Answer:
Explanation:
Focal length f = - 4 cm
Object distance u = - 10 cm
v , image distance = ?
Mirror formula
[tex]\frac{1}{v} +\frac{1}{u} = \frac{1}{f}[/tex]
Putting the given values
[tex]\frac{1}{v} - \frac{1}{10} = - \frac{1}{4}[/tex]
[tex]\frac{1}{v}= - \frac{3}{20}[/tex]
v = - 6.67 cm .
magnification
m = v / u
= - 6.67 / - 10
= .667
so image will be smaller in size in comparison with size of object .
Characteristics will be that ,
1 ) it will be inverted and
2 ) it will be real image .
The primary of an ideal transformer has 100 turns and its secondary has 200 turns. If the input current at the primary is 100 A, we can expect the output current at the secondary to be
Answer:
Explanation:
For current in ideal transformer the formula is
I₁ / I₂ = N₂ / N₁
I₁ and I₂ are current in primary and secondary coil respectively and N₁ and N₂ are no of turns in primary and secondary coil .
Putting the given values
100 / I₂ = 200 / 100 = 2
I₂ = 50 A .
output current = 50 A .
PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton box 0.8 meters. 2. A boy lifts a 5-newton box 0.8 meters. 3.A boy lifts a 8-newton box 0.2 meters. 4.A boy lifts a 10-newton box 0.2 meters.
Explanation:
Work done is given by the product of force and displacement.
Case 1,
1. A boy lifts a 2-newton box 0.8 meters.
W = 2 N × 0.8 m = 1.6 J
2. A boy lifts a 5-newton box 0.8 meters.
W = 5 N × 0.8 m = 4 J
3. A boy lifts a 8-newton box 0.2 meters.
W = 8 N × 0.2 m = 1.6 J
4. A boy lifts a 10-newton box 0.2 meters.
W = 10 N × 0.2 m = 2 J
Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.
If radio waves were used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, Earth would receive their signals at a speed of
Answer:
Explanation:
speed of alien spaceship = .1 c
We shall apply formula of relativistic mechanics to solve the problem
relative velocity =
[tex]\frac{v+v_1}{1 -\frac{v\times v }{c^2} }[/tex]
Here v = v₁ = .1 c
relative velocity = .1c + .1 c / 1 - .1²
= .2 c / .99
= .202 c
The earth would receive the signal at the speed of .202 c .
The resistance of a 0.29 m long piece of wire is measured to be 0.31 Ohms. The wire has a cross-sectional area of 0.003 m2. What is the resistivity of the wire?
Answer:
3.21×10⁻³ Ωm
Explanation:
Applying,
R = Lρ/A................... Equation 1
Where R = Resistance of the wire, L = Length of the wire, ρ = Resistivity of the wire, A = cross sectional area of the wire.
Make ρ the subject of the equation
ρ = RA/L................... Equation 2
Given: R = 0.31 Ohms, A = 0.003 m², L = 0.29 m
Substitute into equation 2.
ρ = 0.31(0.003)/0.29
ρ = 3.21×10⁻³ Ωm
An appliance with a 20.0-2 resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Answer:
0.866 A
Explanation:
From the question,
P = I²R............................. Equation 1
Where P = power, I = maximum current, R = Resistance.
Make I the subject of the equation
I = √(P/R).................... Equation 2
Given: P = 15 W, R = 20 Ω
Substitute these values into equation 2
I = √(15/20)
I = √(0.75)
I = 0.866 A
Hence the maximum current that can flow safely through the appliance = 0.866 A
A flat loop of wire consisting of a single turn of cross-sectional area 8.60 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.40 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.80
Answer:
The induced current is [tex]I = 5.72*10^{-4 } \ A[/tex]
Explanation:
From the question we are told that
The cross-sectional area is [tex]A = 8.60 \ cm^2 = \frac{8.60 }{10000} = 8.60 *10^{-4} \ m[/tex]
The initial value of magnetic field is [tex]B_1 = 0.500 \ T[/tex]
The value of magnetic field at time t is [tex]B_f = 2.40 \ T[/tex]
The number of turns is N = 1
The time taken is [tex]dt[/tex]= 1.02 \ s
The resistance of the loop is [tex]R = 2.80\ \Omega[/tex]
Generally the induced emf is mathematically represented as
[tex]e = - \frac{d \phi}{dt }[/tex]
Where [tex]d \phi[/tex] is the change n the magnetic flux which is mathematically represented as
[tex]d \phi = N *A * d B[/tex]
Where [tex]dB[/tex] is the change in magnetic field which is mathematically represented as
[tex]d B = B_f - B_i[/tex]
substituting values
[tex]d B = 2.40 - 0.500[/tex]
[tex]d B = 1.9 \ T[/tex]
So
[tex]d \phi = 1 * 1.9 * 8.60 *10^{-4}[/tex]
[tex]d \phi = 1.63*10^{-3} \ T[/tex]
So
[tex]e = - \frac{1.63 *10^{-3}}{ 1.02 }[/tex]
[tex]e = - 1.60*10^{-3} \ V[/tex]
Here the negative only indicates that the emf is acting in opposite direction of the motion producing it so the magnitude of the emf is
[tex]e = 1.60*10^{-3} \ V[/tex]
Now the induced current is evaluated as follows
[tex]I = \frac{e}{R }[/tex]
substituting values
[tex]I = \frac{1.60 *10^{-3}}{2.80 }[/tex]
[tex]I = 5.72*10^{-4 } \ A[/tex]
A block and tackle having a velocity ratio of 5 is used to raise a load of 400N through a distance of 10m. If the work done against friction is 100J. Calculate 1. Efficiency of the machine 2. The effort applied
Answer:
Explanation:
Load will be moved by 4L when effort moves by distance L .
4L = 10 m ( given )
L = 2.5 m
work output = work input = 400 x 10 = 4000 J
work by friction = 100 J
net work output = 3900 J .
efficiency = net output of work / work input
= (3900 / 4000) x 100
= 97.5 %
2 )
work input = 4000 J
distance moved by effort = 2.5 m
If effort be F
F X 2.5 = 4000
F = 1600 N .
When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the diameter of this resistor is now tripled, the current will be
Answer:
The current will be 18 A
Explanation:
Given;
potential difference, V = 10 V
current between the resistor, I = 2 A
Apply ohm's law;
V = IR
R = V / I
R = 10 / 2
R = 5Ω
Resistance is given as;
[tex]R = \frac{\rho l}{A}[/tex]
where;
ρ is resistivity
l is length
A is area
[tex]R = \frac{\rho l}{A} \\\\R = \frac{\rho l}{\pi r^2} = \frac{\rho l}{\pi (\frac{d}{2}) ^2} = \frac{\rho l}{\pi (\frac{d^2}{4}) }\\\\R = \frac{4*\rho l}{\pi d^2} \\\\R = (\frac{4*\rho l}{\pi } )\frac{1}{d^2} \\\\R = (k)\frac{1}{d^2} \\\\k = Rd^2\\\\R_1d_1^2 = R_2d_2^2\\\\R_2 = \frac{R_1d_1^2}{d_2^2}[/tex]
When the diameter of the resistor is tripled
d₂ = 3d₁
[tex]R_2 = \frac{5*d_1^2}{(3d_1)^2} \\\\R_2 = \frac{5d_1^2}{9d_1^2} \\\\R_2 = 0.556 \ ohms[/tex]
The current is now calculated as;
Apply ohms law;
V = IR
I = V / R
I = 10 / 0.556
I = 17.99 A
I = 18 A
Therefore, the current will be 18 A
A spherical shell has inner radius 1.5 m, outer radius 2.5 m, and mass 850 kg, distributed uniformly throughout the shell. What is the magnitude of the gravitational force exerted on the shell by a point mass particle of mass 2.0 kg a distance 1.0 m from the center
Answer:
The magnitude of the gravitational force is 4.53 * 10 ^-7 N
Explanation:
Given that the magnitude of the gravitational force is F = GMm/r²
mass M = 850 kg
mass m = 2.0 kg
distance d = 1.0 m , r = 0.5 m
F = GMm/r²
Gravitational Constant G = 6.67 × 10^-11 Newtons kg-2 m2.
F = (6.67 × 10^-11 * 850 * 2)/0.5²
F = 0.00000045356 N
F = 4.53 * 10 ^-7 N
Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the electric field between the two plates?
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
A positively charged particle has a velocity in the negative z direction at a certain point P. The magnetic force on the particle at this point is in the negative y direction. Which one of the following statements about the magnetic field at point P can be determined from this data?
a. Bx is positive
b. Bz is positive
c. By is negative
d. By is positive
e. Bx is negative
Answer:
a. Bx is positive
Explanation:
See attached file
What is the wave length if the distance from the central bright region to the sixth dark fringe is 1.9 cm . Answer in units of nm.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The wavelength is [tex]\lambda = 622 nm[/tex]
Explanation:
From the question we are told that
The distance of the slit to the screen is [tex]D = 5 \ m[/tex]
The order of the fringe is m = 6
The distance between the slit is [tex]d = 0.9 \ mm = 0.9 *10^{-3} \ m[/tex]
The fringe distance is [tex]Y = 1.9 \ cm = 0.019 \ m[/tex]
Generally the for a dark fringe the fringe distance is mathematically represented as
[tex]Y = \frac{[2m - 1 ] * \lambda * D }{2d}[/tex]
=> [tex]\lambda = \frac{Y * 2 * d }{[2*m - 1] * D}[/tex]
substituting values
=> [tex]\lambda = \frac{0.019 * 2 * 0.9*10^{-3} }{[2*6 - 1] * 5}[/tex]
=> [tex]\lambda = 6.22 *10^{-7} \ m[/tex]
[tex]\lambda = 622 nm[/tex]
Your friend just challenged you to a race. You know in order to beat him, you must run 15 meters within 20 seconds in a northern direction. What does your average velocity need to be to win the race? .5 meters per second, north .75 meters per second, north 1.3 meters per second, north 300 meters per second, north
The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 1.8 AA , how many turns of wire would you need
Answer:
The number of turns of wire needed is 3536 turns.
Explanation:
Given;
length of the wire, L = 8 cm = 0.08 m
magnetic field on the wire, B = 0.1 T
current in the wire, I = 1.8 A
The magnetic field produced by a solenoid is calculated as;
B = μ₀ n I
where;
n is the number of turns per length = N / L
μ₀ is permeability of free space = 4π x 10⁻⁷ N/A²
[tex]B = \frac{\mu_o N I}{L} \\\\N = \frac{BL}{\mu_o I} \\\\N = \frac{0.1 *0.08}{4\pi*10^{-7} *1.8} \\\\N = 3536.32 \ turns[/tex]
Therefore, the number of turns of wire needed is 3536 turns.
A wire carries current in the plane of this screen toward the top of the screen. The wire experiences a magnetic force toward the right edge of the screen. Is the direction of the magnetic field causing this force
Answer:
The direction of the magnetic field causing this force is
In the plane of the screen and towards the bottom of the egde
Explanation:
This is by applying Fleming s right hand rule which explains that
When a conductor such as a wire attached to a circuit moves through a magnetic field, an electric current is induced in the wire due to Faraday's law of induction. The current in the wire can have two possible directions. Fleming's right-hand rule gives which direction the current flows.
The right hand is held with the thumb, index finger and middle finger mutually perpendicular to each other (at right angles), as shown in the diagram.[1]
The thumb is pointed in the direction of the motion of the conductor relative to the magnetic field.
The first finger is pointed in the direction of the magnetic field. (north to south)
Then the second finger represents the direction of the induced or generated current within the conductor (from the terminal with lower electric potential to the terminal with higher electric potential, as in a voltage source)
Two uniform solid balls are rolling without slipping at a constant speed. Ball 1 has twice the diameter, half the mass, and one-third the speed of ball 2. The kinetic energy of ball 2 is 37.0 J.
Part A What is the kinetic energy of ball 1?
Express your answer with the appropriate units.
K7 = Value Units
Answer:
The kinetic energy of the ball 1 is 2.06 J
Explanation:
The kinetic energy of a rolling object K = 1/2Iω² + 1/2mv² where I is its rotational inertia, ω its angular speed, m its mass and v = its velocity of center of mass.
Let m₁, I₁, v₁, d₁ represent the mass, rotational inertia, speed and diameter of solid ball 1. and Let m₂, I₂, v₂, d₂ represent the mass, rotational inertia, speed and diameter of solid ball 2.
Since both objects are spheres, I =2/5mr²
Let r₁ = radius of ball 1 and r₂ = radius of ball 2. Since d₂ = 2d₁
⇒ 2r₂ = 4r₁ ⇒ r₂ = 2r₁
Now, the the kinetic energy of sphere 1 is
K₁ = 1/2I₁ω₁² + 1/2m₁v₁² ω₁ = v₁/r₁ which is the angular speed of solid ball 1.
K₁ = 1/2(2/5mr²)v₁²/r₁² + 1/2m₁v₁²
K₁ = 1/5m₁v₁² + 1/2m₁v₁²
K₁ = 7/10m₁v₁²
Also, the the kinetic energy of sphere 2 is
K₂ = 1/2I₂ω₂² + 1/2m₂v₂² ω₂ = v₂/r₂ which is the angular speed of solid ball 2.
K₂ = 1/2(2/5m₂r₂²)v₂²/r₂² + 1/2m₂v₂²
K₂ = 1/5m₂v₂² + 1/2m₂v₂²
K₂ = 7/10m₂v₂²
Now, m₁ = m₂/2 and v₁ = v₂/3
Substituting these into K₁, we have
K₁ = 7/10(m₂/2)(v₂/3)²
K₁ = 7/10 × 1/18m₂v₂²
K₁ = (1/18)(7/10m₂v₂²)
K₁ = K₂/18
K₂ = 37.0 J/18
K₂ = 2.06 J
So, the kinetic energy of the ball 1 is 2.06 J
If a 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft), how long could she power her 0.4 watt flashlight
Answer: 3217.79 hours.
Explanation:
Given, A 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft).
Power = 0.4 watt
Mass of climber = 140 lb
= 140 x 0.4535 kg [∵ 1 lb= 0.4535 kg]
⇒ Mass of climber (m) = 63.50 kg
Let [tex]h_1=29,029\ ft= 8848.04\ m\ \ \ \ [ 1 ft=0.3048\ m ][/tex] and [tex]h_2= 4,600 ft = 1402.08\ m[/tex]
Now, Energy saved =[tex]mg(h_1-h_2)=(63.50)(9.8)(8848.04-1402.08)=4633620.91\ J[/tex]
[tex]\text{Power}=\dfrac{\text{energy}}{\text{time}}\\\\\Rightarrow 0.4=\dfrac{4633620.91}{\text{time}}\\\\\Rightarrow\ \text{time}=\dfrac{4633620.91}{0.4}\approx11584052.28\text{ seconds}\\\\=\dfrac{11584052.28}{3600}\text{ hours}\ \ \ [\text{1 hour = 3600 seconds}]\\\\=3217.79\text{ hours}[/tex]
Hence, she can power her 0.4 watt flashlight for 3217.79 hours.
0.25-kg block oscillates on the end of a spring with a spring constant of 200 N/m. If the oscillations is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is A :
Answer:
5.2m/s
Explanation:
Plss see attached file
If 50 km thick crust having an average density of 3.0 g/cm3 has a surface elevation of 2.5 km above sea level, what would you predict about the surface elevation for 50 km thick crust with an average density of 2.8 g/cm3
Answer:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
Explanation:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
What would you estimate for the length of a bass clarinet, assuming that it is modeled as a closed tube and that the lowest note that it can play is a D b whose frequency is 69 Hz
Answer:
1.24m
Explanation:
See attached file
A single slit 1.4 mmmm wide is illuminated by 460-nmnm light. Part A What is the width of the central maximum (in cmcm ) in the diffraction pattern on a screen 5.0 mm away
Answer:
1.643*10⁻⁴cmExplanation:
In a single slit experiment, the distance on a screen from the centre point is expressed as y = [tex]\frac{\delta m \lambda d}{a}[/tex] where;
[tex]\delta m[/tex] is the first two diffraction minima = 1
[tex]\lambda[/tex] is light wavelength
d is the distance of diffraction pattern from the screen
a is the width of the slit
Given [tex]\lambda[/tex] = 460-nm = 460*10⁻⁹m
d = 5.0mm = 5*10⁻³m
a = 1.4mm = 1.4*10⁻³m
Substituting this values into the formula above to get width of the central maximum y;
y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³
y = 2300*10⁻¹²/1.4*10⁻³
y = 1642.86*10⁻⁹
y = 1.643*10⁻⁶m
Converting the final value to cm,
since 100cm = 1m
x = 1.643*10⁻⁶m
x = 1.643*10⁻⁶ * 100
x = 1.643*10⁻⁴cm
Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is 1.643*10⁻⁴cm
A person takes a trip, driving with a constant speed of 98.5 km/h, except for a 20.0-min rest stop. The person's average speed is 68.8 km/h. (a) How much time is spent on the trip? h (b) How far does the person travel? km
Answer:
Total time taken(T) = 1.1 hour
Distance = 75.68 km
Explanation:
Given:
Average speed = 68.8 km/h
Constant speed = 98.5 km/h
Rest time = 20 min = 20 / 60 = 0.3333 hour
Find:
Total time taken(T)
Total distance (D)
Computation:
Distance = speed × time
D = 68.8 × t.........Eq1
and
D = 98.5 × [t-0.33]
D = 98.5 t - 32.8333.........Eq2
From Eq1 and Eq2
68.8 t = 98.5 t - 32.83333
29.7 t = 32.83333
t = 1.1
Total time taken(T) = 1.1 hour
Distance = speed × time
Distance = 68.8 × 1.1
Distance = 75.68 km
A flat slab of material (nm = 2.2) is d = 0.45 m thick. A beam of light in air (na = 1) is incident on the material with an angle θa = 46 degrees with respect to the surface's normal.
Numerically, what is the displacement, D, of the beam when it exits the slab?
Answer:
Explanation:
Formula of lateral displacement
[tex]S_{lateral}=\frac{t}{cosr} \times sin(i-r)[/tex]
t is thickness of slab , i and r are angle of incidence and refraction respectively .
Given t = .45 m
sin i / sin r = 2.2
sin 46 / sin r = 2.2
sin r = .719 / 2.2 = .327
r = 19°
[tex]S_{lateral}=\frac{t}{cosr} \times sin(i-r)[/tex]
[tex]S_{lateral}=\frac{.45}{cos19} \times sin(46-19)[/tex]
= .45 x .454 / .9455
= .216 m
= 21.6 cm .
The displacement, D, of the beam when it exits the slab is; 21.65 cm.
We are given;
Refractive index of slab material; nm = 2.2
Thickness of slab; t = 0.45 m
Refractive index of air; na = 1
Angle of incidence; θa = 46°
From snell's law, we can calculate the angle of refraction from;
na × sin θa = nm × sin θm
Thus;
1 × sin 46 = 2.2 × sin θm
0.7193 = 2.2 × sin θm
sin θm = 0.7193/2.2
θm = sin^(-1) 0.32695
θm = 19.08°
Formula for the displacement of the beam is;
D = (t/cos θm) × sin (θa - θm)
Plugging in the relevant values gives;
D = (0.45/cos 19.08) × sin (46 - 19.08)
D = 0.4783 × 0.4527
D = 0.2165m = 21.65 cm
Read more at; https://brainly.com/question/24875145
Following a collision between a large spacecraft and an asteroid, a copper disk of radius 28.0 m and thickness 1.20 m, at a temperature of 850°C, is floating in space, rotating about its axis with an angular speed of 20.0 rad/s. As the disk radiates infrared light, its temperature falls to 20.0°C. No external torque acts on the disk.
A) Find the change in kinetic energy of the disk.
B) Find the change in internal energy of the disk.
C) Find the amount of energy it radiates.
Answer:
A. 9.31 x10^10J
B. -8.47x10 ^ 12J
C. 8.38x 10^12J
Explanation:
See attached file pls
A car and a truck, starting from rest, have the same acceleration, but the truck accelerates for twice the length of time. Compared with the car, the truck will travel:_____.
a. twice as far.
b. one-half as far.
c. three times as far.
d. four times as far.
e. 1.4 times as far.
Answer:
d. four times as far
Explanation:
Initial velocity of car and truck, u = 0
let acceleration of both the truck and car = a
let the length of time for the acceleration = t
Let the time the truck accelerated = 2t
The distance traveled by the car is calculated as;
s = ut + ¹/₂at²
s₁ = 0(t) + ¹/₂at²
s₁ = ¹/₂at²
The distance traveled by the truck is calculated as;
s = ut + ¹/₂at²
s₂ = 0(2t) + ¹/₂a (2t)²
s₂ = ¹/₂a x 4t²
s₂ = 4 (¹/₂at²)
s₂ = 4(s₁)
Truck distance = four times car distance
Therefore, Compared with the car, the truck will travel four times as far
d. four times as far
shows a mixing tank initially containing 2000 lb of liquid water. The tank is fitted with two inlet pipes, one delivering hot water at a mass flow rate of .8 lb/s and the other delivering cold water at a mass flow rate of 1.2 lb/s. Water exits through a single exit pipe at a mass flow rate of 2.5 lb/s. Determine the amount of water, in lb, in the tank after one hour
Answer:
the water that remain in the tank in one hour will be 200 lb
Explanation:
Initial mass of water in the tank = 2000 lb
hot water is delivered through the first inlet pipe at a rate of = 0.8 lb/s
cold water is delivered through the second inlet pipe at a rate of = 1.2 lb/s
exit pipe flow rate = 2.5 lb/s
amount of water in the tank after one hour = ?
In one hour, there are 60 x 60 seconds = 3600 sec, therefore
the water through the first inlet pipe in one hour = 0.8 x 3600 = 2880 lb
the water through the second inlet pipe in one hour = 1.2 x 3600 = 4320 lb
the water through the exit in one hour = 2.5 x 3600 = 9000 lb
The total amount of water in the tank = 2000 + 2880 + 4320 = 9200 lb
The total amount of water that leaves the tank = 9000 lb
therefore, in one hour, the water that remain in the tank will be
==> 9200 lb - 9000 lb = 200 lb
A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x
A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.
Answer:
1.6nT [in the negative z direction]
Explanation:
The magnetic field, B, due to a distance of finite value b, is given by;
B = (μ₀IL) / (4πb[tex]\sqrt{b^2 + L^2}[/tex]) -----------(i)
Where;
I = current on the wire
L = length of the wire
μ₀ = magnetic constant = 4π × 10⁻⁷ H/m
From the question,
I = 20A
L = 2.0cm = 0.02m
b = 5.0m
Substitute the necessary values into equation (i)
B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 [tex]\sqrt{5.0^2 + 0.02^2}[/tex])
B = (10⁻⁷ x 20 x 0.02) / (5.0 [tex]\sqrt{5.0^2 + 0.02^2}[/tex])
B = (10⁻⁷ x 20 x 0.02) / (5.0 [tex]\sqrt{25.0004}[/tex])
B = (10⁻⁷ x 20 x 0.02) / (25.0)
B = 1.6 x 10⁻⁹T
B = 1.6nT
Therefore, the magnetic field at the point x = 5.0m on the x-axis is 1.6nT.
PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.
In a fluorescent tube of diameter 3 cm, 3 1018 electrons and 0.75 1018 positive ions (with a charge of e) flow through a cross-sectional area each second. What is the current in the tube
Answer:
The current in the tube is 0.601 A
Explanation:
Given;
diameter of the fluorescent, d = 3 cm
negative charge flowing in the fluorescent tube, -e = 3 x 10¹⁸ electrons/second
positive charge flowing in the fluorescent tube, +e = 0.75 x 10¹⁸ electrons/ second
The current in the fluorescent tube is due to presence of positive and negative charges to create neutrality in the conductor (fluorescent tube).
Q = It
I = Q/t
where;
I is current in Ampere (A)
Q is charge in Coulombs (C)
t is time is seconds (s)
1 e = 1.602 x 10⁻¹⁹ C
3 x 10¹⁸ e/ s = ?
= (3 x 10¹⁸ e/s x 1.602 x 10⁻¹⁹ C) / 1e
= 0.4806 C/s
negative charge per second (Q/t) = 0.4806 C/s
positive charge per second (Q/t) = (0.75 x 10¹⁸ e/s x 1.602 x 10⁻¹⁹ C) / 1e
positive charge per second (Q/t) = 0.12015 C/s
Total charge per second in the tube, Q / t = (0.4806 C/s + 0.12015 C/s)
I = 0.601 A
Therefore, the current in the tube is 0.601 A
Suppose a point charge is located at the center of a spherical surface. The electric field at the surface of the sphere and the total flux through the sphere are determined. Now the radius of the sphere is halved. What happens to the flux through the sphere and the magnitude of the electric field at the surface of the sphere
Answer:
The magnitude of flux remains the same, and the field increases.
Explanation:
This is because the number of field lines leaving the sphere remains constant and the electric field increases because the line density increases
A disk of radius 25.0cm turns about an axis through the center. The pull on the string produces a linear acceleration a(t)=At on the ball The disk starts from rest and after 3 seconds, linear a(3)=1.80m/s2. Find A and then write an expression for the angular acceleration α(t).
Answer:
The value for A is A= 0.6
The angular acceleration is [tex]\alpha (t) = 2.4 \ t \ m/s^2[/tex]
Explanation:
From the question we are told that
The radius of the disk is [tex]r = 25.0 \ cm = 0.25 \ m[/tex]
The linear acceleration is [tex]a(t) = At[/tex]
At time [tex]t = 3 \ s[/tex]
[tex]a(3) = 1.80 \ m/s^2[/tex]
Generally angular acceleration is mathematically represented as
[tex]\alpha(t) = \frac{a(t)}{r}[/tex]
Now at t = 3 seconds
a(3) = A * 3
=> 1.80 = A * 3
=.> A = 0.6
So therefore
a(t) = 0.6 t
Now substituting this into formula for angular acceleration
[tex]\alpha (t) = \frac{0.6 t }{R}[/tex]
substituting for r
[tex]\alpha (t) = \frac{0.6 t }{0.25}[/tex]
[tex]\alpha (t) = 2.4 \ t \ m/s^2[/tex]