Answers
Answer:
Find answers below.
Explanation:
A graph can be defined as the graphical representation of data (informations) on horizontal and vertical lines i.e x-axis and y-axis respectively.
On the x-coordinates, we have the following points or values;
1. Smiley face = 2
2. Diamond = 2
3. Sun = 0
4. Heart = 3.75
On the y-coordinates, we have the following points or values;
1. Smiley face = 1
2. Diamond = 3.75
3. Sun = 3.75
4. Heart = 0
Note: to read the points on a graph, you would look at the exact points marked on the x-coordinates (x-axis) and y-coordinates (y-axis) respectively.
The scale used for this graph is 0.75 for each unit on the x-coordinates and y-coordinates respectively.
Related Questions
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery.
Answers
Complete question:
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.
Answer:
The value of its capacitance is 1.027 x 10⁻¹² F
Explanation:
Given;
area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²
separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m
voltage of the battery, V = 18 V
The value of its capacitance is calculated as;
[tex]C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F[/tex]
Therefore, the value of its capacitance is 1.027 x 10⁻¹² F
Does the same battery always deliver the same amount of flow to any circuit? Mention two observations of any circuits in this lab that support your answer. Explain.
Answers
Answer:
Yes
Explanation:
Given that the battery is the same the PD ( potential difference ) in the circuit will also be the same likewise the flow of charge in the circuit,
Hence the same amount of charge flow is delivered to any circuit.
attached below are examples
What are 3 artificial and 2 natural sources of electromagnetic radiation?
Answers
Answer: its b bro
Explanation:
ajafa'jfbA'FJ
Radiation exerts pressure on surfaces on which it lalls (radintion pressure). Will this pressure be greater on a shiny surface or a dark surface
Answers
Answer:
Shiny surface.
Explanation:
We know that radiation pressure is the pressure over a surface exposed to electromagnetic radiation.
Where if the radiation is absorbed by the material (like in the case of a dark surface), the pressure is the energy density flux divided by the speed of light, while if the radiation is totally reflected (idealized case, but we can suppose that this happens for a shiny surface) the pressure is twice pressure for the absorbed case.
This is a simplification for the radiation pressure but is enough to conclude that the radiation pressure is always greater on reflective surfaces, then for this case, the pressure will be greater on a shiny surface than in a dark surface,
A photon with a frequency of 5.02 × 1014 hertz is absorbed by an excited hydrogen atom. This causes the electron to be ejected from the atom, forming an ion. Calculate the energy of this photon in joules. [Show all work, including the equation and substitution with units.] Determine the energy of this photon in electron-volts. What is the number of the lowest energy level (closest to the ground state) of a hydrogen atom that contains an electron that would be ejected by the absorption of this photon?
Answers
Answer:
Explanation:
An atom emits a photon (particle of light) when transitioning from a ground state to its excited state. To obey conservation of energy, the energy gained by the atom when an electron moves to a lower energy level is equal to the energy it loses in emitting the photon. (The energy of a photon is E = hf, where E is the energy, h is Planck's constant, and f is the frequency of the photon.) Conversely, when an atom absorbs a photon (as is the case in absorption spectra), the electron absorbing the photon moves to a higher energy level.
quick answer please
.
.
.Five identical capacitor plates each of area A are arranged such that adjacent plates are at a distance d apart . The plates are connected to a potential difference of EMF V as shown in the figure. The capacitance :
Answers
Answer:
tq1=EoA/D
-2q1=-2EoAV/d
Option C.
Answer:
tq1=EoA/D
-2q1=-2EoAV/D
Explanation:
option C is correct
A mass of 4 kg is traveling over a quarter circular ramp with a radius of 10 meters. At the bottom of the incline the mass is moving at 21.3 m/s and at the top of the incline the mass is moving at 2.8 m/s. What is the work done by all non-conservative force in Joules?
Answers
Answer:
499.7 J
Explanation:
Since total mechanical energy is conserved,
U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.
So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂
mgh₁ + 1/2mv₁² + W₁ = mgh₂ + 1/2mv₂² + W₂
Substituting the values of the variables into the equation, we have
mgh₁ + 1/2mv₁² + W₁ = mgh₂ + 1/2mv₂² + W₂
4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)² + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)² + W₂
0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s² + W₂
907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s² + W₂
907.38 kgm²/s² = 407.68 kgm²/s² + W₂
W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²
W₂ = 499.7 kgm²/s²
W₂ = 499.7 J
Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J
15. Muous produced by snails help them move along the ground How does the mucus help maits
move?
A Mucus is cold
C. Mucus reduces friction
B. Mucus leaves a trail D. Mucus makes the snail lighter
16. Which of the following statements does NOT tell about the effects of pravitational force
on objects?
A. The more distant the body from the earth, the lesser is the gravitational force of attraction
B. Gravitational force pushes the objects upward
C. Gravitational force is greater when the objects are closer together
D. The farther you are from earth, the lesser your weight
17. Why is it difficult for us to go up in a mountain than go down?
A We are moving against gravity
B. We are pulled by the wind
C. We are moving towards gravity
D. None of the above
18. Which of these will have the strongest gravitational pull on Earth'?
A a baby
C. a drinking glass
B. a half sack of nice
D. an elephant
19. Without air resistance, all objects fall at the same rate.
D. Neither True or False
C Maybe
B. False
20. The gravity between two objects increases as the distance between them
D stabilizes
C neutralizes
B. increases
A. decreases
A True
Answers
Answer:
15.C Mucus reduces friction
I NEED HELP WITH THIS PLEASE
Answers
Answer:
D
Explanation:
One of the factors of increasing the rate of a reaction is increasing concentration. Therefore adding more people increases the number of people on the dance floor , therefore increasing the concentration increases the rate of reaction.
Bola bermassa 200 gram dilempar
ke bawah dari ketinggian 20 m
dengan kecepatan 2 m/s. Jika
percepatan gravitasi bumi 10
m/s2 energi kinetik pada
ketinggian 8 m adalah ......
Answers
Answer:
0.4
Explanation:
[tex] \frac{1}{2} mv ^{2} [/tex]
kinetic energy formula , potential energy is not considered
0.5×0.2×2×2
what is the suitable way of using social media
Answers
Answer:
not using it too much and getting addicted
Explanation:
How long does take for a freely falling object to reach 4.0 m/s
Answers
Answer :
Considering initial velocity is 0,
It takes about 0.4 seconds.
Use the equation v-u =at
4-0 = 9.8×t
t = 4/9.8 = 0.4 seconds approximately.
Find the weight of a man whose mass is 40 kg on earth.
(also
write complete data plus proper formula).
Answers
I am sure it help you with that much ☺️
Explanation:
pleasae give me some thanks please good morning sister
The tires of a car make 60 revolutions as the car reduces its speed uniformly from 95.0 km/h to 60.0 km/h. The tires have a diameter of 0.88 m. If the car continues to decelerate at this rate, how far does it go
Answers
Answer:
-2.869 rad/s2
Explanation:
Data given:
speed, vi at 95.0 km/h = 95 X (1 hour /3600 seconds) X (1000m / 1km)
Note that, for every 1 hour, there will be 60sec X 60sec = 3600 seconds
And for every 1km, there will be 1000m.
So, speed of 95.0 km/h = 26.389 m/s
speed, vi = r ω (radius X angular velocity)
angular velocity, ωi = v/r
ωi = 26.389 m/s ÷ half of 0.88 m diameter
= 59.975 rad/s
decelerating to speed, vf at 60.0 km/h = 60 X X (1 hour /3600 seconds) X (1000m / 1km)
= 16.667m/s
The angular velocity for this speed = 16.667m/s ÷ half of 0.88 m diameter
= 37.879rad/s
How far the car goes is equivalent to the angular acceleration which equals to (ωf^2 - ωi^2) ÷ 2θ
= (37.879rad/s)^2 - (59.975 rad/s)^2 ÷ 2 (60 rev X 2π rad/rev)
= -2.869 rad/s2
With neat circuit diagrams where possible show any two (2) ways of direct current motor
excitations,
Answers
A ball is thrown vertically upward at 24.0 ms can reach a height of 28.8m ( neglecting air resistance).The speed,in m/s,when it is halfway to its highest point is (using g= 10 ms ^2)
Answers
Answer:
The answer is "[tex]16.79\ \frac{m}{s}[/tex]"
Explanation:
In this question, the halfway indicates the height that is [tex]\frac{28.8}{2}=14.4 \ m[/tex]
Using formula:
[tex]v^2=u^2+2as\\\\v^2=24^2+2(-10)(14.4)\\\\[/tex]
[tex]v^2=576-288\\\\v^2=288\\\\v=\sqrt{288}\\\\v=16.97 \ \frac{m}{s}[/tex]
A circle has a radius of 20 cm and a central angle that measures 216° find the length of the arc defined by the central angle
Answers
Answer:
4320cm
Explanation:
(length of the arc)/(length of the radius)=central angle
lengthof the arc/20=216
lengthof the arc=216×20
length of the arc=4320cm
hope this helps, let me know if I'm right!
The length measured along the arc is called length of the arc. The length of the arc defined by the central angle is 4320cm.
What is central angle?Central angle is the angle subtended by the arc at the center of circle.
central angle = length of the arc / length of the radius
216 = length of the arc/20
length of the arc = 216×20
length of the arc = 4320cm
Thus, the length of the arc defined by the central angle is 4320cm.
Learn more about central angle.
https://brainly.com/question/15698342
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a man is standing near the edge of a cliff 85 meters high. he throws a stone upward vertically with an intial velocity of 10 m/s. the stone clears the cliff edge on the way down and falls all the way to the ground. what is the maximum height of the stone above the ground
Answers
Answer:
h = 90.10 m
Explanation:
Given that,
A man is standing near the edge of a cliff 85 meters high, h₀ = 85 m
The initial speed of the stone, u = 10 m/s
The path followed by the projectile is given by :
[tex]h(t)=-4.9t^2+10t+85[/tex] ....(1)
For maximum height,
Put dh/dt = 0
So,
[tex]\dfrac{dh}{dt}=-9.8t+10=0\\\\t=\dfrac{10}{9.8}\\\\=1.02\ s[/tex]
Put the value of t in equation (1).
[tex]h(t)=-4.9(1.02)^2+10(1.02)+85\\\\=90.10\ m[/tex]
So, the maximum height of the stone is equal to 90.10 m.
In the Biomedical and Physical Sciences building at MSU there are 135 steps from the ground floor to the sixth floor. Each step is 16.6 cm tall. It takes 5 minutes and 30 seconds for a person with a mass of 73.5 kg to walk all the way up. How much work did the person do?
Answers
Answer:
W = 16.4 kJ
Explanation:
Given that,
There are 135 steps from the ground floor to the sixth floor.
Each step is 16.6 cm tall.
The mass of a person, m = 73.5 kg
We need to find the work done by the person. We know that,
Work done = Fd
Where
d is the displacement, d = 135 × 0.166 = 22.41
So,
W = 73.5 × 10 × 22.41
= 16471.35 J
or
W = 16.4 kJ
So, 16.4 kJ is the work done by the person.
79. If the takeoff velocity of an airplane on a runway is 300 km/hr with an acceleration of 2 m/s. What is the takeoff time
of the airplane.
Answers
Answer:
The take off time is equal to 41.66s.
Explanation:
Given that,
The initial speed of the airplane, u = 0
Acceleration, a = 2 m/s²
Let v is the take off time. Using first equation of motion,
v = u +at
Put v = 300 km/hr = 83.3 m/s
So,
[tex]t=\dfrac{v}{a}\\\\t=\dfrac{83.33}{2}\\\\t=41.66\ s[/tex]
So, the take off time is equal to 41.66s.
all
What is Velocity
Answers
Answer:
noun
the speed of something in a given direction.
"the velocities of the emitted particles"
(in general use) speed.
"the tank shot backwards at an incredible velocity"
Similar:
speed
pace
rate
tempo
momentum
impetus
swiftness
swift/fast pace
fastness
quickness
speediness
rapidity
briskness
expeditiousness
expedition
dispatch
acceleration
clip
fair old rate
fair lick
steam
nippiness
fleetness
celerity
ECONOMICS
the rate at which money changes hands within an economy.
noun: velocity of circulation; plural noun: velocities of circulation
Answer:
i hope this helps you
Explanation:
hii
the velocity of a ship in the unit of m/s moving initially along west is given by V(t) = 5-2t. It’s acceleration at t=1 s is given by:
1. 0 m/s^2
2. 2m/s^2 along west
3. 2m/s^2 along east
4. None
Whis one is correct?
Answers
Answer:
4. None
Explanation:
Applying,
a(t) = dV(t)/dt
Where a(t) = Acceleration of the ship at a given time.
From the question,
Given: V(t) = 5-2t
Therefore,
a(t) = dV(t)/dt = 5 m/s²
Hence it's acceleration is 5 m/s²
From the question,
The right option is 4. None
2 Lights slows down when it enters water from air.
a What happens to its speed?
b What happens to its wavelength?
c What happens to its frequency?
Answers
Light will travel in straight line unless it meets any object (medium) in its path. Light changes its speed when it passes from one medium to other and the direction of light changes. This bending of light is called refraction.
This will happen in both when light passes through lighter to denser medium and also while comes from denser medium to lighter medium.
The speed of the light is increases when it comes from denser to less dense medium and decreases when it passes through lighter to denser. Light travels faster in vacuum compared with any other medium.
Refractive index defines the speed of light in a medium.
Refractive index (n)= v1/v2, v1 and v2 are the speed of light in vacuum and medium, respectively.
More the refractive index, slower the speed of light in that medium.
Coming to the wavelength of light, λ=v / f and here the frequency (f) of incident will not change when light passes through different medium.
Wavelength is directly proportional to the speed (v), then the wavelength increases if speed of light increases and vice versa.
The wavelength of light decreases when it goes from air to water because the speed of light decreases.
Mark brainliest
___________________ uses the heat energy from the Earth. This type of alternative energy can be installed basically anywhere in North Carolina.
A) wave energy
B) solar energy
C) nuclear energy
D) geothermal energy
Answers
Answer:
Geothermal!!
Explanation:
Geo means earth! Just like Hydro means water, and Pyro means fire :)
Answer:
The Answer is D). Geothermal energy
Explanation:
Hope this helps!!!!!
You throw a glob of putty straight up toward the ceiling, which is 3.50 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.10 m/s.
1. What is the speed of the putty just before it strikes the ceiling? Express your answer with the appropriate units.
2. How much time from when it leaves your hand does it take the putty to reach the ceiling? Express your answer with the appropriate units.
Answers
Answer:
Explanation:
Given that:
the putty initial speed (u) = 9.10 m/s
distance (s) between hand and the ceiling = 3.50 m
the speed of the putty prior to the time it hits the ceiling can be determined by considering the second equation of motion.
v² - u² = 2as
Since the putty is moving in a vertical motion(i.e. in an upward direction)
v² - u² = -2gs
v² = u² - 2gs
[tex]v = \sqrt{u^2 - 2gs}[/tex]
[tex]v = \sqrt{(9.10)^2 -( 2* 9.8) (3.50 -0)}[/tex]
[tex]v = \sqrt{82.81 -19.6 (3.50)}[/tex]
[tex]v = \sqrt{82.81 -68.6}[/tex]
[tex]v = \sqrt{14.21}[/tex]
v = 3.77 m/s
2.
The time it takes to reach the ceiling from the moment it leaves your hand can be calculated by using the first equation of motion:
v = u + at
In an upward direction
v = u - gt
making time t the subject;
v - u = -gt
[tex]t = \dfrac{v-u}{-g}[/tex]
[tex]t = \dfrac{3.77 - 9.10}{-9.8}[/tex]
[tex]t = \dfrac{-5.33}{-9.8}[/tex]
t = 0.54s
an object is 70 um long and 47.66um wide. how long and wide is the object in km?
Answers
Answer:
length = 7*10^(-8)km
width = 4.666*10^(-8) km
Explanation:
We know that:
1 μm = 1*10^(-6) m
and
1km = 1*10^3 m
or
1m = 1*10^(-3) km
if we replace the meter in the first equation, we get:
1 μm = 1*10^(-6)*1*10^(-3) km
1 μm = 1*10^(-6 - 3)km
1 μm = 1*10^(-9)km
Now with this relationship we can transform our measures:
Length: 70 μm is 70 times 1*10^(-9)km, or:
L = 70*1*10^(-9)km = 7*10^(-8)km
And for width, we have 47.66um, this is 46.66 times 1*10^(-9)km, or:
W = 46.66*1*10^(-9)km = 4.666*10^(-8) km
A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed.
(a) What is the mass of the other body?
(b) What is the speed of the two-body center of mass if the initial speed of the 2.0 kg body was 4.0 m/s?
Answers
Answer:
(a) the mass of the second body is 1.2 kg
(b) the speed of the two-body center of mass 2.5 m/s
Explanation:
Given;
mass of the body, m₁ = 2 kg
let the mass of the second body = m₂
let the initial speed of the first body, = u₁ = 4 m/s
then, the final speed of the first body, v₁= ¹/₄u₁ = 0.25u₁
initial speed of the second body, u₂ = 0
let the final speed of the second body = v₂
Apply principle of conservation of linear momentum to determine the mass of the second body;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
2u₁ + 0(m₂) = 2(0.25u₁) + m₂v₂
2u₁ = 0.5u₁ + m₂v₂
1.5(4) = m₂v₂
6 = m₂v₂
Apply one-directional velocity
u₁ + v₁ = u₂ + v₂
u₁ + (0.25u₁) = 0 + v₂
1.25u₁ = v₂
1.25(4) = v₂
5 = v₂
Then, the mass of the second body is calculated as;
m₂v₂ = 6
5m₂ = 6
m₂ = 6/5
m₂ = 1.2 kg
(b) the speed of the two-body center of mass after collision;
[tex]V_c_m = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} \\\\V_c_m = \frac{2(0.25\times 4) \ + \ 1.2(5)}{2\ + \ 1.2} \\\\V_c_m = 2. 5 \ m/s[/tex]
A pressure sensor was used to measure the unsteady pressure in a cylinder. The sensor output was acquired for 15 seconds at a rate of 202 Hz and spectral analysis was performed using FFT. What is the maximum frequency that will be displayed on the power spectrum plot
Answers
Answer:
Maximum frequency on power spectrum plot = 101 Hz
Explanation:
Given:
Time taken for output = 15 seconds
Frequency rate = 202 Hz
Find:
Maximum frequency on power spectrum plot
Computation:
Maximum frequency = Given frequency rate / 2
Maximum frequency on power spectrum plot = Frequency rate / 2
Maximum frequency on power spectrum plot = 202 / 2
Maximum frequency on power spectrum plot = 101 Hz
An air-filled pipe is found to have successive harmonics at 700 Hz , 900 Hz , and 1100 Hz . It is unknown whether harmonics below 700 Hz and above 1100 Hz exist in the pipe. What is the length of the pipe
Answers
Answer:
the length of the pipe is 0.85 m or 85 cm
Explanation:
Given the data in the question;
The successive harmonics are; 700 Hz , 900 Hz , and 1100 H
Now, for a closed pipe,
length of pipe (L) = λ/4
Harmonics; 1x, 3x, 5x, 7x, 9x, 11x
1100Hz - 900Hz = 200Hz
⇒ 2x = 200Hz
x = 100Hz ( fundamental frequency )
λ = V/f = 340 /100 = 3.4 m
Now
Length L = λ / 4
L = 3.4 / 4
L = 0.85 m or 85 cm
Therefore, the length of the pipe is 0.85 m or 85 cm
what are the symptoms of hepatitis 'b'
Answers
Fatigue is a common occurrence.
Appetite loss is common.
Nausea is a feeling of nausea.
Pain in the abdomen.
Urine that is dark in color.
Bowel motions that are clay-colored.