Answer:
The second (abbreviation, s or sec) is the Standard International ( SI ) unit of time. One second is the time that elapses during 9,192,631,770 (9.192631770 x 10 9 ) cycles of the radiation produced by the transition between two levels of the cesium 133 atom. ... One second is equal to 1/86,400 of a mean solar day.
Explanation:
Why would you have trouble breathing at high altitudes?
A. it is colder at the top of a mountain
B. the oxygen molecules are spread farther apart
C. the oxygen molecules are closer together
Answer:
A. It is colder at the top of a mountain
Explanation:
Two waves, each with an amplitude of 0.5m are superimposed with constructive interference such that they are in phase. Describe the resultant amplitude
The resultant amplitude of the two waves is 1.0m
Interference occurs when two or more waves are superimposed over each other. These waves can either be destructive or constructive in nature. If the waves are superimposed with constructive interference such that they are in phase, the resultant amplitude will be the sum of the amplitude of both waves.Resultant amplitude = 0.5m + 0.5m
Resultant amplutude = 1.0m
Hence the resultant amplitude of the two waves is 1.0m
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What specific type of tide has the smallest difference between high and low tide?
Answer: Neap tides
Explanation: Neap tides are tides that have the smallest tidal range, and they occur when the Earth the Moon, and the Sun form a 90o angle. They occur exactly halfway between the spring tides when the Moon is at first or last quarter.
How do you find the velocity of a moving car?
Answer:
Velocity (v) is a vector quantity that measures displacement (or change in position, Δs) over the change in time (Δt), represented by the equation v = Δs/Δt. Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (Δt), represented by the equation r = d/Δt.
Explanation:
hope this helps a little bit sorry if i got it wrong
Please Help Me!!!
Water is a colorless and odorless liquid. It can exist in solid, liquid, and gas states. It boils at 100 degrees C and melts at 0 degrees C.
Which option best describes this information?
A.
These are the physical properties of water.
B.
These are the chemical properties of water.
C.
These are the physical changes water undergoes.
D.
These are the chemical changes water undergoes.
E.
These are the molecular changes water undergoes.
Answer:
There are the molecular changes water undergoes
PLS HELP ME
Secondary waves CANNOT travel through
a.
rock.
b.
liquids.
c.
Earth’s mantle.
d.
Earth’s crust.
Answer:
B. Liquids
Explanation:
Just had this question on my test.
Hope this helps :)
Compare and contrast hydrogen fuel cell vehicles with vehicles that have intemal combustion engines.
A hydrogen fuel cell vehicle is different from internal combustion engine as in fuel cell, hydrogen combines with oxygen whereas in internal combustion engines, gasoline is used.
What are hydrogen fuel cell vehicles and internal combustion engines?
In a internal combustion engine or fuel cell vehicle, the hydrogen fuel is combined with oxygen atom. While the combustion in a hydrogen- fuel or in a gasoline-powered engine is converted into mechanical energy, however in the hydrogen fuel cell vehicle, the chemical energy from the hydrogen and oxygen is converted into electrical energy.
Both hydrogen internal combustion engines and the hydrogen fuel cells can power vehicles using hydrogen, a zero-carbon fuel. Hydrogen engines burn hydrogen gas in an internal combustion engine, in the same way as the gasoline is used in a power engine. Examples of hydrogen internal combustion engines include gasoline engines, diesel engines, gas-turbine engines, and rocket-propulsion systems.
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A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. ten second later, he is moving at 15 m/s. what is his acceleration
Answer:
1 m/s²
Explanation:
From the question,
Using
a = (v-u)/t.................... Equation 1
Where a = accelartion of the bicycle, v = Final velocity, u = initial velocity, t = time.
Given: v = 15 m/s, u = 5 m/s, t = 10 s
Substitute these values into equation 1
a = (15-5)/10
a = 10/10
a = 1 m/s²
Hence the acceleration of the bicycle is 1 m/s²
Two resistors, 2.50 Ω and 4.95 Ω, are wired in series to a 12.00 V battery. What is the equivalent resistance of the circuit?
Answer:
7.45 Ω
Explanation:
From the question given above, the following data were obtained:
Resistor 1 (R₁) = 2.50 Ω
Resistor 2 (R₂) = 4.95 Ω
Voltage (V) = 12 V
Equivalent Resistance (Rₑq) =?
Since the resistors are in series arrangement, the equivalent resistance can be obtained as follow:
Rₑq = R₁ + R₂
Rₑq = 2.5 + 4.95
Rₑq = 7.45 Ω
Therefore, the equivalent resistance is 7.45 Ω
A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.48 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1840 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 5.19 V/m, (b) in the negative z direction and has a magnitude of 5.19 V/m, and (c) in the positive x direction and has a magnitude of 5.19 V/m
Answer:
a) F = 15.6 10⁻¹⁹ k^ N, b) F = -1 10⁻¹⁹ k^ N,
c) F = (8.3 i^ + 73 k^) 10⁻¹⁹ N
Explanation:
In this exercise we calculate the magnetic and electric force separately.
Let's start by calculating the magnetic force
F_m = q v x B
where bold letters indicate vectors, the modulus of this expression is
F_m = q v B
the direction is given by the right hand rule, where the thumb points in the direction of velocity, the fingers are extended in the direction of the magnetic field and the palm is in the directional ne force if the charge is positive
let's calculate the magnitude
F_m = 1.6 10⁻¹⁹ 1840 2.48 10⁻³
F_m = 7.3 10⁻¹⁹ N
we calculate the direction, the thumb is in the direction of + y,
fingers extended in the direction of -ax
the palm remains + z
therefore the magnetic force is in the direction of the positive side of the z axis
The electric force is
F_e = q E
give several possibilities for the electric field
a) electric field E = 5.19 V / m in the direction of + z
we calculate
F_e = 1.6 10⁻¹⁹ 5.19
F_e = 8.3 10⁻¹⁹ N
the total force is
F = F_m + F_e
F = (8.3 k ^ + 7.3 k ^) 10⁻¹⁹
F = 15.6 10⁻¹⁹ k^ N
b) in the -z direction
F_e = -8.3 10⁻¹⁹ k^ N
F = Fm + Fe
F = (-8.3 k ^ + 7.3 k ^) 10⁻¹⁹
F = -1 10⁻¹⁹ k^ N
c) in the + x direction
F_e = 8.3 10⁻¹⁹ i^ N
the total force is
F = (8.3 i^ + 73 k^) 10⁻¹⁹ N
Which statement is true of an alpha particle?
A. It is electromagnetic radiation, not a true particle.
B. It is an electron.
C. It is the largest decay particle
D. It is a particle that travels at the speed of light.
Answer:
Atoms are made up of various parts; the nucleus contains minute particles ... neutrons, and the atom's outer shell contains other particles called electrons. ... Physical Forms of Radiation; Radioactive Decay; Nuclear Fission; Ionizing Radiation ... of high-energy waves that can travel great distances at the speed of light and ...
Explanation:
The true statement of an alpha particle is it is an electron. The correct option is B.
What is an alpha particle?The alpha particle are composite particles which consists of two protons and two neutrons tightly bonded together. They are ejected from the inside the nucleus of heavy radioactive particles, also called alpha decay.
Atoms are made up electrons orbiting in outer shells, protons and neutrons in the nucleus. The radioactive particle emit highly ionized radioactive rays which ionize electrons emitting alpha particle.
Thus, the correct option is B.
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Highlight two factors which shows that heat from the sun does reach the earth's surface by convection.
radiation
when the suns radiation fall on the earth and its objects they receive heat energy and hence get heated. Thus the suns heat reaches the earth by. the process of radiation
Absence of medium is the factor which shows that heat from the sun does reach the earth's surface by convection.
The heat of the sun reaches the earth only through radiation because radiation does not require any medium for the transfer of energy but convection and conduction required a medium to transfer heat energy.
We know that there is no medium present in vacuum so convection and conduction are impossible in a vacuum that's why the sun's heat does not reach the earth by conduction or convection.
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Which statement best summarizes the development of the atomic theory over time?
Answer:
electrons move around nucleus in specific regions
We have that the correct statement that summarizes atoms over time will be
The discovery of new evidence resulted in changes to the atomic theory.
Option B
Generally
The Image shows that in the early 1800 the knowledge of atom was limited and it increased over time and by 1920 it had gone through a 4 stages of modifications to the atomic model
Atoms are solid spheresElectrons are attached to atom surfaceElectrons move around in the nucleus in a specificRapid motion of electrons in the cloudTherefore the correct statement that summarizes atoms over time will be
The discovery of new evidence resulted in changes to the atomic theory
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The 4kg head of a sledge hammer is moving at 6m/s when it strikes a chisel, driving it into a log. The duration of the impact (or the time for the sledge hammer to stop after contact) is 0.0020s. Find the time average of the impact force
Answer:
The average impact force is 12000 newtons.
Explanation:
By Impact Theorem we know that impact done by the sledge hammer on the chisel is equal to the change in the linear momentum of the former. The mathematical model that represents the situation is now described:
[tex]\bar F \cdot \Delta t = m \cdot (v_{2}-v_{1})[/tex] (1)
Where:
[tex]\bar F[/tex] - Average impact force, in newtons.
[tex]\Delta t[/tex] - Duration of the impact, in seconds.
[tex]m[/tex] - Mass of the sledge hammer, in kilograms.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocity, in meters per second.
If we know that [tex]\Delta t = 0.0020\,s[/tex], [tex]m = 4\,kg[/tex], [tex]v_{1} = -6\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], then we estimate the average impact force is:
[tex]\bar F = \frac{m\cdot (v_{2}-v_{1})}{\Delta t}[/tex]
[tex]\bar F = 12000\,N[/tex]
The average impact force is 12000 newtons.
A firework is launched with a force of 700 N and a momentum of 200 kg-m/s. How much time before is explodes?
Answer:
t = 0.28 seconds
Explanation:
Given that,
Force acting on a firework, F = 700 N
The momentum of the firework, p =200 kg-m/s
We need to find the time before it explodes. Ket the time be t. We know that, the rate of change of momentum is equal to external frce. So,
[tex]F=\dfrac{P}{t}\\\\t=\dfrac{P}{F}\\\\t=\dfrac{200}{700}\\t=0.28\ s[/tex]
So, the required time is equal to 0.28 seconds.
A decibel measures the _____________ of sound.
A.pitch
B,frequency
C.loudness
D.amplitude
At am outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 29.8981 cm on spring with a spring constant
Answer:
Hello your question lacks some very vital information hence I will give you a general equation where you can input the missing values and get your answer
answer : mass of Bananas ( m ) = [tex]\frac{A^2}{V^2} * K[/tex]
To calculate for weight of Bananas = m * g where g = 9.8
Explanation:
Given an amplitude ( A )= 29.8981
Spring constant = x ( unknown )
maximum speed of Bananas( v ) = wA ( unknown )
question : calculate the weight of the bananas
First step : calculate the period of oscillation
T = [tex]\frac{2\pi }{vA}[/tex] --- ( 1 )
next step : express T in terms of mass ( m ) of the bananas
= [tex]2\pi \sqrt{\frac{m}{k} } = \frac{2\pi }{vA}[/tex]
∴ m/k = A^2 / v^2
hence mass of Bananas ( m ) = [tex]\frac{A^2}{V^2} * K[/tex]
To calculate for weight = m * g
A machine raised a load of 360N through a distance of 0.2m. The effort, a force of 50N moved 1.8m during the process. Calculate the efficiency of the machine.
Select one:
a. 40%
b. 100%
c. 80%
d. 60%
Answer:
c. 80%
Explanation:
Given;
load raised by the machine, L = 360 N
distance through which the load was raised, d = 0.2 m
effort applied, E = 50 N
distance moved by the effort, e = 1.8 m
The efficiency of the machine is calculated as follows;
[tex]Efficiency = \frac{ 0utput \ work }{1nput \ work} \times 100\% \\\\Efficiency = \frac{Load \ \times \ distance \ moved \ by \ load }{Efort \ \times \ distance \ moved \ by \ effort} \times 100\%\\\\Efficiency = \frac{360 \times 0.2}{50 \times 1.8} \times 100\%\\\\Efficiency =80\%[/tex]
Therefore, the efficiency of the machine is 80%
You are sharing a house with a few roommates. One of them plays on the basketball team, and is exactly 200 cm tall. The heights of you and the other roommates are 188 cm, 175 cm, and 160 cm. You want to buy the minimum height mirror that will allow all of you to see your entire selves, when the mirror is mounted in a fixed position on the wall. Assume each person's eyes are at a level 95% of their height. How tall a mirror should you buy
Answer:
119 cm
Explanation:
Given heights : 200cm , 188 cm , 175 cm, 160 cm
since the eyes are at 95% level
95% of 200 cm = 190 cm
95% of 160 cm = 152 cm
hence the allowable height of mirror = 200 - ( 5 + 76 ) = 119 cm
and The mirror should be hung 75cm from the floor
A toy gun uses a spring to shoot plastic balls (m = 50 g). The spring is compressed by 3.0 cm. Let k=2.22 × 105 N/m. (a) Of course, you have to do some work on the gun to arm it. How much work do you have to do? (b) Suppose you fire the gun horizontally. How fast does the ball leave the gun? (c) Now suppose you fire the gun straight upwards. How high does the ball go?
Answer:
Explanation:
Given that:
mass of the plastic ball = 50 g = 50 × 10⁻³ kg
spring constant (k) = 2.22 × 10⁵ N/m
compression of spring (x) = 3.0 cm = 3 × 10⁻² m
a) Work done:
[tex]= \dfrac{1}{2}kx^2 \\ \\ = \dfrac{1}{2}\times 2.22 \times 10^{5}\times (3 \times 10^{-2})^2 \\ \\ = 99.9 \ J[/tex]
b) When the gun is fired horizontally;
In the spring, the potential energy is changed to kinetic energy.
i.e.
[tex]\implies \dfrac{1}{2}kx^2 = \dfrac{1}{2}mv^2 \\ \\ 2.22 \times 10^5 \times (3\times 10^{-2}) ^2 =50 \times 10^{-3} \times v^2\\ \\ \mathbf{v= 63.2 m/s}[/tex]
c) The max height the bullet will reach if the gun is being shot upward is:
[tex]H_{max}= \dfrac{v^2}{2g} \\ \\ H_{max}= \dfrac{63.2^2}{2\times 9.8}[/tex]
[tex]\mathbf{H_{max}= 203.9 \ m}[/tex]
Which of these statements are true about hydroelectric power? A.It is the most widely used renewable energy source in the world. B.It never disrupts ecosystems. C.It is not always available. D.A hydroelectric dam controls the force of moving water to generate electricity.
HELP pleaseee which nucleus completes the following equation?
Answer:
its D
Explanation:
i just got it correct on a. p. e. x. :)
Is this right? Please help me ITS SOCIOLOGY
An object has an acceleration of 18.0 m/s/s. If the net force acting upon this object were doubled, then its new acceleration would be m/s/s.
Answer:
36.m/s/s
Explanation:
Which best summarizes a concept related to the work-energy theorem?
- When work is positive, the environment does work on an object.
- When work is negative, the environment does work on an object.
- When work is positive, the kinetic energy in a system remains constant.
- When work is negative, the kinetic energy in a system remains constant.
this is times pls help!
Answer:
When work is positive, the environment does work on an object.
Explanation:
According to the work-energy theorem, the net work done by the forces on a body or an object is equal to the change produced in the kinetic energy of the body or an object.
The concept that summarizes a concept related to the work-energy theorem is that ''When work is positive, the environment does work on an object.''
Answer:
Your answer is A!
Have a good day!
Define the types of friction and give FOUR examples of each
Static Friction
Rolling Friction
Sliding Friction
Fluid Friction
A 1.80-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 15.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations.
(a) Find the force constant of the spring.
N/m
(b) Find the frequency of the oscillations.
Hz
(c) Find the maximum speed of the object.
m/s
(d) Where does this maximum speed occur?
x = ±
m
(e) Find the maximum acceleration of the object.
m/s2
(f) Where does the maximum acceleration occur?
x = ±
m
(g) Find the total energy of the oscillating system.
J
(h) Find the speed of the object when its position is equal to one-third of the maximum value.
m/s
(i) Find the magnitude of the acceleration of the object when its position is equal to one-third of the maximum value.
m/s2
Answer:
Explanation:
(a) Force constant = force/displacement = 15/0.2
=75 N/m
(b) For SHM with spring, the angular frequency = sqrt ( force constant / mass )
= sqrt ( 75 / 1.8 )
= 6.455
Frequency = Angular frequency / 2pi
= 1.03Hz
(c) Maximum speed is when object's potential energy is at minimum.
1/2*mass*speed^2 = 1/2*force constant*displacement^2
speed^2 = 75*0.2^2/1.8 = 1.67
speed = sqrt (1.67)
= 1.29 m/s
(d) It occurs at x=0
(e) Maximum acceleration = Force / m = 15/1.8
= 8.33 m/s^2
(f) It occurs at x=0.2m
Answer:
Explanation:
given:
The mass of the object attached to the spring: m = 1.80 kg
The magnitude of the maximum horizontal force exerted: F = 15.0 N
The amplitude of the spring-object oscillation: A = 0.200 m
a. k=15/.2=75N/m
b. (2pi*frequency)^2 = k/m = 75/1.8 = 41.67
frequency = 1.03Hz
c. max speed^2 = amplitude^2*k/m = 0.2^2 * 41.67
max speed = 1.29 m/s
d. x=0m
e. max accel = F/m = 15/1.8 = 8.33 m/s^2
f. x=0.2m
e. energy = 1/2*k*A^2
= 1/2*75*0.2^2
= 1.5J
Which of the following ways of writing 1000w is incorrect?
1) 1 kW
2) 1 x 10³ W
3) 10 x 10³ W
4) 1.0 x 10³ W
Answer:
the third one is incorrect
Explanation:
10 x 10³= 10^1 x 10^3 = 10^4
10. A pitcher in a professional baseball game throws a fastball, giving the baseball
a momentum of 5.83 kg. m/s. Given that the baseball has a mass of 0.145 kg,
what is its speed?
Answer: 40.2m/sec
Work: v=p/m= 5.83kg/0.145k=40.2cm/sec
A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriendly dog is running across the roof toward her. Next to her is a large wheel mounted on a horizontal axle at its center. The wheel, used to lift objects from the ground to the roof, has a light crank attached to it and a light rope wrapped around it; the free end of the rope hangs over the edge of the roof. The student radius 0.300 m and a moment of inertia of 9.60 kg m^2 for rotation about the axle, how long does it take her to reach the side walk, and how fast will she be moving just beofre she lands?
Answer:
The speed of the student just before she lands, v₂ is approximately 8.225 m/s
Explanation:
The given parameters are;
The mass of the physic student, m = 43.0 kg
The height at which the student is standing, h = 12.0 m
The radius of the wheel, r = 0.300 m
The moment of inertia of the wheel, I = 9.60 kg·m²
The initial potential energy of the female student, P.E.₁ = m·g·h₁
Where;
m = 43.0 kg
g = The acceleration due to gravity ≈ 9.81 m/s²
h = 12.0 h
∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J
The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;
[tex]K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2[/tex]
The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0
∴ K₁ = 0 + 0 = 0
The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0
Where;
h₂ = 0 m
The final kinetic energy, K₂, of the wheel and student is give as follows;
[tex]K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2[/tex]
Where;
v₂ = The speed of the student just before she lands
ω₂ = The angular velocity of the wheel just before she lands
By the conservation of energy, we have;
P.E.₁ + K₁ = P.E.₂ + K₂
∴ m·g·h₁ + [tex]\dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2[/tex] = m·g·h₂ + [tex]\dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2[/tex]
Where;
ω₂ = v₂/r
∴ 5061.96 J + 0 = 0 + [tex]\dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2[/tex]
5,061.96 J = 21.5 kg × v₂² + 53.[tex]\overline 3[/tex] kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²
v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²
v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s
The speed of the student just before she lands, v₂ ≈ 8.225 m/s.