Answer:
white noise is a bunch of frequencies at the same time :)
Explanation:
Science question, can someone please help ? I’ll give brainliest !!
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]
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
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
Which of these describes a real image?
what 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
30. A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pushing the box with a horizontal force of 35N. What is the magnitude of the force of static friction between the box and floor
Answer:
84.05
Explanation:
F=mg×0.25F=20x9.81×0.25f=49.05NF=35NF=f+F
F=49.05+35
=84.05
please help
a girl pulls a wheeled suitcase with a force of 3N. If the suitcase has a mass of 6 kg, what is the acceleration?
Explanation:
Start with what you know and list your knowns and unknowns
F = ma
F= 3N
m = 6kg
a =?
3N = 6kg x a
solve for a
3N / 6kg = a
_____ 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
plzzzz urgent
solve this
Answer:
[tex] \large{ \tt{☄ \: EXPLANATION}} : [/tex]
Before solving , You'll have to know - When an object starts from the state of rest , in this case , initial velocity ( u ) = 0Notice that we're provided the time ( t ) in minutes. So , first thing we have to do is convert the minutes into seconds. It would be - Time ( t ) = 5 minutes = 5 × 60 sec = 300 sec [ 1 min = 60 sec ]Here , We're provided - Initial velocity ( u ) = 0 , Final velocity ( v ) = 60 m / s , Time taken ( t ) = 300 seconds & We're asked to find out the acceleration ( a ) & distance covered by the jeep ( s ) .[tex] \large{ \tt{♨ \:LET'S \: START}} : [/tex]
Acceleration is defined as the rate of change of velocity. We know :[tex] \large{ \boxed{ \tt{❁ \: ACCELERATION \: (a) = \frac{FINAL \: VELOCITY(v) - INITIAL \: VELOCITY(u)}{TIME \: TAKEN \: ( \: t \: )}}}} [/tex]
- Plug the values & then simplify !
[tex] \large{ \bf{↬a = \frac{60 - 0}{300} = \frac{60}{300} = \boxed{ \bold{ \bf{0.2 \: m {s}^{ - 2} }}} }}[/tex]
The acceleration of the jeep is 0.2 m/s²[tex] \large{ \tt{۵ \: AGAIN, \: USING\: SECOND \: EQUATION \: OF \: MOTION}} : [/tex]
[tex] \boxed{ \large{ \bf{✾ \: s = \frac{u + v}{2} \times t}}}[/tex]
- Plug the values & then simplify !
[tex] \large{ \bf{↦s = \frac{0 + 60}{2} \times 300 = \boxed{ \bold{ \bf{9000 \: m}}}}}[/tex]
The distance covered by the jeep is 9000 m .❃ The days that break you are the days that make you ! ♪
♡ Hope I helped! ツ
☃ Have a wonderful day / evening ! ☼
# StayInAndExplore ! ☂
▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁
Answer:
What she said is right I know it is
A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, causing it to reverse its direction and giving it a velocity of +25 m/s the impulse the stick applies to the park is most nearly
Answer:
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Explanation:
The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:
[tex]I = m\cdot (\vec{v}_{2} - \vec{v_{1}})[/tex] (1)
Where:
[tex]I[/tex] - Impulse, in kilogram-meters per second.
[tex]m[/tex] - Mass, in kilograms.
[tex]\vec{v_{1}}[/tex] - Initial velocity of the hockey park, in meters per second.
[tex]\vec{v_{2}}[/tex] - Final velocity of the hockey park, in meters per second.
If we know that [tex]m = 0.2\,kg[/tex], [tex]\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right][/tex] and [tex]\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right][/tex], then the impulse applied by the stick to the park is approximately:
[tex]I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right][/tex]
[tex]I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right][/tex]
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10 nm. Assume that the walls act like a parallel plate capacitor, each with a charge density of 10-5 C/m2, and the outer wall is positively charged. Although unrealistic, assume that the space between cell wall is filled with air. What is the magnitude of the electric field between the membranes
Answer:
E = 1.1 10⁶ N / C
Explanation:
In this case they indicate that we can approximate the membrane as a parallel plate capacitor, we can use
E = [tex]\frac{\sigma}{\epsilon_o }[/tex]
note that in this case the electric field created by each plate goes in the same direction, they are added
let's calculate
E = [tex]\frac{10^{-5}}{8.85 \ 10^{-12}}[/tex]
E = 1.1 10⁶ N / C
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".
A bike, a truck, and a train—all without passengers, motors, or engines—roll down the same hill. Put the vehicles in order from the least amount of motion energy to the most.
Answer:
Train Bike Truck
Explanation:
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-
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
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 wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same material and length stretch if its cross-sectional area is 8.00 mm2 and the same force is used to stretch it
Answer: [tex]0.05\ mm[/tex]
Explanation:
Given
Cross-sectional area of wire [tex]A_1=4\ mm^2[/tex]
Extension of wire [tex]\delta l=0.1\ mm[/tex]
Extension in a wire is given by
[tex]\Rightarrow \delta l=\dfrac{FL}{AE}[/tex]
where, [tex]E=\text{Youngs modulus}[/tex]
[tex]\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)[/tex]
for same force, length and material
[tex]\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)[/tex]
Divide (i) and (ii)
[tex]\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm[/tex]
car moves a distance of 420 m. Each tire on the car has a diameter of 42 cm. Which shows how many revolutions each
tire makes as they move that distance?
Plzzzz help asap
Answer:
10 is the correct answer
Answer:
Total Distance: 420 meters
Diameter: 42 cm
Notice the units meters vs cm
420÷ 42 = 10 total revolutions
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
what do you mean by work?
Work is transfer of energy in an object when it travel some distance by external force,
Work= force × displacement
what is velocity Write its formula
Answer:
Explanation:
what is the direction of acceleration due to gravity ?
The direction of acceleration due to gravity is always towards earth, going downwards.
mark me brainliesttt :))
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.
A truck starts from rest with an acceleration of 0.3 m/ S^2 find its speed in km/h when it has moves through 150 m
Answer:
9.5 m/s
Explanation:
Distance, S = 150m
Acceleration, a = 0.3 m/s^2
Initial velocity, u = 0 m/s
Final velocity, v
Use kinematics equation
v^2 - u^2 = 2aS
v^2 - 0 = 2*0.3*150 = 90
v = sqrt(90) = 9.49 m/s
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
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]
'
'
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.
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.
A train accelerates from 30 km/h to 45 km/h in 15.0 second. Find its acceleration and the distance it travels during this time
Answer:
a. Acceleration, a = 0.28 m/s²
b. Distance, S = 156 meters
Explanation:
Given the following data;
Initial velocity = 30 km/h
Final velocity = 45 km/h
Time = 15 seconds
a. To find the acceleration;
Conversion:
30 km/h to m/s = 30*1000/3600 = 8.33 m/s
45 km/h to m/s = 45*1000/3600 = 12.5 m/s
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
Substituting into the equation;
[tex]a = \frac{12.5 - 8.3}{15}[/tex]
[tex]a = \frac{4.2}{15}[/tex]
Acceleration, a = 0.28 m/s²
b. To find the distance travelled, we would use the second equation of motion given by the formula;
[tex] S = ut + \frac {1}{2}at^{2}[/tex]
Where;
S represents the displacement or height measured in meters.
u represents the initial velocity measured in meters per seconds.
t represents the time measured in seconds.
a represents acceleration measured in meters per seconds square.
Substituting into the equation, we have;
[tex] S = 8.3*15 + \frac {1}{2}*(0.28)*15^{2}[/tex]
[tex] S = 124.5 + 0.14*225[/tex]
[tex] S = 124.5 + 31.5 [/tex]
S = 156 meters