Answer:
Radius, r is equal to 229.16×10^6m
Explanation:
Given the following parameters;
Centripetal force on the satellite, Fc = 190N.
Mass of the satellite, M = 1,550-kg.
Speed of the satellite, V = 5,300m/s.
The relationship between a satellite of mass (m) moving in a circular orbit of radius (r) with a speed (v) and a centripetal force (Fc) is given by the equation;
Fc = (MV²)/r
Since, we're solving radius, r; we make "r" the subject of formula;
Thus, r = (MV²)/Fc
Substituting into the above equation;
r = (1550 × [5300]²)/190
r = (1550 × 28090000)/190
r = 43539500000/190
r = 229155263.16
Radius, r is equal to 229.16 × 10^6m
Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equationF=Gm1m2r2where F is the magnitude of the gravitational attraction on either body, m1 and m2 are the masses of the bodies, r is the distance between them, and G is the gravitational constant. In SI units, the units of force are kgâ‹…m/s2, the units of mass are kg, and the units of distance are m. For this equation to have consistent units, the units of G must be which of the following?Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equation,where is the magnitude of the gravitational attraction on either body, and are the masses of the bodies, is the distance between them, and is the gravitational constant. In SI units, the units of force are , the units of mass are , and the units of distance are . For this equation to have consistent units, the units of must be which of the following?A. kg3mâ‹…s2B. kgâ‹…s2m3C. m3kgâ‹…s2D. mkgâ‹…s2
Answer:
The S.I unit of G must be m³/kg.s² to keep the equation consistent.
Explanation:
We have the equation:
F = Gm₁m₂/r²
where,
F = Gravitational Force between Two Bodies
G = Gravitational Constant
m₁ = Mass of 1st Body
m₂ = Mass of 2nd body
r = Distance between the Bodies
The S.I Units of the quantities are:
F = Newton = kg.m/s²
m₁ = kg
m₂ = kg
r = meter = m
G = ?
Therefore, to find the units of Gravitational Constant (G), we substitute the known units in the formula:
kg.m/s² = G(kg)(kg)/m²
G = m³/kg.s²
Explain the force responsible for the formation of our solar system.
Answer:
gravity
Approximately 4.5 billion years ago, gravity pulled a cloud of dust and gas together to form our solar system.
Explanation:
Which best explains why making a pancake from batter is an example of a chemical change?
-The pancake that forms is a different state of matter.
-The change from batter to pancake can be reversed.
-A new substance forms when the batter is cooked.
-The batter changes shape when it is cooked.
The correct answer is C. A new substance forms when the batter is cooked.
Explanation:
When a chemical change occurs the properties, and composition of substances change. This means atoms in the substance re-arrange to form a new substance. This only occurs when there is a chemical change, but not when physical changes occur, indeed a physical change only affects the state of the matter, shape, size, etc.
In the case of the pancake, this is an example of a chemical change because though the process of cooking the pancake changes its composition. Due to this, the properties of the cooked pancake, and the butter are not the same as a new substance forms. Also, in this and most chemical changes, reversibility is not possible, that is why you cannot reverse the process and make the cooked pancake batter once again.
Answer:
C. on edge
Explanation:
Explain why you cannot charge one end of a steel rod and leave the other end uncharged
Answer:
Electrons can be made to move from one object to another. However, protons do not move because they are tightly bound in the nuclei of atoms.Static charge occurs when electrons build up on an object. Static charge:
can only build up on objects which are insulators, eg plastic or wood
cannot build up on objects that act as conductors, eg metals
Conductors allow the electrons to flow away, forming an electric current.
When a static charge on an object is discharged, an electric current flows through the air. This can cause sparks. Lightning is an example of a large amount of static charge being discharged.
Explanation:
A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.220 T/s.
Required:
What is the magnitude of the electric field induced in the ring?
Answer:
Ein: 2.75*10^-3 N/C
Explanation:
The induced electric field can be calculated by using the following path integral:
[tex]\int E_{in} dl=-\frac{\Phi_B}{dt}[/tex]
Where:
dl: diferencial of circumference of the ring
circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m
ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)
The electric field is always parallel to the dl vector. Then you have:
[tex]E_{in}\int dl=E_{in}(2\pi r)=E_{in}(0.157m)[/tex]
Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:
[tex]E_{in}(0.157m)=-A\frac{dB}{dt}=-(1.96*10^{-3}m^2)(-0.220T/s)=4.31*10^{-4}m^2T/s\\\\E_{in}=\frac{4.31*10^{-4}m^2T/s}{0.157m}=2.75*10^{-3}\frac{N}{C}[/tex]
hence, the induced electric field is 2.75*10^-3 N/C
Batman and Robin are attempting to escape that dastardly villain, the Joker, by hiding in a large pool of water (refractive index nwater = 1.333). The Joker stands gloating at the edge of the pool. (His makeup is watersoluble.) He holds a powerful laser weapon y1 = 1.49 m above the surface of the water and fires at an angle of θ1 = 27◦ to the horizontal. He hits the Boy Wonder squarely on the letter "R", which is located y2 = 3.77 m below the surface of the water. θ x y y 1 1 2 R J Batplastic surface Mirrored Surface water B How far (horizontal distance) is Robin from the edge of the pool? (Fear not, Batfans. The "R" is made of laser-reflective material.) Answer in units of m.
Answer:
x_total = 4.29m
Explanation:
To solve this exercise we must work in parts. Let's use the law of refraction to find the angle of the refracted ray and trigonometry to find the distances.
Let's start by looking for the angles that the laser refracts
n₁ sin θ₁ = n₂ sin θ₂
where n₁ is the air refraction compensation n₁ = 1, n₂ the water refractive index n₂ = 1,333
θ₂ = sin⁻¹ (n₁ sin θ₁/n₂)
θ₂ = sin⁻¹ (1 sin 27 / 1,333)
θ₂ = sin⁻¹ 0.34057
θ₂ = 19.9º
now let's find the distance from the edge of the pool to the point where the ₂lightning strikes the water
tan θ₁ = y₁ / x₁
x₁ = y₁ / tan θ₁
x₁ = 1.49 / tan 27
x₁ = 2,924 m
Now let's look for the waterfall in the water as far as Robin
tan θ₂₂ = y₂ / x₂
x₂ = y₂ / tan θ₂
x₂ = 3.77 / tan 19.9
x₂ = 1,364
the distance from the edge of the pool to Robin is
x_total = x₁ + x₂
x_total = 2,924 + 1,364
x_total = 4.29m
Tidal forces are gravitational forces exerted on different parts of a body by a second body. Their effects are particularly visible on the earth's surface in the form of tides. To understand the origin of tidal forces, consider the earth-moon system to consist of two spherical bodies, each with a spherical mass distribution. Let re be the radius of the earth, m be the mass of the moon, and G be the gravitational constant.
(a) Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration ae of the earth due to the gravitational pull of the moon? Express your answer in terms of G, m, and r.
(b) Since the gravitational force between two bodies decreases with distance, the acceleration a_near experienced by a unit mass located at the point on the earth's surface closest to the moon is slightly different from the acceleration a_far experienced by a unit mass located at the point on the earth's surface farthest from the moon. Give a general expression for the quantity a_near - a_far. Express your answer in terms of G, m, r, and re.
Answer:
Explanation:
radius of earth = re
mass of the noon = m
mass of the earth = E
distance between earth and moon = r
acceleration of earth ae
force on earth = GMm / r²
acceleration of the earth
ae = force / mass
= GMm / (r² x M )
= Gm / r²
b ) The point on the earth nearest to moon will be at a distance of r - re
a_near = Gm / ( r - re)²
The point farthest on the earth to moon will be at a distance of r + re
a_ far = Gm / ( r + re )²
How much time is needed to push a 5,000 N car 50 meters if you are using a machine with a power of 4,500 W?
Answer:
55.56
Explanation:
5000N * 50 = 250000/4500= 55.55555555 or 55.56
Revlew Millikan's Photoelectric Experiment Robert A. Mlkan (1868 1953). although best known for his "oil-drop experiment," which measured the charge of an electron, also perfomed pioneering research on the photoelectric effect. In experiments on lithium, for example, Millikan observed a maximum kinetic energy of 0.550 eV when electrons were ejected with 433.9-nm light. When light of 253.5 m was used, he observed a maximum kinetic energy of 2.57 eV.
Part A What is the work function,W, for lithium, as determined from Milikan's results? Express your answer to three significant figures and include appropriate units.
Part B What maximum kinetic energy do you expet illikan found when he used light with a wavelength of 362.4 TIm? Express your answer to three significant figures and include appropriate units Value Units
Answer:
A.) Work function = 2.3 eV
B.) Max. K.E observed = 1.1 eV
Explanation:
A.) Millikan observed a maximum kinetic energy of 0.550 eV when electrons were ejected with 433.9-nm light. When light of 253.5 m was used, he observed a maximum kinetic energy of 2.57 eV.
work function (f) is the minimum energy required to remove an electron from the surface of the material.
hf = Ø + K.E (maximum)
Where
h = Plank constant 6.63 x 10-34 J s
Ø = work function
hc/λ = Ø + K.E (max)
(6.63×10^-34 × 3×10^8)/433.9×10^-9 = Ø + 0.550 × 1.6×10^-19
4.58×10^-19 = Ø + 8.8×10^-20
Ø = 4.58×10^-19 - 8.8×10^-20
Ø = 3.7 × 10^-19 J
Converting Joule to eV
Ø = 3.7 × 10^-19/1.6×10^-19
Ø = 2.3 eV
B.) When light of wavelength 362.4 m is used
The maximum K.E observed = incident light K.E - (the work function).
Incident K.E = hf = hc/λ
Incident K.E =
(6.63×10^-34 × 3×10^8)/362.4
Incident K.E = 5.5 × 10^-28J
Let's convert joule to eV
Incident K.E = 5.5×10^-28/1.6×10^-19
Incident K.E = 3.4 × 10^-9
Max. K.E observed = 3.4 - 2.3
Max. K.E observed = 1.1 eV
A ball thrown downward by initial speed of 3m/s. It hits the ground after 5 second.
How high is throwing point?
If the frequency of a wave is tripled, what happens to the period of the wave?
Answer:
if the frequency of the wave if tripled then period of wave gets tripled
When the wave frequency is tripled, the period of the wave becomes one-third of its original.
Relationship between frequency and period of a waveThe frequency (f) of a wave is inversely proportional to the period (T) of the wave.
Mathematically,
[tex]$f \propto \frac{1}{T}$[/tex]
Thus, when frequency increases, the period decrease and when frequency decreases, the period of the wave increases.
So when frequency of a wave increases by three time, the period of the wave decreases by three times.
Hence, when the frequency is tripled, the period of the wave becomes one-third of its original value.
Learn more about "frequency and period" here :
https://brainly.com/question/18657169
A string, fixed at both ends and 2.14 m long, is vibrating in its second harmonic. It excites a 0.5-m long pipe closed on one end into its third harmonic (first overtone). Use a speed of sound in air of 345 m/s if needed. (a) Sketch the normal modes given above for both the string and the closed-open pipe. [8] (b) What is the wave speed on the string? [7] (c) Write down the wave function y(x,t) for the 2nd harmonic wave on the string. Assume an arbitrary amplitude. [10] (d) If the mass density of the string is 0.55 g/m, what is the tension in the string? [5]
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
Which shows evidence of active transport?
A scientist places four identical cells into four different
liquids, each with different concentrations of magnesiuni.
Celll
w
Description of Liquid
Slightly more magnesium than the
cell
The least amount of magnesium
O
O
O
O
cells W and Z
cell W only
cell Y only
cells X and Y
Result
Took in
magnesium
Took in
magnesium
Took in
magnesium
Took in
magnesium
Slightly less magnesium than the
cell
The most amount of magnesium
Answer: D
X and Y
Explanation:
X The least amount of magnesium Took in magnesium
Y Slightly less magnesium than the cell Took in magnesium
Because active transport occurs when ions or molecules move from less concentration region to high concentration region through semi membrane with the help of some energy.
Answer:
Cells X and Y
Explanation:
Active transport occurs when a substance moves across a membrane against its concentration gradient.
Cells W and Z were placed in a liquid containing more magnesium than the cells. Magnesium therefore, diffuses passively down it's concentration gradient into the cells.
However, cells X and Y were placed in a solution containing less magnesium than the cells, yet these cells took in magnesium against this concentration gradient. This, shows that active transport had taken place.
Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiempo alcanzará el coche a la furgoneta? ¿A qué distancia se producirá el encuentro?
Answer:
t = 0.33h = 1200s
x = 18.33 km
Explanation:
If the origin of coordinates is at the second car, you can write the following equations for both cars:
car 1:
[tex]x=x_o+v_1t[/tex] (1)
xo = 10 km
v1 = 55km/h
car 2:
[tex]x'=v_2t[/tex] (2)
v2 = 85km/h
For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:
[tex]x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}[/tex]
[tex]t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s[/tex]
The position in which both cars coincides is:
[tex]x=(55km/h)(0.33h)=18.33km[/tex]
When you turn down the volume on the television, you reduce the _______ carried by the sound waves, so you also reduce their ________
Answer:
when you turn down the volume on the television, you reduce the intensity carried by the sound waves, so you also reduce their amplitude.
Explanation:
When you turn down the volume of the television, you are actually reducing the intensity of the sound wave, which is directly proportional to the amplitude of the sound. Amplitude is height of the sound wave.
Therefore, when you turn down the volume on the television, you reduce the intensity carried by the sound waves, so you also reduce their amplitude.
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
Answer:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
b) Determine the magnitude and direction of current in wire 3
Explanation:
a) [tex]F_{net} \text {on wire }3=0[/tex]
[tex]\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm[/tex]
position of wire = 50 - 1.2
= 48.8cm
b) [tex]F_{net} \text {on wire }1=0[/tex]
[tex]\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A[/tex]
Direction ⇒ downward
Let’s discuss some particle physics. In 1897, physicist J.J. Thomson conducted a seminalexperiment with a cathode ray tube. In the experiment, a beam of an unknown particle of chargeqandmassmis shot through the tube onto a fluorescent screen, at some speedvand in thex-direction. Uponimpact the beam leaves a bright spot on the screen. Sincevis very fast, effects of gravity are negligible andthe beam forms essentially a straight line normal to the screen.
To eliminate this deflection, an external uniform magnetic field of magnitudeBis turned on as well. What should be the direction and magnitudeBof this field such that ∆z= 0?
Answer:
The direction of the magnetic field should be in the x and y coordinates alone, and their should be no change towards the z coordinate. Also the magnitude of the magnetic field in the z direction would be zero if there is no magnetic field in that coordinate.
The coordinate should be:
B = bi + bj
Thinking about the winter we missed out on this year. Calvin and his tiger go sledding down a snowy hill. There is friction between the snow and the sled, and there is air resistance. Looking at the entire time that they spend on the hill (from when they start from rest at the top of the hill until they reach the bottom moving fast), indicate whether each of the following quantities is positive, negative, or zero, by writing either, -, or 0 for the questions below. We take Calvin, the tiger, and the sled as our system.
a. The work done by the normal force that the snow exerts on the system?
b. The work is done by the frictional force that the snow exerts on the system?
c.The change in the kinetic energy of the system?
d.The change in the gravitational potential energy of the system?
e. The total work is done on the system?
Answer:
a) W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d) the variation of the potential enrgy is negative,
e) total work is positive
Explanation:
Work in physics is defined by the scalar scalar product of force by displacement
W = F. dx
The bold are vectors; this can be written in the form of the mules of the quantities
W = F dx cos θ
where θ is the angle between force and displacement.
a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is
θ = 90 cos 90 = 0
W=0
In conclusion the work is zero
b) The friction force opposes the displacement whereby the angle is
θ = 180 cos 190 = -1
W = - fr d
Work is negative
c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy
m g sin θ L = ΔK
this magnitude is positive since the angle is zero cos 0 = 1
how the system starts from rest ΔK = Kf -K₀= + Kf -0
work is positive and scientific energy variation is positive
d) change in potential energy
The potential energy is is ΔU = Uf -U₀
we fix the reference system in the bases of the plane so Uf = 0
ΔU = -U₀
the variation of the potential enrgy is negative
e) The total work is formed by the work of the weight component, the work of the friction force
W_Total = W_weight - W_roce
as the body moves down
W_Total> 0
Therefore the total work is positive
if the power developed in an electric circuit is doubled the energy used in one second is
Answer:
Energy is doubled.
Explanation:
Power developed in an electric is the rate of change in time of electric energy travelling throughout circuit. The most common units is the amount of energy used in a second. Therefore, if power is double, the energy used in one second is also doubled.
Which forms of energy are this a blender?
Answer:
Electrical energy.
Explanation:
Electrical energy comes into the blender, which turns into magnetic energy in the electric motor. This magnetism repels permanent magnets. The motor spins as well as the blade. So the magnetism is converted to kinetic energy.
In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors
A =(1,0,−3), B =(−2,5,1), and C =(3,1,1). Calculate the following, and express your answers as ordered triplets of values separated by commas.
a. A- B=______
b. B-C=_______
c. -A +B- C=______
d. 3A- 2C=______
e. -2A+3B- C=_____
f. 2A- 3(B-C)
Answer:
a)A-B =(1,0,-3) +(2,-5,-1) =(3, -5, -4);
b)B-C =(-2-3, 5-1, 1-1) = (-5,4,0);
c)-A +(B –C) = (-1,0,3) +(-5,4,0) =(-6,4,3);
d) =(3,0,-9) –(6,2,2) = (-3, -2, -11);
e) = (-2,0,6) +(-6,15,3) +(-3,-1,-1) =
d) = (2,0,-6) –(-5*3, 4*3, 0) =
Explanation:
1)C=k*A = (k*a1, k*a2, k*a3);
2)C=-A = -1*A;
3)C= A+B = (a1+b1, a2+b2, c2+c2);
4)C=A-B =A +(-B);
5)A+B=B+A;
6)A+B+C =(A+B)+C =A+(B+C
What does the speed of sound depend on ?
Answer:
As with any wave the speed of sound depends on the medium in which it is propagating.
Explanation:
What was the revolution in atomic theory produced by the discovery of the electron?
Explanation:
the atom are comprised of particle. this atomic theory also proves that atoms cannot be destroyed nor created and it composed of many particles. it is also said that atoms of the same element can be identical.
A car is travelling at 16.7m/s. If the car can slow at a rate of 21.5m/s^2, how much time does the driver need in order to stop the red light?
Answer:
About 0.7767 seconds
Explanation:
[tex]\dfrac{16.7m/s}{21.5m/s^2}\approx 0.7767s[/tex]
Hope this helps!
Which type of power plant uses the movement of air in nature to generate
electricity?
A. Radiant
B. Wind
C. Coal
D. Hydroelectric
Answer:
Wind
Explanation:
Which factors affect the resistance of a material?
Explanation:
The resistance of a material is given by the formula as follows :
[tex]R=\rho \dfrac{L}{A}[/tex]
Here,
[tex]\rho[/tex] = resistivity of a material
L is the length of the material
A is the area of cross section of the material
So, the factors on which the resistance of a material depends are :
Length and area of cross section
how much external energy is required to bring three identical point charges (20uc) from infinity and place them at the corners of an equilateral triangle with side of 2 meter length:
Answer:
U = 269.4 kJ
Explanation:
The energy required to place the three charges from infinity is given by:
[tex]U=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}[/tex]
In this case, you have that
q1 = q2 = q3 = q = 20uC
r12 = r13 = r23 = r = 2m
k: Coulomb constant = 8.98*10^9 NM^2/C^2
Then, you replace the values of q, r and k in the equation for the energy U:
[tex]U=3k\frac{q^2}{r}\\\\U=3(8.98*10^9Nm^2/C^2)\frac{(20*10^{-6}C)^2}{2m}=269400\ J=269.4\ kJ[/tex]
hence, the required energy is 269.4 kJ
What is the relationship Between frequency and sound?
Answer:
The sensation of a frequency is commonly referred to as the pitch of a sound. A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave.
Explanation:
Hope this helps : )
Consider a spinning plate is dropped onto a stationary plate (which is itself at rest on a frictionless surface). Both plates have a radius of 30cm and a mass of 1kg. The spinning plate is initially spinning at a rate of 0.7 revolutions per second. Hint: This is like a totally-inelastic collision.
Required:
a. After a sufficiently long time, what is the angular velocity of the initially-spinning plate? What about the initially-stationary plate?
b. Assume that the period of velocity matching happens over a course of 2 seconds. Further, assume that the torque exerted by each plate on the other is constant over time. In that case, what is the magnitude of the acceleration that each plate feels during those two seconds? Hint: Use the rotational impulse-momentum theorem.
Answer:
The final angular velocity is [tex]w_f = 2.1994 rad/sec[/tex]
The angular acceleration is [tex]\alpha = 1.099 \ rad/sec^2[/tex]
Explanation:
From the question we are told that
The radius of each plate is [tex]r = 30 \ cm = \frac{30}{100} = 0.3 \ m[/tex]
The mass of each plate is [tex]m_p = 1 \ kg[/tex]
The angular speed of the spinning plate is [tex]w = 0.7 \ rev \ per \ sec = 0.7 * 2 \pi = 4.3988 \ rad/sec[/tex]
From the law of conservation of momentum
[tex]L_i = L_f[/tex]
Where [tex]L_i[/tex] is the initial angular momentum of the system (The spinning and stationary plate ) which is mathematically represented as
[tex]L_i = I_1 w + 0[/tex]
here [tex]I_1[/tex] is the moment of inertia of the spinning plate which mathematically represented as
[tex]I_1 = \frac{m_pr^2}{2}[/tex]
and the zero signify that the stationary plate do not have an angular momentum as it is at rest at the initial state
[tex]L_f[/tex] is the final angular momentum of the system (The spinning and stationary plate) , which is mathematically represented as
[tex]L_f = (I_1 + I_2 ) w_f[/tex]
Where
[tex]I_2[/tex] is the moment of inertia of the second plate (This was stationary before but now it spinning due to the first pate ) and is equal to [tex]I_1[/tex]
and [tex]w_f[/tex] is the final angular speed
So we have
[tex]I_1 w = (I_1 + I_2)w_f[/tex]
[tex]\frac{m_p r^2}{2} * w = 2 * \frac{m_p r^2}{2} * w_f[/tex]
[tex]w = 2 * w_f[/tex]
substituting values
[tex]4.3988 = 2 * w_f[/tex]
[tex]w_f = \frac{4.3988 }{2}[/tex]
[tex]w_f = 2.1994 rad/sec[/tex]
The the rotational impulse-momentum theorem can be mathematially represented as
[tex]\tau * \Delta t = 0.09891[/tex]
Where [tex]\tau[/tex] is the torque and [tex]\Delta t[/tex] is the change in time
So at [tex]\Delta t = 2 \ sec[/tex]
[tex]\tau = \frac{0.09891}{2}[/tex]
[tex]\tau = 0.0995 \ Nm[/tex]
now the angular acceleation is mathematically represented as
[tex]\alpha = 2 * \frac{\tau}{m_p * r^2 }[/tex]
substittuting values
[tex]\alpha = 2 * \frac{0.0995}{1 * 0.3^2}[/tex]
[tex]\alpha = 1.099 \ rad/sec^2[/tex]
An electricity-generating station needs to deliver 20 MW of power to a city 1.0 km away. A common voltage for commercial power generators is 22 kV, but a step-up transformer is used to boost the voltage to 230 kV before transmission.
(a) If the resistance of the wires is 2.0 ohms and the electricity costs about 10 cents/kWh, estimate what it costs the utility company to send the power to the city for one day?
Answer:
Explanation:
Power to be transferred = 20 x 10⁶ W .
Voltage at which power is transferred V = 230 x 10³ V .
If current in the carrying wire is I
V x I = Power
230 x 10³ x I = 20 x 10⁶
I = 86.9565 A
Power loss in the transmission line
I² R , I is current and R is resistance
= 86.9565² x 2
= 15122.86 W
= 15.123 KW
In one day power loss
= 15.123 x 24 kWH .
= 363 kWH .
Cost = 363 x 10
= 3630 cent
= 36.30 dollar .