Answer:Alternativa A. Damos o nome de interferência a superposição de efeitos que ocorre ao ser produzido dois pulsos de onda, que serão propagados e acabarão inevitavelmente por se encontrar. No instante em que os pulsos se cruzarem, há então, uma superposição de efeitos individuais de cada um deles. Se durante o cruzamento, houver um reforço das ondas, estará ocorrendo a este fenômeno.
what is the difference between VELOCITY and SPEED?
Answer:
Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement. Put another way, speed is a scalar value, while velocity is a vector. ... In its simplest form, average velocity is calculated by dividing change in position (Δr) by change in time (Δt).
Explanation:
A force of 3 newtons moves a 10 kilogram mass horizontally a distance of 3 meters. The mass does not slow down or speed up as it moves. Which of the following must be true?
a) 9 joules of kinetic energy were produced
b) 9 joules of gravitational potential energy were produced
c) 9 joules of heat energy were produced
d) 9 joules of kinetic energy and heat were produced
Answer:
9 joules of heat energy was produced
Explanation: there is no acceleration therefore its not a kinetic energy
Energy= force × distance
= 3×3
=9
A spring with a 10-kg mass and a damping constant 15 can be held stretched 2 meters beyond its natural length by a force of 6 newtons. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity.
Required:
Find the position of the mass at any time t.
Answer:
Explanation:
Given that:
mass = 10kg
damping constant C = 15 kg/s
length = 2 m
Force F = 6N
Using the Hooke's law:
F = kx
6 = 15x
k = 6 N /2 m
spring constant k = 3 N/m
For the critical damping
C² - 4k*m= 0
m = C²/4k
m = (15)²/4(3) kg
m = 225/12 kg
m = 18.75 kg
What happens in the gray zone between solid and liquid?-,-
What recommendations would you give to the global government to help Decrease the global effects of human impact on the environment mystery recommendations and how they will positively impact our planet
Answer:
We can help to keep it magnificent for ourselves, our children and grandchildren, and other living things besides us.
Explanation:
5 ways our governments can confront climate change
PROTECT AND RESTORE KEY ECOSYSTEMS
SUPPORT SMALL AGRICULTURAL PRODUCERS
PROMOTE GREEN ENERGY
COMBAT SHORT-LIVED CLIMATE POLLUTANTS
BET ON ADAPTATION, NOT JUST MITIGATION
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
A) 8.03 x 10^16 nuclei
B) 4.01 x 10^16 nuclei
C) 2.02 x 10^16 nuclei
D) 1.61 x 10^17 nuclei
OPTION C is the correct answer.
The radioactive decay follows the first order kinetics. The number of atoms decaying at any time is proportional to the number of atoms present at that instant. The amount of sample left is 2.02 x 10¹⁶nuclei. The correct option is C.
What is half-life?The time required for the decay of one half of the amount of the species is defined as the half-life period of a radionuclide. The half-life period is a characteristic of a radionuclide. The half lives can vary from seconds to billions of years.
The isotope decay of an atom is given by the equation:
ln [A] = -kt + ln [A]₀
The rate constant, k is:
k = ln 2 / Half life
k = ln 2 / 4.96 x 10³
k = 1.40 × 10⁻⁴ s⁻¹
t = 1.98 x 10⁴
[A]₀ = 3.21 x 10¹⁷
ln [A] = -1.40 × 10⁻⁴ × 1.98 x 10⁴ + ln [3.21 x 10¹⁷] = 37.538
[A] = 2.02 x 10¹⁶ nuclei
Thus the correct option is C.
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A gymnast of mass 70.0 kgkg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81m/s29.81m/s2 for the acceleration of gravity.
PART A Calculate the tension T in the rope if the gymnast climbs the rope at a constant rate.
PART B Calculate the tension TTT in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.00 m/s2
PART C Calculate the tension TTT in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.00 m/s2m/s2 .
Answer:
43994
Explanation:
Hope this helps!
A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal energy of the box, determine the magnitude of the increase.
Answer:
Explanation:
From the given information:
The initial PE [tex](PE)_i[/tex] = m×g×h
= 5 kg × 9.81 m/s² × 10 m
= 490.5 J
The change in Potential energy P.E of the box is:
ΔP.E = [tex]P.E_f -P.E_i[/tex]
ΔP.E = 0 - [tex](PE)_i[/tex]
ΔP.E = [tex]-P.E_i[/tex]
If we take a look at conservation of total energy for determining the change in the internal energy of the box;
[tex]\Delta P.E + \Delta K.E + \Delta U = 0[/tex]
[tex]\Delta U = -\Delta P.E - \Delta K.E[/tex]
this can be re-written as:
[tex]\Delta U =- (-\Delta P.E_i) - \Delta K.E[/tex]
Here, K.E = 0
Also, 70% goes into raising the internal energy for the box;
Thus,
[tex]\Delta U =(70\%) \Delta P.E_i-0[/tex]
[tex]\Delta U =(0.70) (490.5)[/tex]
ΔU = 343.35 J
Thus, the magnitude of the increase is = 343.35 J
A 56 kg pole vaulter falls from rest from a height of 5.1 m onto a foam rubber pad. The pole vaulter comes to rest 0.29 s after landing on the pad.
Required:
a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision
a. The athlete's velocity just before reaching the pad is [tex]35.21m/s[/tex]
b. The constant force exerted on the pole vaulter is 6799.52 N
a. We use Newton's equation of motion,
[tex]v=u+at\\\\S=ut+\frac{1}{2}at^{2}[/tex]
Where u is initial velocity, v is final velocity, a is acceleration , t is time and S represent distance.
Given that, s = 5.1 m , t = 0.29s, u = 0
Substitute in above equation.
[tex]5.1=\frac{1}{2}*a*(0.29)^{2} \\\\a=\frac{5.1*2}{0.084}=121.42m/s^{2}[/tex]
the athlete's velocity, [tex]v=0+121.42*(0.29)=35.21m/s[/tex]
b. The constant force exerted on the pole vaulter due to the collision is given as, [tex]Force=mass*acceleration[/tex]
[tex]Force=56*121.42=6799.52N[/tex]
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how do you use the coefficient to calculate the number of atoms in each molecule?
Answer:
To find out the number of atoms: MULTIPLY all the SUBSCRIPTS in the molecule by the COEFFICIENT. (This will give you the number of atoms of each element.)
Explanation:
Calculate the current flowing when the voltage across is 35V and the resistance is 7ohms.
Explanation:
V= IR
35=I×7
I=35/7
I=5amperes
pls give brainliest
A lumberjack is trying to drag a small tree that he cut down. If the static
coefficient of friction of the tree on the ground is 0.5 and the tree weighs 430
N, what is the minimum amount of horizontal force that he will need to apply
so that the tree will start moving?
A. 215 N
B. 430 N
C. 365 N
D. 500 N
Answer:
A
Explanation:
weight of the tree =normal force
Horizontal force =coefficient of friction x Fnormal
0.5×430=215
An electric device, which heats water by immersing a resistance wire in the water, generates 20 cal of heat
per second when an electric potential difference of 6 V is placed across its leads. What is the resistance in Ω
of the heater wire? (Note: 1 cal = 4.186 J)
Select one:
a. 0.86
b. 0.17
c. 0.29
d. 0.43
Answer:
1 cal/s =4.184w
p=50 cal/s =2093w
v=12v
P = V*I
I =P/V
I = 17.43 A
P =1²*R
R = P/I²
R = 0.68What country first colonised Ghana
Answer: Colonialism is a big topic, but it can only be understood by looking at human experiences. Formal colonialism first came to the region we today call Ghana in 1874, and British rule spread through the region into the early twentieth century. The British called the territory the “Gold Coast Colony”.
Explanation: hey, hope this hlps! oh, btw you picked the wrong subject for this question it should have been history insteat of phiscics!
At 20 ◦C a copper wire has a resistance of 4×10−3 Ω and a temperature coefficient of resistivity of 3.9×10−3 (C◦)−1, its resistance at 100 ◦C is
A.
52.5 × 10-3 Ω
B.
5.25 × 10-3 Ω
C.
5.25 × 10-4 Ω
D.
5.25 × 10-2 Ω
E.
25.5 × 10-3 Ω
Answer:
[tex]R _{t} = R _{0}( \alpha t + 1) \\ = 4 \times {10}^{ - 3} (3.9 \times {10}^{ - 3} \times 20 + 1) \\ = 4 \times {10}^{ - 3} (1.078) \\ = 4.312 \times {10}^{ - 3} \: Ω[/tex]
What is the energy equivalent of an object with a mass of 1.05g?
Answer:
The equivalent energy of an object given its mass is calculated through the equation,
E = mc²
where c is the speed of light (3 x 10^8 m/s)
Substituting the known values,
E = (1.05 g/ 1000) (3 x 10^8 m/s)²
E = 9.45x10^13 J
Explanation:
B. Complete the lists:
Things that I must do for my family
Things I must never do to my family
1.
2.
2.
3.
3.
4.
5.
5.
Answer:
Things you should do for your family
help your parentstreat them kindlylisten and obey themappreciate them for anything they do for you talk softlythings you shouldn't
backanswering them Disobey And anything that's harsh or make it parents sadAccording to Newton’s law of universal gravitation, which statements are true?
A body is dropped from a height H. In how much time will it reach the ground?
Answer:
[tex]s = ut + \frac{1}{2} g {t}^{2} \\ t(u + \frac{1}{2} gt) = H \\ u + \frac{1}{2} gt = H \\ t = 2(H - u) \div g \\ t = \frac{(H - u)}{5} \\ u \: is \: speed \: or \: velocity[/tex]
Trong máy phát điện xoay chiều ba pha khi tổng điện áp tức thời của cuộn 1,2 là e1+e2=120V thì điện áp tức thời của cuộn 3 là
Answer:
I just noticd i dont speak this launguage
Explanation:
reviews the general principles in this problem. A projectile is launched from ground level at an angle of 13.0 ° above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?
Answer: θ would equal approximately 28.7°
This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.
Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°
Now if we multiply the range by 2, we get:
2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:
2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ
Thus, θ = 28.67780425
It's been awhile since I did this; though I hope it helped!
A positively charged plastic ruler is brought close to a piece paper resting on the desk. The piece of paper was initially neutral. When the ruler was brought closer, the paper is attracted to the ruler. The surface of the paper became charged through:_________
Answer: static electricity
Explanation:
When the plastic ruler is rubbed, friction opposes the motion and causes the transfer of electron from one surface to another such that plastic becomes negatively charged. When ruler is brought nearer to the paper, it induces the positive charge in the piece of paper.
A 35 kg child slides down a playground slide at a constant speed. The slide has a height of 3.8 m and is 8.0 m long. Find the magnitude of the kinetic friction force acting on the child.
Answer:
The magnitude of the kinetic frictional force acting on the child is 162.93 N
Explanation:
Given;
mass of the child, m = 35 kg
height of the slide, h = 3.8 m
length of the slide, d = 8.0 m
The change in thermal energy associated with the kinetic frictional force is calculated as follows;
[tex]\Delta E_{th} + \Delta K.E + \Delta U = 0\\\\\Delta E_{th} + (\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2) + (mgh_f - mgh_i) =0\\\\since \ the \ speed \ is \ constant, \ v_f = v_i \ and \ \Delta K.E = 0\\\\Also, \ final \ height \ , h _f= 0\\\\\Delta E_{th} - mgh_i = 0\\\\\Delta E_{th} = mgh_i\\\\\Delta E_{th} = 35 \times9.8 \times 3.8\\\\\Delta E_{th} = 1303.4 \ J[/tex]
The magnitude of the kinetic frictional force that produced this thermal energy is calculated from the work done by frictional force;
[tex]\Delta E_{th} = F \times d\\\\F = \frac{\Delta E_{th} }{d} \\\\F = \frac{1303.4}{8} \\\\F = 162.93 \ N[/tex]
Therefore, the magnitude of the kinetic frictional force acting on the child is 162.93 N
The following 1H NMR absorptions were obtained on a spectrometer operating at 200 MHz and are given in Hz downfield from TMS. Convert the absorptions to δ units. a) 416 Hz = δ b) 1.97×103 Hz = δ c) 1.50×103 Hz = δ
Answer:
For (a): The chemical shift is [tex]2.08\delta[/tex]
For (b): The chemical shift is [tex]9.85\delta[/tex]
For (c): The chemical shift is [tex]7.5\delta[/tex]
Explanation:
To calculate the chemical shift, we use the equation:
[tex]\text{Chemical shift in ppm}=\frac{\text{Peak position (in Hz)}}{\text{Spectrometer frequency (in MHz)}}[/tex]
Given value of spectrometer frequency = 200 MHz
For (a):Given peak position = 416 Hz
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{416Hz}{200MHz}\\\\\text{Chemical shift in ppm}=2.08\delta[/tex]
For (b):Given peak position = [tex]1.97\times 10^3 Hz[/tex]
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{1.97\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=9.85\delta[/tex]
For (c):Given peak position = [tex]1.50\times 10^3 Hz[/tex]
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{1.50\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=7.5\delta[/tex]
What effect does increased blood flow have on the body when performing exercises? A. delivers more sugar to organs B. delivers more energy to muscles C. delivers more oxygen to the body D. delivers more protein to muscles Please select the best answer from the choices provided. O A . OB ос OD Next Submit Save and Exit Mark this and return
Consider a swimmer that swims a complete round-trip lap of a 50 m long pool in 100 seconds. What is the swimmers average speed and average velocity?
Answer:
The average speed is 1 m/s
The average velocity is 0
Explanation:
Given;
length of the pool, L = 50 m
time taken for the motion, t = 100 s
The total distance = 50 m + 50 m
The total distance = 100 m
The average speed = total distance / total time
= 100 / 100
= 1 m/s
The average velocity = change in displacement / change in time
change in displacement = 50 m - 50 m = 0
The average velocity = 0 / 100
The average velocity = 0
Let’s use these equations to compare the electrostatic force and the gravitational force in a few different situations. In each case, calculate the strength of the electrostatic force and the strength of the gravitational force.
Two electrons separated by 1 cm
q = 1.6 x 10-19 C
m = 9.1 x 10-31 kg
d = 1 cm
What is the electric force?
What is the gravitational force?
Which force will dominate the motion of the electrons?
Answer:
Electric force is the attractive force between the electrons and the nucleus. It works the same way for a negative charge, you also have an electric field around it. ... Now, like charges repel each other and opposite charges attract.
The gravitational force is a force that attracts any two objects with mass. We call the gravitational force attractive because it always tries to pull masses together, it never pushes them apart. ... This is called Newton's Universal Law of Gravitation.
electric force
This table shows clearly that the electric force dominates the motion of electrons in atoms. However, on a macroscopic scale, the gravitational force dominates. Since most macroscopic objects are neutral, they have an equal number of protons and electrons.
Explanation:
Ingrid lives in a cold country, that sometimes gets a lot of snow. when that happens people can enjoy lot pf skiing. Ingrid goes outside to see if the snow is fir for skiing. she sinks into the snow, but when she puts her ski on, she can move over it without sinking. Why?
Answer:
Because the surface area of her skis are greater than the surface are of her shoes
Explanation:
the reason for this is that the weight per in is too heavy crushing the snow blower but with the skies the weight is distributed to the point were the snow can support her weight
how to calculate sound of an echo
by an echo meter
please flw me and thank my answers
#Genius kudi
g Is a nucleus that absorbs at 4.13 δ more shielded or less shielded than a nucleus that absorbs at 11.45 δ? _________ Does the nucleus that absorbs at 4.13 δ require a stronger applied field or a weaker applied field to come into resonance than the nucleus that absorbs at 11.45 δ?
Answer: A nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded and a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field
Explanation:
While interpreting the data in NMR, the positions of signals are studied.
The nucleus/ protons having a higher value of [tex]\delta[/tex] are said to be less shielded. They are said to be upfield.
The nucleus/protons having a lower value of [tex]\delta[/tex] are said to be more shielded. They are said to be downfield.
So, a nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded by the nucleus that absorbs at [tex]4.13\delta[/tex]
Also, the less shielded nucleus/protons will require a weak applied field to come into resonance than the more shielded nucleus/protons
So, a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field to come into resonance than the nucleus that absorbs at [tex]11.45\delta[/tex]
