Answer:
WEATHERING is represented by the scenario (As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock).
Erosion is represented by the scenario (As the waves recede, they carry the sediment away).
Explanation:
A wave is a disturbance which travels through a medium and transfers energy from one point to another. When wind blows over a water body like the ocean, ocean waves are formed. As the generated energy from the wind is transported through the water by the waves, the can hit against rocks on the shores leading to its break down with time. WEATHERING occurs when tiny bit of sediments from rocks are loosened due to the impact of ocean waves.
Erosion can be described as the wearing away of the earth's surface due to the impact of wind, rainfall ( water) or waves. There are different types of erosion which is classified according it's cause of formation.
Wave erosion occurs when sediments such as sand, shell and other materials are carried to the shoreline by ocean waves. This erodes the shore over time as the sediments act like sandpapers.
If the speed of an object does NOT change, the object is traveling at a
constant speed
increasing speed
decreasing speed
Answer:
If the speed does not change at all, the object would be moving at a constant speed.
According to Newton's law of universal gravitation, which statements are true?
As we move to higher altitudes, the force of gravity on us decreases.
O As we move to higher altitudes, the force of gravity on us increases,
O As we gain mass, the force of gravity on us decreases.
O Aswe gain mass, the force of gravity on us increases.
DAs we move faster, the force of gravity on us increases.
A 35 kg box initially sliding at 10 m/s on a rough surface is brought to rest by 25 N
of friction. What distance does the box slide?
Answer:
the distance moved by the box is 70.03 m.
Explanation:
Given;
mass of the box, m = 35 kg
initial velocity of the box, u = 10 m/s
frictional force, F = 25 N
Apply Newton's second law of motion to determine the deceleration of the box;
-F = ma
a = -F / m
a = (-25 ) / 35
a = -0.714 m/s²
The distance moved by the box is calculated as follows;
v² = u² + 2ad
where;
v is the final velocity of the box when it comes to rest = 0
0 = 10² + (2 x - 0.714)d
0 = 100 - 1.428d
1.428d = 100
d = 100 / 1.428
d = 70.03 m
Therefore, the distance moved by the box is 70.03 m.
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a driving speed of 50 mi/h. When the driver is sober, a stop can be made just in time to avoid hitting an object that is first visible 385 ft ahead. After a few drinks under exactly the same conditions, the driver fails to stop in time and strikes the object at a speed of 30 mi/h. Determine the driver's perception/reaction time before and after drinking. (Assume practical stopping distance.)
Answer:
a. 10.5 s b. 6.6 s
Explanation:
a. The driver's perception/reaction time before drinking.
To find the driver's perception time before drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)
a = - 499.52 m²/s²/234.7 m
a = -2.13 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (0 m/s - 22.35 m/s)/-2.13 m/s²
t = - 22.35 m/s/-2.13 m/s²
t = 10.5 s
b. The driver's perception/reaction time after drinking.
To find the driver's perception time after drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)
a = 179.83 m²/s² - 499.52 m²/s²/234.7 m
a = -319.69 m²/s² ÷ 234.7 m
a = -1.36 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²
t = - 8.94 m/s/-1.36 m/s²
t = 6.6 s
A 2028 kg Oldsmobile traveling south on Abbott Road at 14.5 m/s is unable to stop on the ice covered intersection for a red light at Saginaw Street. The car collides with a 4146 kg truck hauling animal feed east on Saginaw at 9.7 m/s. The two vehicles remain locked together after the impact. Calculate the velocity of the wreckage immediately after the impact. Give the speed for your first answer and the compass heading for your second answer. (remember, the CAPA abbreviation for degrees is deg) -1.75
Answer:
v = 8.1 m/s
θ = -36.4º (36.4º South of East).
Explanation:
Assuming no external forces acting during the collision (due to the infinitesimal collision time) total momentum must be conserved.Since momentum is a vector, if we project it along two axes perpendicular each other, like the N-S axis (y-axis, positive aiming to the north) and W-E axis (x-axis, positive aiming to the east), momentum must be conserved for these components also.Since the collision is inelastic, we can write these two equations for the momentum conservation, for the x- and the y-axes:We can go with the x-axis first:[tex]p_{ox} = p_{fx} (1)[/tex]
⇒ [tex]m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx} (2)[/tex]
Replacing by the givens, we can find vfx as follows:[tex]v_{fx} = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} } = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)[/tex]
We can repeat the process for the y-axis:[tex]p_{oy} = p_{fy} (4)[/tex]
⇒[tex]m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy} (5)[/tex]
Replacing by the givens, we can find vfy as follows:[tex]v_{fy} = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} } = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)[/tex]
The magnitude of the velocity vector of the wreckage immediately after the impact, can be found applying the Pythagorean Theorem to vfx and vfy, as follows:[tex]v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s (7)[/tex]
In order to get the compass heading, we can apply the definition of tangent, as follows:[tex]\frac{v_{fy} }{v_{fx} } = tg \theta (8)[/tex]
⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)
⇒ θ = tg⁻¹ (-0.738) = -36.4º
Since it's negative, it's counted clockwise from the positive x-axis, so this means that it's 36.4º South of East.Determine the magnitude of the electric field at the point P. Express your answer in terms of Q, x, a, and k. Express your answer in terms of the variables Q, x, a, k, and appropriate constants.
Complete Question
The question image is in the first uploaded image
Answer:
[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]
Explanation:
From the question we are told that
Distance b/w Q mid point and P is given as x
Generally the equation for magnitude of the electric field at the point P is given as
[tex]E=\frac{kQ}{d^2}[/tex]
where
[tex]k=\frac{1}{4\pi e_0}[/tex]
[tex]d=x^2-a^2[/tex]
Therefore
[tex]E= \frac{1}{4\pi e_0} \frac{Q}{(x^2-a^2)^2}- \frac{1}{4\pi e_0} \frac{Q}{(x^2+a^2)^2}[/tex]
[tex]E= \frac{Q}{4\pi e_0} (\frac{1}{(x^2-a^2)^2}- \frac{1}{(x^2+a^2)^2})[/tex]
Therefore equation for magnitude of the electric field at the point P is
[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]
How does speed and mass effect kinetic energy ?
Answer:
in fact, kinetic energy is directly proportional to mass: if you double the mass, then you double the kinetic energy. Second, the faster something is moving, the greater the force it is capable of exerting and the greater energy it possesses. ... Thus a modest increase in speed can cause a large increase in kinetic energy.
Explanation:
Answer: The more mass of an object has, the more Kinetic energy it has.
Explanation:
Kinetic energy is comparable to mass. If you double the mass then you double the kinetic energy. The faster the object is moving the greater the energy possesses. A large increase in speed can have a large increase in kinetic energy.
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
Answer:
it will option option A hope it helps
What energy store is in the human
BEFORE he/she lifts the hammer?
I believe the answer would be protentional because they have the potential energy in them to lift the hammer.
How much kinetic energy does a 0.104 kg hamster have if it is moving at 24.0 m/s?
Answer:
30J
Explanation:
Given parameters:
Mass of hamster = 0.104kg
Velocity = 24m/s
Unknown:
Kinetic energy = ?
Solution:
Kinetic energy is the energy due to the motion of a body. It is mathematically derived by;
Kinetic energy = [tex]\frac{1}{2}[/tex] m v²
m is the mass
v is the velocity
Kinetic energy = [tex]\frac{1}{2}[/tex] x 0.104 x 24² = 30J
The radius of the Sun is 6.96 x 108 m and the distance between the Sun and the Earth is roughtly 1.50 x 1011 m. You may assume that the Sun is a perfect sphere and that the irradiance arriving on the Earth is the value for AMO, 1,350 W/m2. Calculate the temperature at the surface of the Sun.
Answer:
5766.7 K
Explanation:
We are given that
Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]
Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]
Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]
We have to find the temperature at the surface of the Sun.
We know that
Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]
Where [tex]K_{sc}=1350 W/m^2[/tex]
[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]
Using the formula
[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]
T=5766.7 K
Hence, the temperature at the surface of the sun=5766.7 K
Anyone can help me out with this question ? Just number 2,
Answer:
- 21⁰C .
Explanation:
Speed of jet = 2.05 x 10³ km /h
= 2050 x 1000 / (60 x 60 ) m /s
= 569.44 m / s
Mach no represents times of speed of sound , the speed of jet
1.79 x speed of sound = 569.44
speed of sound = 318.12 m /s
speed of sound at 20⁰C = 343 m /s
Difference = 343 - 318.12 = 24.88⁰C
We know that 1 ⁰C change in temperature changes speed of sound
by .61 m /s
So a change in speed of 24.88 will be produced by a change in temperature of
24.88 / .61
= 41⁰C
temperature = 20 - 41 = - 21⁰C .
Four cylindrical wires of different sizes are made of the same material. Which of the following combinations of length and cross-sectional area of one of the wires will result in the smallest resistance?
a. Length Area
3L 3a
b. Length Area
3L 6a
c. Length Area
6L 3a
d. Length Area
6L 6a
Answer:
Explanation:
For resistance of a wire , the formula is as follows .
R = ρ L/S
where ρ is specific resistance , L is length and S is cross sectional area of wire .
for first wire resistance
R₁ = ρ 3L/3a = ρ L/a
for second wire , resistance
R₂ = ρ 3L/6a
= .5 ρ L/a
For 3 rd wire resistance
R₃ = ρ 6L/3a
= 2ρ L/a
For fourth wire , resistance
R₄ = ρ 6L/6a
= ρ L/a
So the smallest resistance is of second wire .
Its resistance is .5 ρ L/a
Un autobús en una autopista lleva una magnitud de la velocidad de 95 km/h, el conductor observa que debido a un derrumbe la carretera está cerrada, en ese instante acciona los frenos, deteniendo el autobús después de recorrer 60 m. a) ¿Cuál es el valor de la aceleración en el autobús?
Answer:
La aceleración del autobús es -5.80 m/s².
Explanation:
Podemos encontrar la aceleración del autobús usando la siguiente ecuación:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
Where:
[tex]v_{f}[/tex]: es la velocidad final = 0 (se detiene al final)
[tex]v_{0}[/tex]: es la velocidad inicial = 95 km/h
d: es la distancia recorrida = 60 m
Por lo tanto, la aceleración es:
[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} = \frac{0 - (95 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2*60 m} = -5.80 m/s^{2} [/tex]
El signo negativo se debe a que el autobús está desacelerando (hasta que se detiene).
Entonces, la aceleración del autobús es -5.80 m/s².
Espero que te sea de utilidad!
what is momentum of a train that is 60,000 kg that is moving at velocity of 17m/s?
explain your answer
Energy from the Sun is transferred from the Earth’s surface to the atmosphere, resulting in
atmospheric convection currents that produce winds. How do physical properties of the air
contribute to convection currents?
a -The warmer air sinks because it is more dense than cooler air.
b -The warmer air rises because it is more dense than cooler air.
c- The warmer air sinks because it is less dense than cooler air.
d -The warmer air rises because it is less dense than cooler air.
What is the period of an objects motion?
A particle with charge Q and mass M has instantaneous speed uy when it is at a position where the electric potential is V. At a later time, the particle has moved a distance R away to a position where the electric potential is V2 ) Which of the following equations can be used to find the speed uz of the particle at the new position?
a. 1/2M(μ2^2-μ1^2)=Q (v1-v2)
b. 1/2M(μ2^2-μ1^2)^2=Q(v1-v2)
c. 1/2Mμ2^2=Qv1
d. 1/2Mμ2^2=1/4πx0 (Q^2/R)
Answer:
A
Explanation:
Ke = 1/2 MV^2
A rocket burns fuel to create hot gases that explode violently out of the rocket engine. This explosion creates thrust. Thrust is a force that pushes the rocket upward. What force must thrust overcome in order to send a rocket up into space?
Answer:
Thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.
Explanation:
From the concept of Escape Velocity, derived from Newton's Law of Gravitation, definition of Work, Work-Energy Theorem and Principle of Energy Conservation, which is the minimum speed such that rocket can overcome gravitational forces exerted by the Earth, and according to the Tsiolkovski's Rocket Equation, which states that thrust done by the rocket is equal to the change in linear momentum of the rocket itself, we conclude that thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.
2.
Which is the value of a vector quantity?
A 200V
B 100kg/m
C 20m/s, east
D 50J/(kg°C)
А
B
C
D
3.
The diagrams show three uniform beams P Q and Reach pivoted at its centre
Answer:
c
Explanation:
a vector quantity has both magnitude and direction
The value of 20m/s, east is a vector quantity is Hence, option (C) is correct.
What is vector quantity?A physical quantity that has both directions and magnitude is referred to as a vector quantity.
A lowercase letter with a "hat" circumflex, such as "û," is used to denote a vector with a magnitude equal to one. This type of vector is known as a unit vector.
Given values 200V, 100kg/m, 50J/(kg°C) are denoting magnitude of different physical quantity. Hence, they and scalar quantity ( Physical quantities with merely magnitude and no direction are referred to as scalar quantities. These physical quantities can be explained just by their numerical value without any further guidance.).
But The value of 20m/s, east has a magnitude of 20 m/s and a direction along east. Hence, 20m/s, east denotes a vector quantity is Hence, option (C) is correct.
Learn more about vector quantity here:
https://brainly.com/question/774036
#SPJ2
F=9 N, a=3 m/s², m=?
Answer:
3kg
Explanation:
Given parameters:
Force = 9N
Acceleration = 3m/s²
Unknown:
Mass = ?
Solution:
From Newton's second law of motion:
Force = mass x acceleration
So;
9 = mass x 3
mass = 3kg
A 5-kg object is moving with a speed of 4 m/s at a height of 2 m. The potential energy of the object is approximately
J.
Answer:
P.E = 98 Joules
Explanation:
Given the following data;
Mass = 5kg
Speed = 4m/s
Height = 2m
We know that acceleration due to gravity is equal to 9.8m/s²
To find the potential energy;
Potential energy can be defined as an energy possessed by an object or body due to its position.
Mathematically, potential energy is given by the formula;
[tex] P.E = mgh[/tex]
Where, P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
Substituting into the equation, we have;
[tex] P.E = 5*9.8*2[/tex]
P.E = 98 Joules
A car is traveling on a straight road at a constant 35 m/sm/s, which is faster than the speed limit. Just as the car passes a police motorcycle that is stopped at the side of the road, the motorcycle accelerates forward in pursuit. The motorcycle passes the car 13.5 ss after starting from rest. What is the acceleration of the motorcycle (assumed to be constant)
Answer:
2.59m/s
Explanation:
Using the equation of motion
v = u+at
v is the final velocity = 35ms
u is the initially velocity = 9m/s
t is the time = 13.5s
a is the acceleration
Substitute into the formula
35 = 0+13.5a
a = 35/13.5
a = 2.59m/s²
Hence the acceleration of the motorcycle is 2.59m/s
A group of 25 particles have the following speeds: two have speed 11 m/s, seven have 16 m/s , four have 19 m/s, three have 26 m/s, six have 31 m/s, one has 37 m/s, and two have 45 m/s.
Requiredd:
a. Determine the average speed.
b. Determine the rms speed.
c. Determine the most probable speed.
Answer:
a) Average speed is 24.04 m/s
b) the rms speed is 25.84 m/s
c) the most probable speed is 16 m/s
Explanation:
Given the data in the question;
a) Determine the average speed.
To determine the average speed, we simply divide total some of speed by number of particles;
Average speed = [(2×11 m/s)+(7×16 m/s)+(4×19 m/s)+(3×26 m/s)+(6×31 m/s)+(1×37 m/s)+(2×45 m/s)] / 25
= 601 / 25
= 24.04 m/s
Therefore, Average speed is 24.04 m/s
b) Determine the rms speed
we know that (rms speed)² = sum of square speed / total number of particles
so
(rms speed)² = [(2×11²)+(7×16²)+(4×19²)+(3×26²)+(6×31²)+(1×37²)+(2×45²)] / 25
(rms speed)² = 16691 / 25
(rms speed)² = 667.64
(rms speed) = √ 667.64
(rms speed) = 25.84 m/s
Therefore, the rms speed is 25.84 m/s
c) Determine the most probable speed.
Most particles (7) have velocity 16 m/s
i.e 7 is the maximum number of particle for a particular speed ,
Therefore, the most probable speed is 16 m/s
A person pushes down on a lever with a force of 100 N. At the other end of the lever, a force of 200 N lifts a heavy object. What is the mechanical advantage of the lever?
A. 1/2, because the object will be lifted half the distance
B. -1, because the direction changes
C. 2, because the output force is twice the input force
D. 1, because the same amount of work is done
Answer:
Explanation:
C 200÷100=2
Output ÷ Input= MA
which of the following is used to answer scientific questions?
A. Experiments
B. Intuition
C. Opinion polls
D. Imagination
Choose the element that has a smaller atomic radius :scandium or selenium
A three-phase line, which has an impedance of (2 + j4) ohm per phase, feeds two balanced three-phase loads that are connected in parallel. One of the loads is Y-connected with an impedance of (30 + j40) ohm per phase, and the other is connected with an impedance of (60 - j45) ohm per phase. The line is energized at the sending end from a 60-Hz, three-phase, balanced voltage source of 120 √3V (rms, line-to-line).
Determine:
a. the current, real power and reactive power delivered by the sending-end source
b. the line-to-line voltage at the load
c. the current per phase in each load
d. the total three-phase real and reactive powers absorbed by each load and by the
Answer:
hello your question has a missing information
The other is Δ-connected with an impedance of (60 - j45) ohm per phase.
answer : A) 5A ∠0° ,
p( real power ) = 1800 and Q ( reactive power ) = 0 VAR
B) 193.64 v
C) current at load 1 = 2.236 A , current at load 2 = 4.472 A
D) Load 1 : 450 watts(real power ) , 600 VAR ( reactive power )
Load 2 : 1200 watts ( real power ), -900 VAR ( reactive power )
Explanation:
First convert the Δ-connection to Y- connection attached below is the conversion and pre-solution
A) determine the current, real power and reactive power delivered by the sending-end source
current power delivered (Is) = 5A ∠0°
complex power delivered ( s ) = 3vs Is
= 3 * 120∠0° * 5∠0° = 1800 + j0 ---- ( 1 )
also s = p + jQ ------ ( 2 )
comparing equation 1 and 2
p( real power ) = 1800 and Q ( reactive power ) = 0 VAR
B) determine Line-to-line voltage at the load
Vload = √3 * 111.8
= 193.64 v
c) Determine current per phase in each load
[tex]I_{l1} = Vl1 / Zl1[/tex]
= [tex]\frac{111.8<-10.3}{50<53.13}[/tex] = 2.236∠ 63.43° A hence current at load 1 = 2.236 A
[tex]I_{l2} = V_{l2}/Z_{l2}[/tex]
= [tex]\frac{111.8<-10.3}{25<-36.87}[/tex] = 4.472 ∠ 26.57° A hence current at load 2 = 4.472 A
D) Determine the Total three-phase real and reactive powers absorbed by each load
For load 1
3-phase real power = [tex]3I_{l1} ^{2} R_{l1}[/tex] = 3 * 2.236^2 * 30 = 450 watts
3-phase reactive power = [tex]3I_{l1} ^{2} X_{l1}[/tex] = 3 * 2.236^2 * 40 = 600 VAR
for load 2
3-phase real power = [tex]3I_{l1} ^{2} R_{l2}[/tex] = 1200 watts
3-phase reactive power = [tex]3I_{l1} ^{2} X_{l2}[/tex] = -900 VAR
The sum of load powers and line losses, 1800 W+ j0 VAR and The line voltage magnitude at the load terminal is 193.64 V.
(a) The impedance per phase of the equivalent Y,
[tex]\bar{Z}_{2}=\frac{60-j 45}{3}=(20-j 15) \Omega[/tex]
The phase voltage,
[tex]\bold { V_{1}=\frac{120 \sqrt{3}}{\sqrt{3}}=120 VV }[/tex]
Total impedance from the input terminals,
[tex]\bold {\begin{aligned}&\bar{Z}=2+j 4+\frac{(30+j 40)(20-j 15)}{(30+j 40)+(20-j 15)}=2+j 4+22-j 4=24 \Omega \\&\bar{I}=\frac{\bar{V}_{1}}{\bar{Z}}=\frac{120 \angle 0^{\circ}}{24}=5 \angle 0^{\circ} A\end{aligned} }[/tex]
The three-phase complex power supplied [tex]\bold {=\bar{S}=3 \bar{V}_{1} \bar{I}^{*}=1800 W}[/tex]
P =1800 W and Q = 0 VAR delivered by the sending-end source.
(b) Phase voltage at load terminals will be,
[tex]\bold {\begin{aligned}\bar{V}_{2} &=120 \angle 0^{\circ}-(2+j 4)\left(5 \angle 0^{\circ}\right) \\&=110-j 20=111.8 \angle-10.3^{\circ} V\end{aligned} }[/tex]
The line voltage magnitude at the load terminal,
[tex]\bold{\left(V_{ LOAD }\right)_{L-L}=\sqrt{3} 111.8=193.64 V(V }[/tex]
(c) The current per phase in the Y-connected load,
[tex]\bold {\begin{aligned}&\bar{I}_{1}=\frac{\bar{V}_{2}}{\bar{Z}_{1}}=1-j 2=2.236 \angle-63.4^{\circ} A \\&\bar{I}_{2}=\frac{\bar{V}_{2}}{\bar{Z}_{2}}=4+j 2=4.472 \angle 26.56^{\circ} A\end{aligned} }[/tex]
The phase current magnitude,
[tex]\bold {\left(I_{p h}\right)_{\Delta}=\frac{I_{2}}{\sqrt{3}}=\frac{4.472}{\sqrt{3}}=2.582 }[/tex]
(d) The three-phase complex power absorbed by each load,
[tex]\bold {\begin{aligned}&\bar{S}_{1}=3 \bar{V}_{2} \bar{I}_{1}^{*}=430 W +j 600 VAR \\&\bar{S}_{2}=3 \bar{V}_{2} \bar{I}_{2}^{*}=1200 W -j 900 VAR\end{aligned}}[/tex]
The three-phase complex power absorbed by the line is
[tex]\bold{\bar{S}_{L}=3\left(R_{L}+j X_{L}\right) I^{2}=3(2+j 4)(5)^{2}=150 W +j 300 VAR }[/tex]
Since, the sum of load powers and line losses,
[tex]\bold {\begin{aligned}\bar{S}_{1}+\bar{S}_{2}+\bar{S}_{L} &=(450+j 600)+(1200-j 900)+(150+j 300) \\&=1800 W +j 0 VAR\end{aligned} }[/tex]
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A remote controlled airplane moves 7.2 m in 2.5seconds what is the plane’s velocity
Answer:
2.88m/s
Explanation:
Given parameters:
Displacement = 7.2m
Time taken = 2.5s
Unknown:
Velocity of the plane = ?
Solution:
Velocity is the displacement divided by the time taken.
Velocity = [tex]\frac{displacement}{time taken}[/tex]
So;
Velocity = [tex]\frac{7.2}{2.5}[/tex] = 2.88m/s
g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal radiation from the filament becomes visible. One bulb filament has a surface area of 30 mm2 and emits 60 W when operating. If the bulb filament has an emissivity of 0.8, what is the operating temperature of the filament
Answer:
2577 K
Explanation:
Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.
So, T = ⁴√(P/σεA)
Since P = 60 W, we substitute the vales of the variables into T. So,
T = ⁴√(P/σεA)
= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)
= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)
= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)
= ⁴√(0.00441 × 10¹⁶K⁴)
= 0.2577 × 10⁴ K
= 2577 K