100 POINTS AND BRAINLIEST
Answer:
b
Explanation:
The materials are most likely used to change the surface area of the ramp. For example, the sand paper can change the ramp's texture. The wax paper can also change the surface area of the ramp by texture, it can be made smoother.
Answer:
B.
Explanation:
Hope it helped!
Astronauts on a distant planet set up a simple pendulum of length 1.2 m. The pendulum executes simple harmonic motion and makes 100 complete vibrations in 280 s. What is the magnitude of the acceleration due to gravity on this planet
Answer:
If you use P = 2 * pi * (L / g)^1/2 for the period of the simple pendulum
g = 4 * pi^2 * 1.2 / 2.8^2 = 6.04 m/s2
Note: omega = 2 pi * f = 2 pi / P and omega = (g / L)^1/2
Answer:
6.0426 m/s^2
Explanation:
The period of a simple pendulum is equal to 2pi sqroot L/g
T = 280s/100 rev = 2.8s
Plug in 1.2 m for L and 2.8s for T, then solve for g to get 6.04 m/s^2 (dependent on how you round)
A 5.0-m radius playground merry-go-round with a moment of inertia of 1,630 kg m2 is rotating freely with an angular speed of 1.6 rad/s. Two people, each having a mass of 69.5 kg, are standing right outside the edge of the merry-go-round and step on it with negligible speed. What is the angular speed of the merry-go-round right after the two people have stepped on
Answer:
The right solution is "0.511".
Explanation:
Given:
Initial moment of inertia,
= 1630 kg.m²
Radius,
= 5 m
Angular speed,
= 1.6 rad/s
Now,
The moment of inertia after stepping on will be:
= [tex]1630+2\times (69.5\times (5)^2)[/tex]
= [tex]1630+2\times (69.5\times 25)[/tex]
= [tex]5105 \ Kg.m^2[/tex]
hence,
As per the question, the angular speed is conserved, then
⇒ [tex]1630\times 1.6=5105\times \omega'[/tex]
[tex]2608=5105\times \omega'[/tex]
[tex]\omega'=\frac{2608}{5105}[/tex]
[tex]=0.511[/tex]
BRAINLIST A wave travels at a constant speed. How does the wavelength change if the
frequency is reduced by a factor of 3? Assume the speed of the wave remains
unchanged.
A. The wavelength does not change.
B. The wavelength increases by a factor of 3.
C. The wavelength decreases by a factor of 3.
D. The wavelength increases by a factor of 9.
On a rectangle with length 100 m, width 50m and 2 vehicles stationed together. Find the time that they meet given that car A travels 5 m/s and leaves 3 seconds before car B, and car B is traveling at 3 m/s in the opposite direction. Can you create a generic equation from the previous scenario
Answer:
[tex]T=35.625sec[/tex]
Explanation:
From the question we are told that:
Length [tex]L=100 m[/tex]
Width [tex]W=50m[/tex]
Velocity of Car A [tex]V_A=5m/s[/tex]
Velocity of Car B [tex]V_B=3m/s[/tex]
Distance traveled by car A before car B moves
[tex]d_l=5*3[/tex]
[tex]d_l=15[/tex]
Therefore total distance traveled at same time interval
[tex]D=total\ distance-d_l[/tex]
Where
Total distance=Perimeter of rectangle
[tex]P=2(L+B)[/tex]
[tex]P=2(100+50)[/tex]
[tex]P=300[/tex]
Therefore
[tex]D=total\ distance-d_l[/tex]
[tex]D=300-15\\D=285m[/tex]
Generally the equation for time taken to meet is mathematically given by
[tex]T=\frac{Distance D}{Relative\ speed V_r}[/tex]
Where
Relative speed = Speed of car A +Speed of car B
[tex]V_r=V_A+V_B[/tex]
[tex]V_r=5+3[/tex]
[tex]V_r=8m/s[/tex]
Therefore the time taken to meet
[tex]T=\frac{ D}{ V_r}[/tex]
[tex]T=\frac{ 285}{ 8}[/tex]
[tex]T=35.625sec[/tex]
how dose an exam question outed from text book
Answer:
In which school you are???
Explanation:
How are soil and air similar?
Answer:
The air in the soil is similar in composition to that in the atmosphere with the exception of oxygen, carbon dioxide, and water vapor. In soil air as in the atmosphere, nitrogen gas (dinitrogen) comprises about 78%. In the atmosphere, oxygen comprises about 21% and carbon dioxide comprises about 0.036%.
hope this helps
have a good day :)
Explanation:
A 1460 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 18.0 m/s. They stick together. In what direction and with what speed do they move after the collision?
Answer in degrees north of east
AND
the speed after the collision in m/s
Answer:
Solution given:
North car
mass[m1]=1460kg
velocity[u1]=27 m/s
mass[m2]=2165kg
velocity [u2]=18m/s
let v be velocity after collision
we have
From the principle of conservation of linear momentum
m1u1+m2u2=(m1+m2)v
1460*27+2165*18=(1460+2165)v
v=[tex] \frac{78390}{3625} [/tex]
v=21.6m/s
the speed after the collision in 21.6 m/s.
For angle.
Tan angle =[tex] \frac{m1u1}{m2u2} [/tex]
Tan angle =[tex] \frac{1460*27}{2165*18} [/tex]
Tan angle=327.74
angle=Tan-¹(327.74)=89.82=90°
in degrees north of east is 90°
Which of the following is form of energy:
a) Power
b) Light
C) pressure
d) None
Answer:
Explanation:
b) light
Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9520c as measured in the laboratory. What is the magnitudeof the velocity of one particle relative to the other
Answer:
the magnitude of the velocity of one particle relative to the other is 0.9988c
Explanation:
Given the data in the question;
Velocities of the two particles = 0.9520c
Using Lorentz transformation
Let relative velocity be W, so
v[tex]_r[/tex] = ( u + v ) / ( 1 + ( uv / c²) )
since each particle travels with the same speed,
u = v
so
v[tex]_r[/tex] = ( u + u ) / ( 1 + ( u×u / c²) )
v[tex]_r[/tex] = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )
we substitute
v[tex]_r[/tex] = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )
v[tex]_r[/tex] = 1.904c / ( 1 + 0.906304 )
v[tex]_r[/tex] = 1.904c / 1.906304
v[tex]_r[/tex] = 0.9988c
Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c
At the Olympics, high-diving competition,a diver from the top board curves her body in order to?
A. Dive clearly into the water
B. Spin more
C. Increase her energy
D. Spin more slowly
E. Increase her speed
The correct option is E. Increase her speed
High-diving is the process or way in which a person jump into water or swimming pool from a higher heights.
A Diver is a person who dive or Fall into a water from a top board or diving board .
Therefore she was curving her body from the top board in order to Increase her speed which will enable her to spin or flip more faster.
Inconclusion The correct option is E. Increase her speed.
Learn more about Diving here:
https://brainly.com/question/2784286
An old fashioned string of 80 Christmas lights is wired in series. Each bulb has a resistance of 2 Ohms and the entire string is plugged into a 120V outlet. What is the current passing through each of the bulbs?
The sum of the resistance = 2 ohms x 80 lights = 160 ohms.
Current = Total voltage / total resistance:
Current = 120V / 160 ohms
Current = 0.75 Amps
An electron that has an instantaneous velocity of is moving through the uniform magnetic field (a) Find the force on the electron due to the magnetic field. (b) Repeat your calculation for a proton having the same velocity
Answer:
Explanation:
From the given information:
The missing value are as follows:
The velocity [tex]v ^{\to} = (2.0 \times 10^6 \ m/s) \hat i + (3.0 \times 10^6 \ m.s) \hat j[/tex]
The uniform magnetic field [tex]B^{\to} = (0.030 T)\hat i -(0.15 T) \hat j[/tex]
The force on electron as a result of the Magnetic field is:
[tex]F^\to = q(v^\to \times B^\to)[/tex]
here;
Change of electron [tex]q = -1.6 \times 10^{-19} C[/tex]
Then,
[tex]v^{\to} \times B^{\to} = \left|\begin{array}{ccc}2\times10^6&3\times 10^6&0\\0.03&-0.15&0\end{array}\right|[/tex]
[tex]= \hat i(0-0) -\hat j ( 0-0) +\hat k (-3\times 10^5 -0.9\times 10^5) \\ \\ =-3.9 \times 10^5 \ k[/tex]
∴
[tex]F^\to = q(v^\to \times B^\to)[/tex]
[tex]F^{\to} = -1.6 \times 10^{-19} \times -3.9 \times 10^5 \\ \\ F^{\to} = 6.24 \times 10^{-14 } \ N[/tex]
For the proton with the same velocity:
[tex]q = 1.6 \times 10^{-19 } \ C \\ \\ F = q (v^\to * B^\to) \ \\ \\ F^\to = q(-3.9 \times 10^5) \\ \\ F = (1.6 \times 10^{-19}) (-3.9 \times 10^5) \\ \\ \mathbf{F ^{\to} = -6.24 \times 10^{-14} \ N}[/tex]
What is the speed of a wave if it has a wavelength of
42 m and a frequency of 7 hertz?
Answer:
♕ [tex]\large{ \red{ \tt{Step - By - Step \: Explanation}}}[/tex]
☃ [tex] \underline{ \underline{ \blue{ \large{ \tt{G \: I \: V \: E\: N}}}}} : [/tex]
Frequency ( f ) = 7 HertzWavelength ( λ ) = 42m♨ [tex] \underline {\underline{ \orange{ \large{ \tt{T \: O \: \: F \: I \: N\: D}}}} }: [/tex]
Wave velocity ( v )☄ [tex]\underline{ \underline{ \large{ \pink{ \tt{S\: O \: L \: U \: T\: I \: O \: N}}}}}: [/tex]
✧ [tex] \red{ \boxed{ \large{ \purple{ \sf{Wave \: velocity(v) = Frequency(f) \times Wavelength(λ)}}}}}[/tex]
~Plug the known values and then multiply!
↦ [tex] \large{ \tt{7 \times 42}}[/tex]
↦ [tex] \boxed{ \boxed{ \large{ \bold{ \tt{294 \: m {s}^ {- 1} }}}}}[/tex]
☥ [tex] \large{ \boxed{ \boxed{ \large{ \tt{Our \: Final \: Answer : \underline{ \large{ \tt{294 \: m {s}^{ - 1}}}}}}}}} [/tex]
---------------------------------------------------------------
❁ [tex] \underline{ \large{ \red{ \tt{D\: E\: T \: A \: I \: L\: E \: D \: \: I\: N \: F \: O}}}} : [/tex]
Frequency ( f ) : The number of complete waves , set up in a medium in one second is called frequency of the wave. The SI unit of frequency is Hertz ( Hz ). For example : if a sound wave completes 15 compressions and 15 rarefactions in one second , it's frequency is 15 Hz.Wavelength ( λ ) : The distance between two consecutive troughs or crests in a transverse wave or the distance between two consecutive compressions or rarefactions in a longitudinal wave us called wavelength. It is the distance travelled by a wave in a time equal to it's time period. It's SI unit is metre ( m ).Wave velocity ( v ) : The velocity with which a wave propagates in a medium is called wave velocity. It's SI unit is m/s.# KILL : Excuses
KISS : Opportunities
MARRY : Goals
♪ Hope I helped! ♡
☂ Have a wonderful day / night ! ツ
✎ [tex] \underbrace{ \overbrace{ \mathfrak{Carry \: On \: Learning}}}[/tex] ✔
▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁
Please help!!!!!!!!!
Answer:
try S or Q
Explanation:
During forced exhalation, such as when blowing up a balloon, the diaphragm and chest muscles create a pressure of 72.0 mm Hg between the lungs and chest wall. What force in newtons does this pressure create on the 605 cm2 surface area of the diaphragm
Answer:
F = 580.75 N
Explanation:
The force can be calculated through the following formula:
[tex]F = PA[/tex]
where,
F = Force = ?
P = Pressure = (72 mm Hg)(133.322 Pa/1 mm Hg) = 9599.18 Pa
A = Area = (605 cm²)(1 m²/10⁴ cm²) = 0.0605 m²
Therefore,
[tex]F = (9599.18\ Pa)(0.0605\ m^2)[/tex]
F = 580.75 N
An ideal massless spring with a spring constant of 2.00 N/m is attached to an object of 75.0 g. The system has a small amount of damping. If the amplitude of the oscillations decreases from 10.0 mm to 5.00 mm in 15.0 s, what is the magnitude of the damping constant b
Answer: 0.00693
Explanation:
Given
Spring constant [tex]k=2\ N/m[/tex]
Mass of object [tex]m=75\ g[/tex]
The amplitude of the oscillation decreases from 10 mm to 5 mm in 15 s
Equation of amplitude for the ideal spring-mass system is
[tex]\Rightarrow A=A_oe^{-\frac{bt}{2m}}\quad \quad [\text{b=damping constant}]\\\text{Insert the values}\\\\\Rightarrow 5=10e^{\frac{b\times 15}{2\times 0.075}}\\\\\Rightarrow e^{-\frac{b\times 15}{2\times 0.075}}=0.5\\\\\text{Taking natural log both sides}\\\\\Rightarrow \ln \left(e^{-\frac{b\times 15}{2\times 0.075}}\right)=\ln 0.5\\\\\Rightarrow -\dfrac{15b}{0.15}=-0.693\\\\\Rightarrow b=0.00693[/tex]
The atmospheric features of Neptune are easier to see than those of Uranus because A. Neptune has greater warmth and less haze. B. Neptune has more methane. C. The atmosphere of Uranus rotates differentially. D. Uranus has no significant atmosphere.
Answer:
Option B is the correct answer (Neptune has more methane)Explanation:
From the options given,
The atmospheric features of Neptune are easier to see than those of Uranus because Neptune has more methane
Neptune has small amount of methane and water which gives it blue colour and white patches which distinguish it from uranus
For more information, visit
http://abyss.uoregon.edu/~js/ast121/lectures/lec20.html
4- What force must be applied to a surface area of 0.0025m , to create a pressure ol
200.000Pa?
At what speed does light travel through water. n=1.33
Answer:
"In water, the speed of light is v01.33=2.26 x 108 m/s."
Explanation:
5. Stopping a fast-moving object is harder than stopping a slow-moving
one.
True
False
A retired bank president can easily read the fine print of the financial page when the newspaper is held no closer than arm's length, 59.1 cm from the eye. What should be the focal length of an eyeglass lens that will allow her to read at the more comfortable distance of
Answer:
Explanation:
comfortable distance is 25 cm .
He must be using convex lens . In that case rays coming from object placed at 25 cm appears to be coming from 59.1 cm due to converging nature of convex lens.
object distance u = -25 cm
image distance v = -59.1 cm
Lens formula
1 / v - 1 /u = 1 /f
-1 / 59.1 + 1 / 25 = 1/f
- .0169 + .04 = 1 / f
.0231 = 1 / f
f = 43.3 m
When playing tennis, if you hit the ball off of the top end of the racket, an uncomfortable standing wave vibration is produced in the racket. This wave is
Answer:
Transverse.
Explanation:
Electromagnetic waves is a propagating medium used in all communications device to transmit data (messages) from the device of the sender to the device of the receiver.
An electromagnetic spectrum refers to a range of frequency and wavelength that an electromagnetic wave is distributed or extends. The electromagnetic spectrum comprises of gamma rays, visible light, ultraviolet radiation, x-rays, radio waves, and infrared radiation.
A transverse wave can be defined as a type of wave in which particles of the medium of propagation oscillates or vibrates in a direction that is perpendicular to the direction that the wave moves i.e at right angle to the direction of propagation of the wave.
Basically, sound is a transverse and all transverse wave are the direct opposite of a longitudinal wave that usually travel in the same direction of its oscillation.
Hence, this wave is transverse.
Why is it important for equipment for sport to be strong? To protect us
Answer:
To protect us.
Explanation:
For ex. your dunking on a basketball hoop if that wasn't strong you would fall on your back and get injured.
Can someone help me
Look at the Position vs. Time and Velocity vs. Time plots. What is the person's velocity when his position is at its maximum value (around 6 m )
Answer:
The person's velocity is zero.
Explanation:
An athlete is performing squats in the weight room. The knee is going from anatomical position to 92 degrees and then back to anatomical position each squat. The athlete performs a total of 10 squats. This is done over a time period of 30 seconds. What is the angular acceleration (rad/sec 2) of the knee
Answer:
α = 0.357 ras / s²
Explanation:
This is a rotational kinematics exercise, it tells us that it performs 10 squats in 30 s, for which it performs one squat at t = 3 s, also indicates that the angle of the squat is θ = 92º
θ = θ₀ + w₀ t + ½ α t²
the athlete starts from rest, whereby w₀ = 0 and the initial angle in the vertical position is zero (θ₀=0)
θ = ½ α t²
α = 2 θ /t²
let's reduce the magnitudes to the SI system
θ = 92º (π rad /180º) = 0.511π rad
let's calculate
α = 2 0.5111π /3²
α = 0.1136π rad / s²
α = 0.357 ras / s²
What effect does the Duck Velocity have on the waves seen by the observer?
Towards the boat:
Away from the boat:
Same as the boat:
The average velocity of blood flowing in a certain 4-mm-diameter artery in the human body is 0.28 m/s. The viscosity and density of blood are approximately 4 cP and 1.06 Mg/m3, respectively. Determine the volumetric flow rate of blood in the artery. (m3/s)
Answer:
V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s
Explanation:
The volume flow rate of the blood in the artery can be given by the following formula:
[tex]V = Av[/tex]
where,
V = Volume flow rate = ?
A = cross-sectional area of artery = πd²/4 = π(0.004 m)²/4 = 1.26 x 10⁻⁵ m²
v = velcoity = 0.28 m/s
Therefore,
[tex]V = (1.26\ x\ 10^{-5}\ m^2)(0.28\ m/s)[/tex]
V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s
A 70 kg stunt pilot begins pulling out of a dive into a vertical circle. If the plane's speed at the lowest point of the circle is 80 m/s, what is the apparent
weight of the pilot at the lowest point of the pullout? The pilot experiences a force of 5g from the centripetal acceleration at the bottom of the dive.
O
3880 N
о
3430 N
0
4116 N
3986 N
Answer:
Apparent weight of pilot due to centripetal acceleration:
m v^2 / R = 5 g m = 5 * 70 * 9.8 = 3430 N
Weight of pilot = 70 * 9.8 = 686 N
Total = 3430 + 686 = 4116 N