Answer:
Z = 138.5 Ω
Explanation:
In a series RLC circuit the impedance is
Z = [tex]\sqrt{R^2 + ( X_L - X_C)^2 }[/tex]
the capacitive impedance is
X_C = 1 / wC
the inductive impedance is
X_L = wL
in this exercise indicate that C = 50 10⁻³ F, L = 0.3 H and the frequency is f=60 Hz
angular velocity and frequency are related
w = 2π f
w = 2π 60
w = 376.99 rad / s
let's calculate
Z = [tex]\sqrt{80^2 + ( 376.99 \ 0.3 - \frac{1}{376.99 \ 50 \ 10^{-3}} )^2 }[/tex]
Z = [tex]\sqrt{6400 + ( 113.1 - 0.053)^2}[/tex]
Z = √19179.6
Z = 138.5 Ω
The bus travelled at velocity 15 ms-l for 5 minutes before it came to a stop. By using suitable linear equation, calculate the distance the bus has travelled.
Answer:
ans: 2250 meters
Explanation:
initial velocity (U)= 15 m/s
final velocity (V) = 0m/s , since need to come in rest
total time taken (T) = 5 min= 300 seconds
total distance covered (S)= UT + 1/2 aT^2 ,
a= acceleration
S= 15 × 300 + 0.5 ×(0 - 15) × 300
since a = (V - U)/ T
S = 4500 - 2250
S= 2250 m
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.
To know more about half-life, visit;
https://brainly.com/question/24710827
#SPJ3
You should extending your throwing hand straight up to the sky to follow-through.
O True
O False
False
It's not straight up
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
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
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:
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
The question is incomplete. The complete question is :
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
Solution :
Given that :
The predicted range is 0.3503 m
Time of the fall is :
[tex]$t=\sqrt{\frac{2H}{g}}$[/tex]
[tex]v_1t= 0.35[/tex] ...........(i)
[tex]v_0t= 1.09[/tex] ...........(ii)
Dividing the equation (ii) by (i)
[tex]$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$[/tex]
∴ [tex]v_0=3.11 \ v_1[/tex]
Now loss of energy = change in the kinetic energy
[tex]$W=\frac{1}{2} m [v_0^2-v_1^2]$[/tex]
[tex]$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$[/tex]
[tex]$W=7.307\times 10^{-3} \ v_0^2$[/tex]
If f is average friction force, then
(f)(L) = W
(f) (1) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]
(f) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]
The Average force of friction is ( F ) = 7.307 * 10⁻³ v₀²
Given data:
Predicted range ( v₁t ) = 0.3503 m
Actual range ( v₀t ) = 1.09 m
mass = 16.3 g
First step : Determine the value of V₀
[tex]t = \sqrt{\frac{2H}{g} }[/tex] , v₁t = 0.3503 , ( v₀t ) = 1.09 m
To obtain the value of V₀
Divide ( v₀t ) by ( v₁t ) = 1.09 / 0.3503 = 3.11 v₁
∴ V₀ = 3.11 v₁
Next step : Determine the average force of friction ( f )
given that loss of energy results in a change in kinetic energy
W = [tex]\frac{1}{2} m ( vo^{2} - v1^{2} )[/tex]
= 1/2 * 16.3 * 10⁻³ * [ v₀² - [tex](\frac{v_{0} }{3.11} )^{2}[/tex] ]
∴ W = 7.307 * 10⁻³ v₀²
Average force of friction = W / Actual length
= 7.307 * 10⁻³ v₀² / 1
∴ Average force of friction ( F ) = 7.307 * 10⁻³ v₀²
Hence we can conclude that the average force of friction is 7.307 * 10⁻³ v₀²
Learn more about average force of friction : https://brainly.com/question/16207943
Your question has some missing data below are the missing data related to your question
Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
For a different power plant the power input (not output) is 3570 MW. The coal has an energy content of 28,000 kJ per kg. How much coal must be put into the plant each day
Answer:
31475404.8 kg/day
Explanation:
From the information given:
The power plant capacity W = 3570 MW
Energy content of the coal = 28000 kJ/kg
let assume that the thermal efficiency = 35%
Recall that:
1 kw = 3600 kJ/hr provided that the energy conversion is 100% efficient
But assuming the thermal efficiency = 35%.
Then:
Heat input = 3600/0.35 = 10286 kJ/kw.hr
Now, for producing 1 kw.hr, the quantity of the required coal = 10286/28000
= 0.36736 kg
For 3570 MW, the amount of coal that must be input is:
= 0.36736 kg × 3570000
= 1311475.2 kg/hr
= 1311475.2 × 24 kg/day
= 31475404.8 kg/day
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the junction. The current of wire 2 is 0.65 A out of the junction.
Required:
a. How many electrons per second move past a point in wire 3?
b. In which direction do the electrons move -- into or out of the junction?
Answer:
a. 1.56 × 10¹⁸ electrons per second
b. The electrons in wire 3 flow into the junction.
Explanation:
Here is the complete question
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?
Solution
(a) How many electrons per second move past a point in wire 3?
Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and i₃ = = current in wire 3,
So, i₃ = -(i₁ + i₂)
taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A
So, i₃ = -(i₁ + i₂)
i₃ = -(0.40 A + (-0.65 A))
i₃ = -(0.40 A - 0.65 A)
i₃ = -(-0.25 A)
i₃ = 0.25 A
Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561 × 10¹⁹ electrons per second = 1.561 × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second
(b) In which direction do the electrons move -- into or out of the junction?
Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.
In a certain region of space near earth's surface, a uniform horizontal magnetic field of magnitude B exists above a level defined to be y = 0. Below y = 0 , the field abruptly becomes zero (seethe figure). A vertical square wire loop has resistivity rho mass density rhom, diameter d, and side length l. It is initially at rest with its lower horizontal side at y = 0 and is then allowed to fall under gravity, with its plane perpendicular to the direction of the magnetic field.
a) While the loop is still partially immersed in the magnetic field (as it fallsinto the zero-field region), determine the magnetic "drag" forcethat acts on it at the moment when its speed is v.
b) Assume that the loop achieves a terminal velocity vt before its upper horizontal side exits the field. Determine a formulafor vt
c) If the loop is made of copper and B = 0.80 T find vt
Answer:
a) F = [tex]\frac{\pi d^2B^2lv}{16p}[/tex]
b) attached below
c) 0.037 m/s
Explanation:
a) Determine the magnetic "drag" force acting at the moment
speed = v
first step: determine current in the loop
I = [tex]\frac{\pi d^2}{16pl} B lv[/tex] ----- ( 1 )
given that the current will induce force on the three sides of the loop found in the magnetic field
forces on vertical sides = + opposite
we will cancel out
hence equation 1 becomes
F = [tex]\frac{\pi d^2B^2lv}{16p}[/tex] ( according to Lenz law we can say that the direction of force is upwards and this force will slow down the decrease in flux )
b) Determine the formula for Vt
attached below
c) Find Vt
given :
B = 0.80 T
density of copper = 8.9 * 10^3 kg/m^3
resistivity of copper = 1.68 * 10^-8 Ωm
∴ Vt = 16 ( 8.9 * 10^3 kg/m^3 ) ( 1.68 * 10^-8 Ωm ) ( 9.8 m/s^2 ) / ( 0.08 T)^2
= 0.037 m/s
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
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:
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]
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 much heat energy is required to vaporize a 1 g ice cube at 0 C. The heat of fusion of ice is 3.36 x 10^5 J/kg . The heat of vaporization of water is 2.26 x 10^6 J/Kg. ( specific heat of water = 4200 J/Kg C )
Answer:
Q = M * (Cf + C * 100 + Cv)
Cf and Cf are heats of fusion and vaporization and C is the heat required to heat mass M of water 1 deg
Q = .001 kg ( 3.36 * E5 + 100 deg * 4200 + 2.26 * E6) J
Q = .001 kg ( 3.36 J / kg + 4.2 J / kg + 22.6 J /kg) * 10E5
Q = .001 kg * 30.2 * 10E5 J / kg = 3020 J
What happens in the gray zone between solid and liquid?-,-
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
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!
A dog accelerates at 1.50 m/s2 to reach a velocity of 13.5 m/s while covering a distance of 49.3 m. What was his initial velocity?
Let v be the dog's initial velocity. Then
(13.5 m/s)^2 - v ^2 = 2 (1.50 m/s^2) (49.3 m)
==> v ^2 = (13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m)
==> v = √((13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m))
==> v ≈ 5.86 m/s
If I am going to explore Mars what will I need?
Answer:
uhm oxygen tank, a suit (like an astronaut suit) uh food, and a space ship
Explanation:
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:
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?
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
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:
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 glass of milk has what kind of energy?
A. Chemical Potential Energy
B. Kinetic Energy
C. Elastic Potential Energy
D. Radiant Energy
A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d. A battery fully charges the capacitor and is then disconnected. The plate separation is now increased to a distance of 2d. What would be the change, if any, of the voltage across the capacitor, the electric field between the plates, and the energy stored in the capacitor?
Answer:
The answer is "Option D".
Explanation:
Please find the complete question in the attached file.
As plate separation increased to 2d the capacitance get halred but the change remain same
[tex]\therefore V=\frac{Q}{C}[/tex]
The voltage doubles are now electric field remain same because both the distance and voltage get doubled.
[tex]\to E=\frac{v}{d}\ = \frac{2v}{2d}\\\[/tex]
So,
[tex]energy=\frac{1}{2}\ \frac{Q^2}{C}\\\\c'=\frac{C}{2}\\\\E'=2E[/tex]
how many continents do have in africa
Answer:
There was 7 continents in africa
Can someone help me
Answer:
Explanation:before the phase change the substance is a particle.
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]
