables 5 days –1 month
Aluminium cans 80–100 years
Glass bottles 1 million years
Styrofoam cup 500 years to forever
Plastic bags 500 years to forever
Determine the amount of work done by the applied force when a 87 N force is applied to move a 15 kg object a horizontal
distance of 4.5 meters at a constant speed.
Answer:
391.5 J
Explanation:
The amount of work done can be calculated using the formula:
W = F║d where the force is parallel to the displacementLooking at the formula, we can see that the mass of the object does not affect the work done on it.
Substitute the force applied and the displacement of the object into the equation.
W = (87 N)(4.5 m) W = 391.5 JThe amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.
Mass doesn't matter on amount of work done .We can calculate amount of work done through Force and Displacement
Force=87NDisplacement=4.5m[tex]\boxed{\sf W=Fd}[/tex]
W denotes to work doneF denotes to forced denotes to displacement[tex]\\ \sf\longmapsto Work\:Done=87(4.5)[/tex]
[tex]\\ \sf\longmapsto Work\:done=391.5J[/tex]
Suppose that scientists observe violent gas eruptions on a planet with an acceleration due to gravity of 3.9 m/s2. The jets throw sand and dust about 75 m above the surface. What is the speed i of the material just as it leaves the surface?
Answer:
The velocity with which the sand throw is 24.2 m/s.
Explanation:
acceleration due to gravity, a = 3.9 m/s2
height, h = 75 m
final velocity, v = 0
Let the initial velocity at the time of throw is u.
Use third equation of motion
[tex]v^{2} = u^{2} - 2 g h \\\\0 = u^{2} -2\times 3.9\times 75\\\\u=24.2m/s[/tex]
The velocity with which the sand throw is 24.2 m/s.
In a pinball machine, the launching spring with a spring constant of 30 N/m is compressed 0.15 m. How fast will it launch a 0.20 kg pinball?
Answer:
1.84 m/s
Explanation:
Applying,
The kinetic energy of the pinball = Elastic energy of the spring
mv²/2 = ke²/2
mv² = ke².................. Equation 1
Where m = mass of the pin ball, v = velocity of the pin ball, k = force constant of the spring, e = extension of the spring.
make v the subject of the equation
v = √(ke²/m)................ Equation 2
From the question,
Given: e = 0.15 m, k = 30 N/m, m = 0.20 kg
Substitute these values into equation 2
v = √[(30×0.15²)/0.2]
v = 1.84 m/s
A 5 kg mass compresses a horizontal spring by .06 meters. The spring has a spring constant of 2 N/m. If the surface is frictionless, find the velocity of the mass when the spring is released.
Answer:
Explanation:
The frictionless surface implies that the speed of the spring is at a max. When the speed of the spring is at its max, the potential energy in the spring is 0. Use the equation for the Total Energy in a Spring/Mass System:
KE + PE = [tex]\frac{1}{2}kA^2[/tex] where KE is the Kinetic Energy available to the spring, PE is the potential energy available to the spring, and the sum of those is equal to one-half times the spring constant, k, times the amplitude of the spring's movement away from the equilibrium position. Sometimes this amplitude is the same as the displacement of the spring. This can be tricky. But since we are only given one value for the distance, we are going to use it as an amplitude. Keeping in mind that the PE is 0 when KE is at its max, then the equation becomes
KE + 0 = [tex]\frac{1}{2}kA^2[/tex] or to put it simpler terms:
KE = [tex]\frac{1}{2}kA^2[/tex] We need to find the value for KE before we can fully solve the problem we are being tasked with.
Filling in using the info given:
[tex]KE=\frac{1}{2}(2.0)(.06)^2[/tex] Notice I added another place of significance to the 2 because 1 simply isn't enough and the physics teacher in me can't handle that. Simplifying a bit:
[tex]KE=(.06)^2[/tex] because the k = 2 cancels out the 2 in the denominator of the 1/2. So
KE = 3.6 × [tex]10^{-3[/tex]
Now plug that in for KE and solve for v:
KE = [tex]\frac{1}{2}mv^2[/tex]:
[tex]3.6*10^{-3}=\frac{1}{2}(5.0)v^2[/tex] and
[tex]v=\sqrt{\frac{2(3.6*10^{-3})}{5.0} }[/tex] gives us a velocity of
v= [tex]3.8*10^{-2[/tex]
Q:
The kinetic energy of an object can be determined
using the equation KE = 1/2mv^2, where ke is the
kinetic energy, m is the mass, and v is the speed of
the object. What is the mass of an object moving at
a speed of 12m/s that has 36 J of kinetic energy?
A:
0
2.0 kg
8.0 kg
0.13 kg
0.50 kg
Answer:
Explanation:
We will use the KE equation you wrote here and fill in what we are given:
[tex]36=\frac{1}{2}m(12)^2[/tex] and isolating the m:
[tex]m=\frac{2(36)}{12^2}[/tex] which gives us
m = .50 kg
The relationship between the decay constant (λ) and the half-life (t1⁄2) is given by the equation t(1/2) = 0.693/λ. Use the equation for N to derive this relationship.
Hint: when t = t(1/2), N/No = 0.5 (1 point)
A sample of carbon-14 initially consists of 5 × 1024 particles. Carbon-14 has a half-life of 5730 years.
a. What is the decay constant for carbon-14? (Answer in units of yr–1.) (1 point)
b. How many radioactive particles of the sample remain after 100 years? (1 point)
c. What percentage of the radioactive particles remains after 500 years? (1 point)
d. How many radioactive particles of the sample remain after 1000 years? (1 point)
e. How much time will it take for 50% of the particles to decay? (1 point)
f. How much time will it take for 99% of the particles to decay? (1 point)
g. How many half-lives will it take for 99% of the sample to decay? (1 point)
h. What is the initial decay rate of the sample? (Answer in decays/yr.) (1 point)
i. After 200 years, what is the decay rate of the sample? (1 point)
j. How long will it take for the decay rate to decrease to 1015 decays/year? (1 point)
k. How many half-lives have passed after the time you found in part (j)? (1 point)
Answer: A: 70/2=35
B: 35/2=17
C: 17.5/2=8.75
D: 8.75 of C-14 will be left
E: 5,730 years
F: 5,631
Explanation:
that’s all I got, hope I helped kinda
What kind of waves are present during an earthquake?
a. Transverse
b. Longitudinal
c. Both transverse and longitudinal
Answer:
Explanation:
During an earthquake, there are both transverse and longitudinal wave. The answer is c.
Answer:
Explanation:
ans is c. Both transverse and longitudinal
When a wave steepens until it collapses it becomes a ________. wave of oscillation forced wave breaker wave of translation swell
Answer:
Breaker
Explanation: A wave is a motion or disturbance that transfers energy from one location to another without any permanent displacement of the particles involved in the wave motion.
Wave motion can occur in various media such as water, air etc a wave is described as a breaker when it steepens and before it finally stops or losses its energy/collapses.
what is fire proof plastic
Explanation:
.Flame retar,dant plastic additi,ves are compounds added to plastics and other materials to inhibit, suppress or de,lay combustion. These compounds are useful in impending burning in the ign,ition phase of fire. They do not prevent char,ring or melting nor do they increase the heat resistance of a material
A 56.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.517 and 0.260, respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed
Answer:
Explanation:
Magnitude of frictional force = μ mg
μ is either static or kinetic friction.
To start the crate moving , static friction is calculated .
a ) To start crate moving , force required = μ mg where μ is coefficient of static friction .
force required =.517 x 56.6 x 9.8 = 286.76 N .
b ) to slide the crate across the dock at a constant speed , force required
= μ mg where μ is coefficient of kinetic friction , where μ is kinetic friction
= .26 x 56.6 x 9.8 = 144.21 N .
A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height of 9m. Find the work required to empty the tank by pumping all of the water over the top of the tank. Use the fact that acceleration due to gravity is 9.8 m/sec2 and the density of water is 1000kg/m3. Round your answer to the nearest kilojoule.
Answer:
[tex]W=17085KJ[/tex]
Explanation:
From the question we are told that:
Height [tex]H=16m[/tex]
Radius [tex]R=3[/tex]
Height of water [tex]H_w=9m[/tex]
Gravity [tex]g=9.8m/s[/tex]
Density of water [tex]\rho=1000kg/m^3[/tex]
Generally the equation for Volume of water is mathematically given by
[tex]dv=\pi*r^2dy[/tex]
[tex]dv=\frac{\piR^2}{H^2}(H-y)^2dy[/tex]
Where
y is a random height taken to define dv
Generally the equation for Work done to pump water is mathematically given by
[tex]dw=(pdv)g (H-y)[/tex]
Substituting dv
[tex]dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)[/tex]
[tex]dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy[/tex]
Therefore
[tex]W=\int dw[/tex]
[tex]W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy[/tex]
[tex]W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)[/tex]
[tex]W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0[/tex]
[tex]W=3420.84*0.25[2401-65536][/tex]
[tex]W=17084965.5J[/tex]
[tex]W=17085KJ[/tex]
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_____ Rocks are porous.
A) Sedimentary
B) Metamorphic
C) igneous
Answer:
Sedimentary rocks are porous.
Explanation:
So option A) Sedimentary is your answer. Have a great summer
As a ball is released from 10ft above the ground, it falls freely (without friction)
toward the ground, at what point does the ball have the maximum gravitational
potential energy?
Can't be determined
At the top, at the release point.
Half
way
down the ground
At the floor
Answer:
Explanation:
The formula to find Potential Energy is PE = mgh, where m is the mass of the object, g is gravity, and h is the height from which the object can potentially fall. Because this is linear, then PE will increase as either the mass or the height increase (gravity is constant at 9.8 m/s/s). If the mass of the ball being dropped doesn't change, then the only thing that determines this ball's max PE is the height from which it is dropped; max PE ALWAYS occurs at the highest point from which an object can potentially fall. So your answer is "At the top".
How to keep my family safe during a hurricane?
If you are not advised to evacuate or are unable to do so safely, stay indoors, away from windows, skylights and doors. Continue to monitor weather reports and do not go outside until the storm has passed. And, to protect your children from seeing too many sights and images of the hurricane, limit media exposure.
A 15-watt bulb is connected to a circuit that has a total of 60. Ω of resistance. How many electrons are passing through that bulb every second?
Answer:
3.2075*10^16
Explanation:
Q=P/V just search up a converter and youll get 30V and so you do 15/30 which is a half and a single coulomb is 6.415*10^16 so you half it. I belive this is correct if you dont belive me wait for someone else smarter to answer and compare.
Suppose you have 10 helium-filled party balloons tied to a digital camera, so you can take photos of the core of a storm. With 10 balloons the camera will slowly but surely rise. As a storm approaches the atmospheric pressure drops. Assuming the air temperature and density are the same, the volume of each balloon will increase. When you let go, the speed of ascent will be
Answer:
a = 10 ρ_air g [tex]\frac{\Delta P}{m P_o}[/tex]
Explanation:
Let's solve this problem in parts, with the initial data the camera rises slowly, so we can assume that at constant speed, we apply the equilibrium condition
B - W = 0
The thrust is given by Archimedes' law
B = rho_aire g V_body
The volume of the body can be found from the ideal gas ratio
P V = n R T
let's use the subscript "o" for the initial concisions
V₀ = (nR) T₀/P₀ 1
we substitute
10 ρ_air g (nR) T₀ /P₀ = W
when the storm approaches the pressure decreases, P
If we use the ideal gas equation
V = (nR) T₀ / P
we combine this equation with equation 1
V = V₀P₀ / P
if we write the pressure
P = P₀ -ΔP
we substitute
V = [tex]\frac{V_oP_o}{P_o - \Delta P} =V_o \ ( 1 - \frac{\Delta P}{P_o} )^{-1}[/tex]
we expand serie and eliminate higher order terms
V = V₀ ([tex]1+ \frac{\Delta P}{Po}[/tex])
with this expression we can write the thrust
B = B₀ + ΔB
Newton's second law for the new conditions is
B - W = m a
(B₀ + ΔB) - W = ma
ΔB = m a
a = ΔB / m
a = 10 ρ_air g [tex]\frac{\Delta P}{m P_o}[/tex]
This is the initial acceleration of the camera
Science question, can someone please help ? I’ll give brainliest !!