Answer:
u = 88.54 m/s
Explanation:
Given that,
A discus thrower achieves a high throw of 100 m.
Angle of projection is 30°
We need to find the initial speed of the discuss. It is a cse of projectile motion. The maximum height reached by the discus is given by :
[tex]H=\dfrac{u^2\sin^2\theta}{2g}[/tex]
u is the initial speed of the discus
So,
[tex]u^2=\dfrac{2\times 9.8\times 100}{\sin^2(30)}\\\\u=\sqrt{7840}\\\\u=88.54\ m/s[/tex]
So, the initial speed of the discus is 88.54 m/s.
Capacitor C1 is in series with capacitors C2 and C3 in parallel. Then three capacitor system is connected to battery with V0. Determine the charge stored by C1 when C1 = 20 μF, C2 = 10 μF, C3 = 30 μF, and V0 = 18 V.g
Answer:
Q₁ = 2.4 10⁻⁴ C
Explanation:
We have a circuit with several capacitors, let's find the equivalent capacitor of the parallel
[tex]C_{eq1}[/tex] = C₂ + C₃
C_{eq1} = (10 +30) 10⁻⁶
C_{eq1} = 40 10⁻⁶ F
There remains a series system between C₁ and C_{eq1}, let's find the equivalent capacitor
1/C_{eq2} = 1 / C₁ + 1 / C_{eq1}
1 /C_{eq2} = 1 / 20 10⁻⁶ + 1/40 10⁻⁶
1 / C_{eq2} = 0.075 10⁶
C_{eq2} = 13.33 10⁻⁶ F
let's use the relationship
V = Q / C_{eq2}
Q = V C_{eq2}
Q = 18 13.33 10⁻⁶
Q = 2.4 10⁻⁴ C
In a combination of capacitors in series the charge is constant, so the charge on C₁ is the same
Q₁ = 2.4 10⁻⁴ C
Tech A says voltage drops can be measured as long as current is flowing. Tech B says voltage drops can be measured across components, connectors, or cables. Who is correct?
A. Tech A
B. Tech B
C. Both Techs A and B
D. Neither Tech A nor B
Answer:
C. Both Techs A and B
Explanation:
For voltage drop to be measured in the circuit, then there must be a voltage in the circuit. Once there is a voltage across the circuit, there will be current flowing through the the circuit, hence technician A is correct. Voltage drop is usually measured across components in the circuit. Components in a circuit are consumptive in the circuit, hence their is usually a voltage drop when current flows through them in a circuit. Technician B is correct.
Answer:
C
Explanation:
If VF=Vi+AT and Vi=0,A=3,T=4 find Vf?
Answer: 12
Explanation:
Given: VF=Vi+AT
-------------------------
In this case, substitute all the given values into the equation
VF=Vi+AT
VF=0+(3)(4)
VF=0+12
VF=12
Hope this helps!! :)
Answer:
[tex]\huge \boxed{V_f=12}[/tex]
Explanation:
[tex]V_f=V_i+AT[/tex]
This is the formula for final velocity.
The values are given for initial velocity, acceleration, and time elapsed.
[tex]V_i=0, \ A=3, \ T=4[/tex]
Solve for [tex]V_f[/tex].
[tex]V_f=0+(3)(4)[/tex]
Evaluate.
[tex]V_f=12[/tex]
For a certain experiment, Juan must measure the concentration of a certain substance in a solution over time. He needs to collect a measurement every 0.05 seconds. He then needs to display his data in a graph and place that graph in a text document. Select the best tools to use for this experiment. Check all that apply.
Answer:
Probeware and computer
Explanation:
Computers are more powerful and better than a graphing calculator for this situation.
are the tools he must use.
You have been asked to design a can with a volume of 672cm3 that is shaped like a right circular cylinder. The can will have a closed top. What radius r and height h, in centimeters, would minimize the amount of material needed to construct this can
Answer:
r = 4.747 cm and h = 9.4925 cm
Explanation:
We know that volume of a cylinder is given as:
V = πr²h
Also, surface area is given as;
S = 2πr² + 2πrh
Where r is radius and h is height
Now, we are told that the volume is 672 cm³
Thus, πr²h = 672
Making h the subject gives;
h = 672/πr²
Putting 672/πr² for h in the surface area equation gives;
S = 2πr² + 2πr(672/πr²)
Factorizing gives;
S = 2π[r² + 672/πr]
Differentiating to get first derivative gives;
S' = 2π[2r - (672/πr²)]
Equating to zero gives;
2π[2r - (672/πr²)] = 0
4πr - 1344/r² = 0
4πr = 1344/r²
r³ = 1344/4π
r³ = 106.95212175775
r = ∛106.95212175775
r = 4.747 cm
So, since h = 672/πr²
Then, h = 672/π(4.747)²
h = 9.4925 cm
A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m. If the spring is stretched to a length of 0.5m and released with initial velocity 0, find the position of the mass at any time t. Here damping constant is zero.
Answer:
Explanation:
force constant of spring k = force / extension
= 35.6 / 0.5
k = 71.2 N / m
angular frequency ω of oscillation by spring mass system
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
where m is mass of the body attached with spring
Putting the values
[tex]\omega = \sqrt{\frac{71.2}{5} }[/tex]
ω = 3.77 radian / s
The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2
so the equation for displacement from equilibrium position that is middle point can be given as follows
x = .5 sin ( ω t + π / 2 )
= 0.5 cos ω t
= 0.5 cos 3.77 t .
x = 0.5 cos 3.77 t .
The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The parsec is the radius of a circle for which a central angle of 1 s intercepts an arc of length 1 AU. The light-year is the distance that light travels in 1 y.
(a) How many parsecs are there in one astronomical unit?
(b) How many meters are in a parsec?
(c) How many meters in a light-year? (d) How many astronomical units in a light-year? (e) How many light-years in a parsec?
Answer:
a) How many parsecs are there in one astronomical unit?
[tex]4.85x10^{-6}pc[/tex]
(b) How many meters are in a parsec?
[tex]3.081x10^{16}m[/tex]
(c) How many meters in a light-year?
[tex]9.46x10^{15}m[/tex]
(d) How many astronomical units in a light-year?
[tex]63325AU[/tex]
(e) How many light-years in a parsec?
3.26ly
Explanation:
The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ([tex]1.496x10^{11} m[/tex]) is defined as 1 astronomical unit:
[tex]\tan{p} = \frac{1AU}{d}[/tex]
Where d is the distance to the star.
Since p is small it can be represent as:
[tex]p(rad) = \frac{1AU}{d}[/tex] (1)
Where p(rad) is the value of in radians
However, it is better to express small angles in arcseconds
[tex]p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}[/tex]
[tex]p('') = 2.06x10^5 p(rad)[/tex]
[tex]p(rad) = \frac{p('')}{2.06x10^5}[/tex] (2)
Then, equation 2 can be replace in equation 1:
[tex]\frac{p('')}{2.06x10^5} = \frac{1AU}{d}[/tex]
[tex]\frac{d}{1AU} = \frac{2.06x10^5}{p('')}[/tex] (3)
From equation 3 it can be see that [tex]1pc = 2.06x10^5 AU[/tex]
a) How many parsecs are there in one astronomical unit?
[tex]1AU . \frac{1pc}{2.06x10^5AU}[/tex] ⇒ [tex]4.85x10^{-6}pc[/tex]
(b) How many meters are in a parsec?
[tex]2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}[/tex] ⇒ [tex]3.081x10^{16}m[/tex]
(c) How many meters in a light-year?
To determine the number of meters in a light-year it is necessary to use the next equation:
[tex]x = c.t[/tex]
Where c is the speed of light ([tex]c = 3x10^{8}m/s[/tex]) and x is the distance that light travels in 1 year.
In 1 year they are 31536000 seconds
[tex]x = (3x10^{8}m/s)(31536000s)[/tex]
[tex]x = 9.46x10^{15}m[/tex]
(d) How many astronomical units in a light-year?
[tex]9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}[/tex] ⇒ [tex]63325AU[/tex]
(e) How many light-years in a parsec?
[tex]2.06x10^{5}AU . \frac{1ly}{63235AU}[/tex] ⇒ [tex]3.26ly[/tex]
With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 41 m
Answer:
The speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.
Explanation:
Given;
maximum vertical height of the throw, H = 41 m
Apply the following kinematic equation;
V² = U² + 2gH
where;
V is the final speed with which the ball will rise to a maximum height
U is the initial speed of the ball = 0
g is acceleration due to gravity = 0
V² = U² + 2gH
V² = 0² + 2gH
V² = 2gH
V = √2gH
V = √(2 x 9.8 x 41)
V = 28.35 m/s
Therefore, the speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.
A small spherical body is tied to a string of length 1 m and revolved in a vertical circle such that the tension in the string is zero at the highest point . Find the linear speed of the body in the 1) lowest position & 2) highest position
Explanation:
At the highest point, the tension force is 0, so the only force acting on the sphere is gravity. Sum of forces on the sphere in the centripetal direction:
∑F = ma
mg = mv²/r
v = √(gr)
v = √(9.8 m/s² × 1 m)
v = 3.13 m/s
If the speed is constant, then the linear speed at the lowest point is also 3.13 m/s. Otherwise, we would need to know the tension in the string at that point.
. The Moon has an average distance from the Earth of 384,403 km and takes 27.32166 days to orbit the Earth. What is the velocity of the Moon in kilometers per hour
Answer:
Velocity of moon = 586.23 km/h
Explanation:
We are given;
Distance of moon from the Earth = 384403 km
Time taken to orbit earth;t = 27.32166 days
24 hours make 1 day, thus 27.32166 days = 27.32166 × 24 = 655.72 hours
Formula for velocity is distance/time
Thus,
Velocity of moon = distance from moon to earth/time taken to orbit the earth
Velocity of moon = 384403/655.72 = 586.23 km/h
A photoelectric-effect experiment finds a stopping potentialof 1.93V when light of 200nm is used to illuminate thecathode.
a) From what metal is the cathode made from?
b) What is the stopping potential if the intensity of thelight is doubled?
Answer:
a) Tantalum
b) 1.93 V
Explanation:
The energy of the incident photon= hc/λ
h= Plank's constant=6.63×10^-34 Is
c= speed of light = 3×10^8 ms-1
λ= wavelength of incident photon
E= 6.63×10^-34 × 3×10^8/ 200×10^-9
E= 0.099×10^-17
E= 9.9×10^-19 J
The kinetic energy of the electron = eV
Where;
e= electronic charge = 1.6×10^-19 C
V= 1.93 V
KE= 1.6×10^-19 C × 1.93 V
KE= 3.1 ×10^-19 J
From Einstein's photoelectric equation;
KE= E -Wo
Wo= E -KE
Wo=9.9×10^-19 J - 3.1 ×10^-19 J
Wo= 6.8×10^-19 J
Wo= 6.8×10^-19 J/1.6×10^-19
Wo= 4.25 ev
The metal is Tantalum
b) the stopping potential remains 1.93 V because intensity of incident photon has no effect on the stopping potential.
Suppose a particle of mass m is confined to a one-dimensional box of length L. We can model this as an infinite square well in which the particle's potential energy inside the box is zero and the potential energy outside is infinite. For a particle in its first excited state, what is the probability Prob(center20%) of finding the particle within the center 20% of the box
Answer: P = 4.86 × 10⁻²
Therefore, the particle's quantum number is 4.86 × 10⁻²
Explanation:
The expression of wave function for a particle in one dimensional box is given as;
φ(x) = ( √2/L ) sin ( nπx/L )
now we input our given figures, the limit of the particle to find it within the center of the box is
xₓ = L/2 + 20% of L/2
xₓ = L/2 + (0.2)L/2
xₓ = 3L/5
And the lower limit is,
x₁ = L/2 - 20% of L/2
x₁ = L/2 - (0.2) L/2
x₁ = 2L / 5
The expression for the probability of finding the particle within the center of the box is
P = ∫ˣˣₓ₁ ║φ(x)║² dx
P = ∫ ³L/⁵ ₂L/₅║(√2/L) sin ( nπx/L)║²dx
= 2/L ( ∫ ³L/⁵ ₂L/₅║sin ( nπx/L)║²dx
= 2/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 2πnx/L)/2) dx)
The particle is in its first excited state, then
n =2
Then calculate the particle's quantum number as follows;
= 2/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 2π(2)x/L)/(2)) dx)
= 1/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 4πx/L)/2) dx)
= 1/L ( x - (L/4π)sin (4πx/L)) ³L/⁵ ₂L/₅
= 1/L ((3L/5) - (L/4π) sin (( 4π(3L/5)/L)) - (( 2L/5) - (L/4π)sin ( 4π(2L/5)/L)))
= 1/L ( L/5 + L/4π (sin(8π/5) - sin ( 12π/5)))
Use the trigonometric formula to solve the above equation
sinA - sinB = 2sin ( A-B/2) cos (A+B/2)
Calculate the particle's quantum number as follows
P = 1/L ( L/5 + L/4π (sin(8π/5) - sin ( 12π/5)))
= 1/5 + 1/4π ( 2sin( (8π/5 -12π/5 ) / 2 ) cos ( (8π/5 + 12π/5) / 2 ))
= 1/5 + 1/2π ( -sin(2π/5) cos2π
= 1/5 - 1/2π ( sin (2π/5)(1))
= 0.0486 (10⁻²)(10²)
= 4.86 × 10⁻²
Therefore, the particle's quantum number is 4.86 × 10⁻²
The temperature of a plastic cube is monitored while the cube is pushed 8.6 m across a floor at constant speed by a horizontal force of 19 N. The monitoring reveals that the thermal energy of the cube increases by 120 J. What is the increase in the thermal energy of the floor along which the cube slides
Answer:
Answer:
43.4J
Explanation:
We know that
Work done = total heat energy
But work done is force x distance
=> F = 19 x8.6 = 163.4 J
So the total heat. Will be Heat of cube + heat of floor = 163.4J
So that heat of floor will now be
floor = 163.4 J - 120 J = 43.4 Joules
Explanation:
In the anatomical position, the gluteal and lumbar are on the ___
Answer:
Posterior of the body
Explanation:
Gluteal region is located at the proximal end of the femur and posterior to the pelvic girdle. The gluteal muscles help to move the lower limb at the hip joint. The gluteal region is divided into two groups: Deep lateral rotators and superficial abductors and extenders.
The lumbar is the lower region of the spine commonly known as lower back, it has five vertebrates. The lumbar contain tissue and nerves that control communication between legs and brain. In anatomical terms they are located inferior to the rib cage, at the bottom section of the vertebral column and superior to sacrum and pelvis.
What is the absolute pressure at a depth of 9.91 m below the surface of a deep lake? Assume atmospheric pressure is 1.01×105 Pa .
Answer:
P = 198.118 kPa
Explanation:
Given:
Atmospheric pressure = P[tex]_{atm\\}[/tex] = 1.01×10⁵
depth = h = 9.91 m
To find:
Absolute pressure P[tex]_{abs}[/tex]
Solution:
Density of water = ρ = 1.000x10 ³kg/m ³
acceleration due to gravity = ρ = 9.8 m/s²
P[tex]_{abs}[/tex] = P[tex]_{atm\\}[/tex] + ρgh
= 1.01×10⁵ + 1.000x10 ³x 9.8 x 9.91
= 101000 + 1000(9.8)(9.91)
= 101000 + 97118
= 198118 Pa
= 198.118 kPa
P[tex]_{abs}[/tex] = 198.118 kPa
The absolute pressure at a depth below the surface of this deep lake is 198.118 kPa.
Given the following data:
Atmospheric pressure = [tex]1.01 \times 10^5 \;Pa[/tex]Height (depth) = 9.91 meters.Scientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]Density of water = 1000 [tex]kg/m^3[/tex]To calculate the absolute pressure at a depth below the surface of a deep lake:
Mathematically, absolute pressure is given by this formula:
[tex]P_{abs} = P + \rho gh[/tex]
Substituting the given parameters into the formula, we have;
[tex]P_{abs} = 1.01 \times 10^5 + 1000 \times 9.8 \times 9.91 \\\\ P_{abs} = 101000+ 97118 \\\\[/tex]
Absolute pressure = 198118 Pa
Note: 1 kPa = 1000 Pa
Absolute pressure = 198.118 kPa
Read more on absolute pressure here: https://brainly.com/question/10013312
You are looking down on a single coil in a constant magnetic field B = 1.2 T which points directly into of the screen. The dimensions of the coil go from a = 8 cm and b = 17 cm, to a* = 16 cm and b* = 22 cm in t = 0.04 seconds. If the coil has resistance that remains constant at 1.2 ohms. What would be the magnitude of the induced current in amperes?
Answer:
The current is [tex]I = 0.5425 \ A[/tex]
Explanation:
From the question we are told that
The magnetic field is [tex]B = 1.2 \ T[/tex]
The first length is [tex]a = 8 \ cm = 0.08 \ m[/tex]
The second length is [tex]a^* = 16 \ cm = 0.16 \ m[/tex]
The first width is [tex]b = 17 \ cm = 0.17 \ m[/tex]
The second width is [tex]b^* = 22 \ cm = 0.22 \ m[/tex]
The time interval is [tex]dt = 0.04 \ s[/tex]
The resistance is [tex]R = 1.2 \ \Omega[/tex]
Generally the first area is
[tex]A = a * b[/tex]
=> [tex]A = 0.08 * 0.17[/tex]
=> [tex]A = 0.0136 \ m^2[/tex]
The second area is
[tex]A^* = a^* * b^*[/tex]
=> [tex]A^* = 0.16 * 0.22[/tex]
=> [tex]A^* = 0.0352 \ m^2[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = - \frac{ B * [A^* - A]}{dt}[/tex]
This negative show that it is moving in the opposite direction of the motion producing it
=> [tex]|\epsilon | = \frac{ 1.2 * [ 0.0352-0.0135]}{0.04}[/tex]
=> [tex]|\epsilon | = 0.651 \ V[/tex]
The induced current is
[tex]I = \frac{|\epsilon|}{R}[/tex]
=> [tex]I = \frac{ 0.651}{1.2}[/tex]
=> [tex]I = 0.5425 \ A[/tex]
A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to accelerate across the ice. If drag and the friction from the ice apply a force of 5 N on the boy, what is the magnitude of the net force acting on him?
Answer:
70 N
Explanation:
Draw a free body diagram of the boy. There are four forces:
Weight force mg pulling down,
A 300 N normal force pushing up,
A 75 N applied force pulling right,
and a 5 N friction force pushing left.
The boy's acceleration in the y direction is 0, so the net force in the y direction is 0.
The net force in the x direction is 75 N − 5 N = 70 N.
Define fluid flow. What are the types of fluid flow?
Answer:
The different types of fluid flow are: Steady and Unsteady Flow. Uniform and Non-Uniform Flow. ... Compressible and Incompressible Flow. Rotational and Irrotational Flow.
A plane moves 100 meters to the right in two seconds. What is its velocity?
0 m/s
O 30 m/s
50 m/s
O 100 m/s
200 m/s
Answer:
50 m/s
Explanation:
[tex]Distance = 100m\\Time = 2 secs\\\\Velocity = \frac{Distance}{Time} \\\\V = \frac{100m}{2s} \\\\= 50m/s[/tex]
If the momentum of a system is to be conserved, which must be true of the net external force acting on the system?
A. nonzero but constant.
B. increasing
C. decreasing
D. zero
Answer:
D. zero
Explanation:
For momentum of an isolated or closed system to be conserved (initial momentum must equal final momentum), the net external force acting on the system must be zero.
There is always external forces acting on a system, for this system’s momentum to remain constant, all the external forces acting on the system must cancel out, so that the net external force on the system is zero.
[tex]F_{ext} = 0[/tex]
Therefore, the correct option is "D"
D. zero
Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen into a beaker. After the nitrogen evaporates, how much volume does it occupy if its density is equal to that of the dry air at sea level
Answer:
The value is [tex]V_n = 2.2498 \ m^3[/tex]
Explanation:
From the question we are told that
The volume of liquid nitrogen is [tex]V_n = 3.6 \ L= 3.6 *10^{-3} \ m^3[/tex]
The density of nitrogen at gaseous form is [tex]\rho_n = 1.2929 \ kg/m^3[/tex] = The dry air at sea level
Generally the density of nitrogen at liquid form is
[tex]\rho _l = 808 \ kg/m^3[/tex]
And this is mathematically represented as
[tex]\rho_l = \frac{m}{V_l }[/tex]
=> [tex]m = \rho_l * V_l[/tex]
Now the density of gaseous nitrogen is
[tex]\rho_n = \frac{m}{V_n }[/tex]
=> [tex]m = \rho_n * V_n[/tex]
Given that the mass is constant
[tex]\rho_n * V_n = \rho_l * V_l[/tex]
[tex]1.2929* V_n = 808 * 3.6*10^{-3}[/tex]
=> [tex]V_n = 2.2498 \ m^3[/tex]
A drag racer can reach a top speed of 98 m/s. How long will it take the racer to travel 1500 m?
Answer:
[tex]t=15.3s[/tex]
Explanation:
Hello,
In this case, since the speed is defined in terms of the distance over time:
[tex]V=\frac{x}{t}[/tex]
We can easily solve for the time with the given speed and distance:
[tex]t=\frac{x}{V}=\frac{1500m}{98m/s}\\ \\t=15.3s[/tex]
Regards.
Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of the force on a 3.0-m length of one of the wires is equal to 8.0 μN, what is the greater of the two currents?
Answer:
The greater of the two currents is 0.692 A
Explanation:
Given;
distance between the two parallel wires; r = 6 mm = 6 x 10⁻³ m
let the current in the first wire = I₁
then, the current in the second wire = 2I₁
length of the wires, L = 3.0 m
magnitude of force on the wires, F = 8 μN = 8 x 10⁻⁶ N
The magnitude of force on the two parallel wires is given by;
[tex]F = \frac{\mu_o I_1(2I_1)}{2\pi r}\\\\F = \frac{\mu_o 2I_1^2}{2\pi r}\\\\I_1^2 = \frac{F*2\pi r}{2\mu_o} \\\\I_1^2 = \frac{8*10^{-6}*2\pi (6*10^{-3})}{2(4\pi*10^{-7})}\\\\I_1^2 = 0.12\\\\I_1 = \sqrt{0.12}\\\\ I_1 =0.346 \ A[/tex]
the current in the second wire = 2I₁ = 2 x 0.346 A = 0.692 A
Therefore, the greater of the two currents is 0.692 A
A motorboat starting from rest travels in a straight line on a lake. If the boat achieves a speed of 9.0 m/s in 13 s, what is the boat's average acceleration?
Answer:
Acceleration, [tex]a=0.69\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the motorboat, u = 0
Final speed off the motorboat, v = 9 m/s
Time, t = 13 s
We need to find the boat's average acceleration. It is equal to the change in velocity divided by time taken. SO,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{9-0}{13}\\\\a=0.69\ m/s^2[/tex]
So, the acceleration of the boat is [tex]0.69\ m/s^2[/tex].
A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA. What amount of vars must be added to bring the pf to 0.85
Answer:
[tex]\mathtt{Q_{sh} = 600.75 \ vars}[/tex]
Explanation:
Given that:
A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA
Here:
the initial power factor i.e cos θ₁ = 0.7 lag
θ₁ = cos⁻¹ (0.7)
θ₁ = 45.573°
Active power P = 1500 watts
Apparent power S = 2100 VA
What amount of vars must be added to bring the pf to 0.85
i.e the required power factor here is cos θ₂ = 0.85 lag
θ₂ = cos⁻¹ (0.85)
θ₂ = 31.788°
However; the initial reactive power [tex]Q_1[/tex] = P×tanθ₁
the initial reactive power [tex]Q_1[/tex] = 1500 × tan(45.573)
the initial reactive power [tex]Q_1[/tex] = 1500 × 1.0202
the initial reactive power [tex]Q_1[/tex] = 1530.3 vars
The amount of vars that must therefore be added to bring the pf to 0.85
can be calculated as:
[tex]Q_{sh} = P( tan \theta_1 - tan \theta_2)[/tex]
[tex]Q_{sh} = 1500( tan \ 45.573 - tan \ 31.788)[/tex]
[tex]Q_{sh} = 1500( 1.0202 - 0.6197)[/tex]
[tex]Q_{sh} = 1500( 0.4005)[/tex]
[tex]\mathtt{Q_{sh} = 600.75 \ vars}[/tex]
A long solenoid that has 1 200 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur
Answer:
The current required winding is [tex]2.65*10^-^2 mA[/tex]Explanation:
We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil and
n= N/L (number of turns per unit length)
Given data
The number of turns n= 1200 turns
length L= 0.42 m
magnetic field B= 1*10^-4 T
μ₀= [tex]4\pi*10^-^7 T.m/A[/tex]
Applying the equation B=μ₀*n*I
I= B/μ₀*n
I= B*L/μ₀*n
[tex]I= \frac{1*10^-^4*0.42}{4\pi*10^-^7*1.2*10^3 }[/tex]
[tex]I= 2.65*10^-^2 mA[/tex]
Suppose you spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.625×10−3 m3/s and the diameter of the nozzle you hold is 5.19×10−3 m. At what speed v does the water exit the nozzle?
Answer:
0.153 m/s
Explanation:
The flowrate Q = 0.625 x 10-3 m^3-/s
The diameter of the nozzle d = 5.19 x 10^-3 m
the velocity V = ?
The cross-sectional area of the flow A = [tex]\pi d^{2}/4[/tex]
==> (3.142 x 5.19 x 10^-3)/4 = 4.077 x 10^-3 m^2
From the continuity equation,
Q = AV
V = Q/A = (0.625 x 10-3)/(4.077 x 10^-3) = 0.153 m/s
you are working in a physics lab where you have made a simple circuit with a battery and bulb in which part of your circuit is the current flow maximum through the bulb filament or through the battery if you reverse the polarity would there be any difference in the intensity of the bulb
Answer:
The current moves in the terminal.
Yang can focus on objects 150 cm away with a relaxed eye. With full accommodation, she can focus on objects 20 cm away. After her eyesight is corrected for distance vision, what will her near point be while wearing her glasses?
Answer:
Explanation:
Without wearing glasses , her near point is 20 cm .
for correction of eye
u = infinity ,
v = - 150 cm
f = ?
[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f }[/tex]
[tex]\frac{1}{-150} -0 = \frac{1}{f }\\[/tex]
f = - 150 cm
He must be wearing glass of focal length of 150 cm .
If near point be x after wearing glass ,
u = x
v = - 20 cm
f = - 150 cm
[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f }[/tex]
[tex]\frac{1}{-20} -\frac{1}{x} = \frac{1}{-150 }[/tex]
[tex]\frac{1}{-20} + \frac{1}{150 }= \frac{1}{x}[/tex]
x = 23 cm .
While wearing the glasses, Yang's near point will be 23.08 cm.
Given information:
Yang can focus on objects 150 cm away with a relaxed eye.
With full accommodation, she can focus on objects 20 cm away.
For correction, we have to use a concave lens such that it can make the image of a distant object at 150 cm.
So, the object distance will be infinity, and the image distance will be [tex]v=-150[/tex] cm.
So, the focal length of the lens can be calculated by lens formula as,
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\dfrac{1}{-150}-\dfrac{1}{\infty}=\dfrac{1}{f}\\f=-150\rm\;cm[/tex]
Now, after using the lens, the image distance will be [tex]v=-20[/tex] cm. Let u be the near point.
The near point, after correction, can be calculated as,
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\dfrac{1}{-20}-\dfrac{1}{u}=\dfrac{1}{-150}\\\dfrac{1}{u}=\dfrac{1}{150}-\dfrac{1}{20}\\u=23.08\rm\; cm[/tex]
Therefore, while wearing the glasses, Yang's near point will be 23.08 cm.
For more details, refer to the ink:
https://brainly.com/question/4419161
The length and width of a rectangle are 1.82 cm and 1.5 cm respectively. Calculate area of the rectangle and write in correct significant number.
Answer:
Hey mate ,
Area of rectangle = l×b
1.82×1.5
2.73cm2