Answer:
When one system vibrating at its natural frequency is put closer to a stationary system, the stationary system receives impulses.
At resonant frequency, the system vibrating at its own natural frequency suddenly goes on decreasing in order to cope with neighboring system.
These decrease in frequency is known as damping.
Using your Periodic Table, which of the elements below is most likely to be a solid at room temperature?
A.) potassium, B.) Hydrogen, C.) Neon, D.) Chlorine
The answer is definitely Potassium
12 x cos 50 = ?
Does anyone have the answer ? I forgot my my calculator.
12 x cos 50 = 7.713451316...
Reference frame definitely changes when also changes
Numerical problems:
a. convert the following as instructed:
i) 340 cm into m
ii)86400 seconds into day
Answer:
a=3.4m because of the m
b=1day because 86400=a day
an object is sliding down in clean plane the velocity change at a constant rate from 10 cm to 15 CM in 2 second what is it acceleration ?
Initial velocity=10m/s=u
Final velocity=v=15m/s
Time=t=2s
[tex]\boxed{\sf Acceleration=\dfrac{v-u}{t}}[/tex]
[tex]\\ \sf\longmapsto Acceleration=\dfrac{15-10}{2}[/tex]
[tex]\\ \sf\longmapsto Acceleration=\dfrac{5}{2}[/tex]
[tex]\\ \sf\longmapsto Acceleration=2.5m/s^2[/tex]
Answer: a = 2.5 cm/s²
Explanation:
Acceleration = (Final velocity - Initial Velocity)/time taken
a = v-u/t
Initial velocity = 10 cm/s
Final velocity = 15 cm/s
Time = 2 seconds
a = (15-10)/2
a = 5/2
a = 2.5 cm/s²
Therefore the acceleration is 2.5 cm/s²
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The following arbitrary measurements are made and the errors sited are the aximum errors A = 15.21 +0.01, B = 10.82 +0.05, C = 11.00+ 0.03. If D= A + B + C; (a) Calculate the maximum error in D. (b) if the errors sited are standard errors, calculate the standard error in D.
Maximum error in the result of the sum of measurement is equal to the sum absolute error of the individual observed measurements
(a) The maximum error in D is 0.09
(b) The standard error in D is approximately 0.034
The procedure for arriving at the above values is as follows;
The given measurements and the sited errors are;
A = 15.21 + 0.01
B = 10.82 + 0.05
C = 11.00 + 0.03
D = A + B + C
(a) Required parameter;
To calculate the maximum error in D
The equation for the propagation of error in addition is presented as follows;
Given that we have;
x = a + b
Therefore;
x + ±Δx = (a ± Δa) + (b ± Δb) = a + b ± (Δa + Δb)
∴ Δx = Δa + Δb
From the above formula, we have;
Where;
D = A + B + C
The maximum error in D = The sum of the maximum error in A, B, C
∴ The maximum error in D = 0.01 + 0.05 + 0.03 = 0.09
(b) Required parameter:
To find the standard error in D
The standard error is the sampling distribution's standard deviation, SD
Variance = SD²
The combined variance, SD² = The sum of the squares of individual standard deviations
Given that the standard errors represents the standard deviation, we get;
The combined variance, SD² = 0.01² + 0.05² + 0.03²
The combined variance, SD = √(0.01² + 0.05² + 0.03²) = 0.059
[tex]Standard \ error = \dfrac{SD}{\sqrt{n} }[/tex]
Where n = 3, for the three measurement, we get;
[tex]Standard \ error = \dfrac{\sqrt{0.01^2 + 0.05^2 + 0.03^2} }{\sqrt{3} } \approx 0.034[/tex]
The standard error in D is approximately 0.034
Learn more about maximum error and standard error here:
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pls help me asap with this
Answer:
a) cos30=adj/hyp
cos30= horizontal force/10
horizontal force= 8.66 N
rest of a is in the picture.
b) i believe you can continue.
sl unit of upthrust and SI unit of pressure
Answer:
The SI unit of upthrust is Newton(N).
The SI unit of preesure is Pascal(P).
Thank You
Draw a wave that has a wavelength of 3 cm and an amplitude of 1 cm. Label the wavelength, the amplitude, the rest position, and the crest and trough of your wave.
Answer:
Please find attached, the required wave drawn with MS Excel
Explanation:
Functions that represent waves is given as follows
A general form of the wave equation is A·sin(B·x) + D
Where;
B = 2·π/T
T = The period of the wave = 1/f
D = The vertical shift of the wave = 0
A = The amplitude of the wave = 1 for sine wave
v = The wave velocity
λ = The wavelength of the wave
f = The frequency of the wave
v = f·λ
At constant v, λ ∝ 1/f
∴ λ ∝ T
Where T = 3, we have;
B = 2·π/T
∴ B = 2·π/3
Therefore, we have the wave with an amplitude of 1 cm, and wavelength, 3 cm, given as follows
y = sin((2·π/3)·x)
Plotting the above wave with MS Excel, we can get the attached wave
: Chuyển động Một vật xuất phát từ A chuyển động đều về B cách A 240 m với vận tốc 10 m/s. Cùng lúc đó, một vật khác chuyển động đều từ B về A. Sau 15 s hai vật gặp nhau. Tìm vận tốc của vật thứ hai và vị trí hai vật gặp nhau.
The mass of objects is 4kg and it has a density of 5gcm^-3. what is the volume
Answer:
4kg×5gm^3=60
Explanation:
the object if heavy
5. a. Answer the following questions. What is density? Write a formula by showing the relation among density mass and volume.
Answer:
Density is how compact something is. The relationship is M/V=D (Mass divided by Volume equals Density).
Explanation:
WHAT IS DENSITY:
Density is the degree of compactness of a substance.
EXAMPLE:
"a reduction in bone density"
FORMULA OF DENSITY:
The formula for density is d = M/V, where d is density, M is mass, and V is volume.
this is physics practical
Answer:
well done buddy
Explanation:
15millas a km alguien pliss para ahorita porfa lo sigo
Answer:
X = 24.135 kilometres
Explanation:
Given the following data;
Distance = 15 miles
To convert the value in miles to kilometers;
Conversion:
1 mile = 1.609 kilometres
15 miles = X kilometres
Cross-multiplying, we have;
X = 1.609 * 15
X = 24.135 kilometres
define amplitude & period of the particle performing linear S.H.M
Answer:
This type of a behavior is known as oscillation, a periodic movement between two points. ... Amplitude: The distance from the center of motion to either extreme. Period: The amount of time it takes for one complete cycle of motion.
Explanation:
Amplitude (a):- The maximum displacement of particle from its mean position on either side is called amplitude.
Periodic time:- The time taken by a wave to complete one oscillation is called periodic time.
can uh help in in this question step by step
Convert to m/s
[tex]\\ \sf \longmapsto 72\times \dfrac{5}{18}=5(4)=20m/s[/tex]
Final velocity=v=0m/sTime=2s=t[tex]\\ \sf \longmapsto Acceleration=\dfrac{v-u}{t}[/tex]
[tex]\\ \sf \longmapsto Acceleration=\dfrac{0-20}{2}[/tex]
[tex]\\ \sf \longmapsto Acceleration=\dfrac{-20}{2}[/tex]
[tex]\\ \sf \longmapsto Acceleration=a=-10m/s^2[/tex]
Distance be sUsing second equation of kinematics
[tex]\\ \sf \longmapsto s=ut+\dfrac{1}{2}at^2[/tex]
[tex]\\ \sf \longmapsto s=20(2)+\dfrac{1}{2}(-10)(2)^2[/tex]
[tex]\\ \sf \longmapsto s=40+(-20)[/tex]
[tex]\\ \sf \longmapsto s=40-20[/tex]
[tex]\\ \sf \longmapsto s=20m[/tex]
Now
Mass=m=5000kgUsing newtons second law
[tex]\\ \sf \longmapsto Force=ma[/tex]
[tex]\\ \sf \longmapsto Force=5000(-10)[/tex]
[tex]\\ \sf \longmapsto Force=-50000N[/tex]
Force is in opposite direction so its negative[tex]\\ \sf \longmapsto Force=50kN[/tex]
Answer . The acceleration of the truck is 10m/[tex]s^{2}[/tex], and the distance covered is 40 m. Have attached the picture for solution.
Hope that helps.
A 250–g piece of gold is at 19 °C. 5.192 kJ of energy is added to it by heat. The specific heat of gold is 129 J/(kg·°C). Calculate its final temperature.
We heat a 25–g sample of metal from 10 °C to 100 °C. 1.082 kJ of energy is added to it by heat. Calculate
the specific heat of the metal.
Answer:
A. DT is given by Q= MCs DT
m = mass of the substances
Cs= is it's specific heat capacity
Ck= Q
Mk ×DTk
=250 × 9 × 5
129
=Dt = 180.1085271
answer is 180degree C.
Explanation:
B. = 25×10 ×100
1.082
=2500
1.082
= 23105.360 g/kj.
The final temperature is 180 degree. and the specific heat of the metal is 23105.360 g/kj.
How to calculate the specific heat?Q = m . C . ΔT
Q = heat; m = mass; C is the specific heat and
ΔT = Final T° - Initial T°
Q = C lat . m
Q = Heat
m = mass
C lar = Latent heat of fusion
A) DT is given by Q= M Cs DT
where, m = mass of the substances
Cs= is it's specific heat capacity
Ck= Q
Mk × DTk
=250 × 9 × 5
129 =Dt = 180.1085271
Thus, the final temperature is 180 degree.
B) We heat a 25–g sample of metal from 10 °C to 100 °C. 1.082 kJ of energy is added to it by heat = 25×10 ×100
=2500
1.082
Q = 23105.360 g/kj
Hence, the specific heat of the metal is 23105.360 g/kj.
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What is the connection of H ions at a ph=2?
Answer:
Explanation:
High concentrations of hydrogen ions yield a low pH (acidic substances), whereas low levels of hydrogen ions result in a high pH (basic substances). The overall concentration of hydrogen ions is inversely related to its pH and can be measured on the pH scale
Answer the following questions. 3 A student runs 2 m/s. What does this mean?
Answer:
2ms-¹ means that the body under consideration moves 2m in a second, and may be it will continue to move 2m in every 1 second, if there's no external unbalanced force acting on that body (those forces do include frictional forces). mark its brainlist plz. Kaneppeleqw and 6 more users found this answer helpful. Thanks 3.
Answer:
that the student has travels 2 meters every 1 second that passes
Why are road accidents at high speeds very much worse than road accidents at low speeds?
Answer:
The momentum makes it worse.
Explanation:
The momentum of vehicles running at faster speeds is very high and causes a lot of damage to the vehicles.
A CROW BAR WITH LENGTH 200 CM IS USED TO LIFT A LOAD OF 600N . IF THE DISTANCE BETWEEN FULCRUM AND LOAD IS 0.75. CALCULATE ; a, effort b, MA c, VR
Answer:
a. Effort = 960 Newton
b. Mechanical advantage (M.A) = 0.625
c. Velocity ratio (V.R) = 1.67
Explanation:
Given the following data;
Load = 600 NLength of crowbar = 200 cmLength of load arm = 0.75 mConversion:
100 cm = 1 m
X cm = 0.75 m
Cross-multiplying, we have;
X = 0.75 * 100 = 75 cm
First of all, we would find the effort arm;
Effort arm = length of crow bar - length of load arm
Effort arm = 200 - 75
Effort arm = 125 cm
Next, we would determine the mechanical advantage (M.A) of the crow bar;
[tex] M.A = \frac {Effort \; arm}{Load \; arm} [/tex]
Substituting the values into the formula, we have;
[tex] M.A = \frac {125}{200} [/tex]
M.A = 0.625
To find the effort of the crow bar;
[tex] M.A = \frac {Load}{Effort} [/tex]
Making "effort" the subject of formula, we have;
[tex] Effort = \frac {Load}{M.A} [/tex]
[tex] Effort = \frac {600}{0.625} [/tex]
Effort = 960 Newton
Lastly, we would determine the velocity ratio (V.R);
[tex] V.R = \frac {length \; of \; effort \; arm}{length \; of \; load \; arm} [/tex]
[tex] V.R = \frac {125}{75} [/tex]
V.R = 1.67
to all the physicians please help this is for my assignment
Answer:
Q. 1. Newton's Law of gravitation states that all bodies in the universe exerts a force of attraction on all other bodies in the universe with a proportional force to both the product of the masses of the bodies and inversely proportional to the square of the distance between their centers
Mathematically, we have;
[tex]F = G \times \dfrac{m_1 \times m_2}{R^2}[/tex]
Where;
m₁, and m₂ are the masses of the bodies
R = The distance between their centers
G = The gravitational constant = 6.6743 × 10⁻¹¹ N·m²/kg²
The gravitational constant, G, is the Newton's law of gravitation's constant of proportionality between the force of attraction that exist two bodies and the product of their masses divided by the square of the distance between their centers
Q. 2. Newton's law of gravitation in vector form is presented as follows;
[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12}[/tex]
The above equation gives the gravitational force of attraction of body 1 on body 2, with the negative sign and unit vector indicating that the force of of gravity is towards body 1
The force of gravity of body 2 on 1 is presented as follows;
[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{12}^2} \cdot \hat R_{21}[/tex]
The gravitational force of attraction of body 2 on body 1 is therefore, equal in magnitude and opposite in direction of the gravitational force of body 1 on body 2 (towards body 2)
[tex]-\underset{F_{12}}{\rightarrow} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot -(\hat R_{21}) = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21}[/tex]
[tex]-\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21} = \underset{F_{21}}{\rightarrow}[/tex]
[tex]-\underset{F_{12}}{\rightarrow} = \underset{F_{21}}{\rightarrow}[/tex]
Explanation:
giving me the points are enough
Answer:
the product of mass and velocity
....in my syllabus
can anyone help me to explain theory of relativity???
Answer:
The theory of relativity usually encompasses two interrelated theories by Albert Einstein: special relativity and general relativity, proposed and published in 1905 and 1915, respectively. Special relativity applies to all physical phenomena in the absence of gravity. General relativity explains the law of gravitation and its relation to other forces of nature.It applies to the cosmological and astrophysical realm, including astronomy.
The theory transformed theoretical physics and astronomy during the 20th century, superseding a 200-year-old theory of mechanics created primarily by Isaac Newton. It introduced concepts including spacetime as a unified entity of space and time, relativity of simultaneity, kinematic and gravitational time dilation, and length contraction. In the field of physics, relativity improved the science of elementary particles and their fundamental interactions, along with ushering in the nuclear age. With relativity, cosmology and astrophysics predicted extraordinary astronomical phenomena such as neutron stars, black holes, and gravitational waves
the moon revolves around the earth in a nearly circular orbit kept by gravitational force exerted by the earth work done will be
Answer:
Zero because the applied force is perpendicular to the motion of the object.
No work is done on an object moving is a circular path about a central attractive force.
Any work done in such a case would result in a change in the orbit.
If you run at 1.7 m/s FORWARD ,how does this affect the speed of a ball that you throw?
We have a problem about conservation and velocity, we will find that it does affect the speed of the ball, increasing it by 1.7m/s.
There is something called momentum, which we can define as the "quantity of movement", and we can simply write as the product between velocity and mass.
The momentum is conservative, then we have conservation of momentum.
This means that when you run whit the ball in your hands, the momentum of the ball will be equal to your velocity times the mass of the ball, and this must conserve after you throw the ball.
Now with this idea in mind, this means that if you run with a velocity V, and you throw the ball with a velocity V', the velocity of the ball when it leaves your hand will be:
V + V'.
So, if you run with a velocity of 1.7m/s forward and you throw the ball (assuming in the same direction) the speed of the ball will be 1.7m/s larger than if you were to throw it standing still.
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7. You are using a Bunsen burner to heat a chemical. You need your notebook, which is on the other side of the flame.
Accident:
Prevention:
Accident: Get burned
Prevention: turn off the burner.
calculate the average speed of an athele who runs a distance of 100m in 16 s and an additional of 400m in a 44s
Explanation:
speeds = distance/time
=100/16
=6.25m/s
second speed is;
400/44 =9.09
av. speed = total speed /n
= (6.25+9.09)/2
=7.67 m/s
Distance travelled by a free falling object in the first second is: a) 4.9m b) 9.8m c) 19.6m d) 10m
In free fall
[tex]\boxed{\sf s=-\dfrac{1}{2}gt^2}[/tex]
[tex]\\ \sf\longmapsto s=-\dfrac{1}{2}\times 9.8(1)^2[/tex]
[tex]\\ \sf\longmapsto s=-4.9(1)[/tex]
[tex]\\ \sf\longmapsto s=-4.9m[/tex]
Take it positive[tex]\\ \sf\longmapsto s=4.9m[/tex]
Option a is correctTotal distance between Karachi and Hyderabad is 120 km , if a car speed is 40 km/h, In how many hours it can travel back to Hyderabad
it takes 6 hours to travel back to Hyderabad
