Answer:
28.3°C
Explanation:
Using
T(t) = (T(0) - 20)*(e^(-k*t)) + 20
for some positive number k, and some initial temperature T(0).
Boundary conditions:
T(4) = T(0) - 5 _______ (i)
T(8) = T(0) - 7 _______ (ii)
==> solving for T(0) and k :
(i):
(T(0) - 20)*(e^(-k*4)) + 20 = T(0) - 5 ==>
(T(0) - 20)*(e^(-k*4)) = T(0) - 20 - 5
(T(0) - 20)*(e^(-k*4)) = (T(0) - 20) - 5
5 = (T(0) - 20) - (T(0) - 20)*(e^(-k*4))
5 = (T(0) - 20) * ( 1 - e^(-k*4) )
(ii):
(T(0) - 20)*(e^(-k*8)) + 20 = T(0) - 7
(T(0) - 20)*(e^(-k*8)) = (T(0) - 20) - 7
7 = (T(0) - 20) - (T(0) - 20)*(e^(-k*8))
7 = (T(0) - 20) * (1 - e^(-k*8))
In both results, subsitute x = e^(-4k) and C = (T(0) - 20)
(i): 5 = C * (1 - x)
(ii): 7 = C * (1 - x^2) = C * (1-x)*(1+x)
Substitute C*(1-x) from (i) into (ii):
(ii): 7 = 5*(1+x) ==> (1+x) = 7/5 ==> x = 2/5
back into (i):
(i): 5 = C * (1 - 2/5) ==> 5 = C * 3/5 ==> C = 25/3
C = T(0) - 20 ==>
T(0) = C + 20 = 25/3 + 20 = 25/3 + 60/3 = 85/3
= 28.3°C
3. What are the first steps that you should take if you are unable to get onto the Internet? (1 point)
O Check your router connections then restart your router.
O Plug the CPU to a power source and reboot the computer.
O Adjust the display properties and check the resolution.
Use the Control Panel to adjust the router settings.
Answer:
Check your router connections then restart your router.
Explanation:
Answer:
Check your router connections then restart your router.
Explanation:
Most internet access comes from routers so the problem is most likely the router.
A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5 s. Determine 1. The acceleration of the car. 2. The distance it moves in the third second.
Answer:
Explanation:
Initial velocity , u = 30 m/s
final velocity , v = 10 m/s
time , t = 5 seconds
1. Acceleration = v - u / t
= 10 - 30 / 5
= -20 / 5
= - 4 m/s
Kinetic and
A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is holding the brick in place?
11 point)
O raise the board to a higher angle
O press down on the brick in a direction that is perpendicular to the board
O roughen up the texture of the wooden board
o lower the board so it's level with the ground
Answer:
raise the board to a higher angle
Explanation:
Static friction is the force opposite to the applied force.
Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.
So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.
Hence, the correct option is "raise the board to a higher angle".
A car travels at 45 km/h. If the driver breaks 0.65 seconds after seeing the traffic light turn yellow, how far will the car continue to travel before it begins to slow?
Answer:
8.1 m
Explanation:
Convert km/h to m/s.
45 km/h × (1000 m/km) × (1 h / 3600 s) = 12.5 m/s
Distance = speed × time
d = (12.5 m/s) (0.65 s)
d = 8.125 m
Two long, parallel wires are separated by a distance of 2.60 cm. The force per unit length that each wire exerts on the other is 4.30×10^−5 N/m, and the wires repel each other. The current in one wire is 0.520 A.Required:a. What is the current in the second wire? b. Are the two currents in the same direction or in opposite directions?
Answer:
10.75 A
The current is in opposite direction since it causes a repulsion force between the wires
Explanation:
Force per unit length on the wires = 4.30×10^−5 N/m
distance between wires = 2.6 cm = 0.026 m
current through one wire = 0.52 A
current on the other wire = ?
Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as
[tex]F/l[/tex] = [tex]\frac{u_{0}I_{1} I_{2} }{2\pi r }[/tex]
where [tex]F/l[/tex] is the force per unit length on the wires
[tex]u_{0}[/tex] = permeability of vacuum = 4π × 10^−7 T-m/A
[tex]I_{1}[/tex] = current on the first wire = 0.520 A
[tex]I_{2}[/tex] = current on the other wire = ?
r = the distance between the two wire = 0.026 m
substituting the value into the equation, we have
4.30×10^−5 = [tex]\frac{4\pi *10^{-7}*0.520*I_{2} }{2\pi *0.026}[/tex] = [tex]\frac{ 2*10^{-7}*0.520*I_{2} }{0.026}[/tex]
4.30×10^−5 = 4 x 10^-6 [tex]I_{2}[/tex]
[tex]I_{2}[/tex] = (4.30×10^-5)/(4 x 10^-6) = 10.75 A
The current is in opposite direction since it causes a repulsion force between the wires.
6. You push an object, initially at rest, across a frictionless floor with a constant force for a time interval t, resulting in a final speed of v for the object. You then repeat the experiment, but with a force that is twice as large. What time interval is now required to reach the same final speed v?
Answer:
t = t₀ / 2
Explanation:
In this exercise we must use Newton's second law
F = m a
a = F / m
now we can use kinematics
as in object part of rest (v₀ = 0)
v =a t₀
t₀ = v / a
these results are with the first experiment
now repeat the experiment, but F = 2F₀
a = 2F₀ / m = 2 a₀
v = 2 a₀ t
t = v / 2a₀
t = t₀ / 2
The time interval that is required to reach the same final speed (V) is equal to [tex]t=\frac{\Delta t}{2}[/tex].
Given the following data:
Initial speed = 0 m/s (since the object is at rest)Final speed = VTime = [tex]\Delta t[/tex]Speed = VTo find the time interval that is now required to reach the same final speed (V), we would apply Newton's Second Law of Motion:
Mathematically, Newton's Second Law of Motion is given by this formula;
[tex]F = \frac{M(V-U)}{t}[/tex]
Where:
F is the force.V is the final velocity.U is the initial velocity.t is the time.Substituting the given parameters into the formula, we have;
[tex]F = \frac{M(V-0)}{\Delta t}\\\\F = \frac{MV}{\Delta t}[/tex]
When the experiment is repeated, the magnitude of the force is doubled:
[tex]F = 2F[/tex]
Now, we can find the time interval that is required to reach the same final speed (V):
[tex]F = \frac{M(V-0)}{t}\\\\t=\frac{MV}{F}[/tex]
Substituting the value of F, we have:
[tex]t=\frac{MV}{2F} \\\\t=\frac{MV}{\frac{2MV}{\Delta t}} \\\\t=MV \times \frac{\Delta t}{2MV} \\\\t=\frac{\Delta t}{2}[/tex]
Read more here: https://brainly.com/question/24029674
Sammy is 5 feet and 5.3 inches tall.tall.what is sammy's height in metres?
Answer:
65.3
Explanation:
1 foot = 12 inches
Sammy is 5 feet tall.
5 feet = ? inches
Multiply the feet value by 12 to find in inches.
5 × 12
= 60
Add 5.3 inches to 60 inches.
60 + 5.3
= 65.3
Answer:
It will be 》》》》1.664716m
The highest mountain on mars is olympus mons, rising 22000 meters above the martian surface. If we were to throw an object horizontaly off the mountain top, how long would it take to reach the surface? (Ignore atmospheric drag forces and use gMars=3.72m/s^2
a. 2.4 minutes
b. 0.79 minutes
c. 1.8 minutes
d. 3.0 minutes
Answer:
t = 1.81 min , the correct answer is c
Explanation:
This is a missile throwing exercise
The object is thrown horizontally, so its vertical speed is zero (voy = 0), let's use the equation
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
the final height is y = 0 and the initial height is y₀ = 22000 m
0 = y₀ + 0 - ½ g t²
t = √y 2y₀ / g
let's calculate
t = √(2 22000 / 3.72)
t = 108.76 s
let's reduce to minutes
t = 108.76 s (1 min / 60 s)
t = 1.81 min
The correct answer is c
A dentist uses a concave mirror (focal length 2 cm) to examine some teeth. If the distance from the object to the mirror is 1 cm, what is the magnification of the tooth
Answer: 2
Explanation:
1/2=1/1 +1/x
x=-2
magnification= 2/1
magnification=2
A plano-convex glass lens of radius of curvature 1.4 m rests on an optically flat glass plate. The arrangement is illuminated from above with monochromatic light of 520-nm wavelength. The indexes of refraction of the lens and plate are 1.6. Determine the radii of the first and second bright fringes in the reflected light.
Given that,
Radius of curvature = 1.4 m
Wavelength = 520 nm
Refraction indexes = 1.6
We know tha,
The condition for constructive interference as,
[tex]t=(m+\dfrac{1}{2})\dfrac{\lambda}{2}[/tex]
Where, [tex]\lambda=wavelength[/tex]
We need to calculate the radius of first bright fringes
Using formula of radius
[tex]r_{1}=\sqrt{2tR}[/tex]
Put the value of t
[tex]r_{1}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]
Put the value into the formula
[tex]r_{1}=\sqrt{2\times(0+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]
[tex]r_{1}=0.603\ mm[/tex]
We need to calculate the radius of second bright fringes
Using formula of radius
[tex]r_{2}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]
Put the value into the formula
[tex]r_{1}=\sqrt{2\times(1+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]
[tex]r_{1}=1.04\ mm[/tex]
Hence, The radius of first bright fringe is 0.603 mm
The radius of second bright fringe is 1.04 mm.
How much work is needed to pump all the water out of a cylindrical tank with a height of 10 m and a radius of 5 m
Answer:
Explanation:
volume of water being lifted
= π r² h , where r is radius of cylinder and h is height of cylinder
= 3.14 x5² x 10
= 785 m³
mass of water = 785 x 10³ kg
mass of this much of water is lifted so that its centre of mass is lifted by height
10 / 2 = 5m .
So work done = mgh , m is mass of water , h is displacement of centre of mass and g is acceleration due to gravity
= 785 x 10³ x 9.8 x 5
= 38.465 x 10⁶ J
The positron has the same mass as an electron, with an electric charge of +e. A positron follows a uniform circular motion of radius 5.03 mm due to the force of a uniform magnetic field of 0.85 T. How many complete revolutions does the positron perform If it spends 2.30 s inside the field? (electron mass = 9.11 x 10-31 kg, electron charge = -1.6 x 10-19 C)
Answer:
5.465 × 10^10 revolutions
Explanation:
Formula for Magnetic Field = m. v/ q . r
M = mass of electron = mass of positron = 9.11 x 10^-31 kg,
radius of the positron = 5.03 mm
We convert to meters.
1000mm = 1m
5.03mm = xm
Cross multiply
x = 5.03/1000mm
x = 0.00503m
q = Electric charge = -1.6 x 10^-19 C
Magnetic field (B) = 0.85 T
Speed of the positron is unknown
0.85 = 9.11 x 10^-31 kg × v/ -1.6 x 10^-19 C × 0.00503
0.85 × 1.6 x 10^-19 C × 0.00503 = 9.11 x 10^-31 kg × v
v = 0.85 × -1.6 x 10^-19 C × 0.00503/9.11 x 10^-31 kg
v = 6.8408 ×10-22/ 9.11 x 10^-31 kg
v = 750911086.72m/s
Formula for complete revolutions =
Speed × time / Circumference
Time = 2.30s
Circumference of the circular path = 2πr
r =0.00503
Circumference = 2 × π × 0.00503
= 0.0316044221
Revolution = 750911086.72 × 2.30/0.0316044221
= 1727095499.5/0.0316044221
= 546541562294 revolutions
Approximately = 5.465 × 10^10 revolutions
Four charges each of magnitude 15 µC are arranged on the corners of a square of side 5 cm. What is the total potential energy of the system?
Answer:
-105J
Explanation:
See attached file
Reading glasses with a power of 1.50 diopters make reading a book comfortable for you when you wear them 1.8 cmcm from your eye. Part A If you hold the book 28.0 cmcm from your eye, what is your nearpoint distance
Answer:
The near point is [tex]n =44.8 \ cm[/tex]
Explanation:
From the question we are told that
The power is [tex]P = 1.50[/tex]
The distance from the eye is [tex]k = 1.8 \ cm[/tex]
The distance of the book from the eye is [tex]z = -28 \ cm[/tex]
Generally the focal length of the glasses is
[tex]f = \frac{1}{P}[/tex]
=> [tex]f = \frac{1}{1.50 }[/tex]
=> [tex]f = 0.667 \ m[/tex]
=> [tex]f = 66.7 \ cm[/tex]
The object distance is evaluated as
[tex]u = z + k[/tex]
=> [tex]u = -28 + 1.8[/tex]
=> [tex]u = -26.2 \ cm[/tex]
The image distance is evaluated from lens formula as
[tex]\frac{1}{v} = \frac{1}{f} + \frac{1}{u}[/tex]
=> [tex]\frac{1}{v} = \frac{1}{66.7} + \frac{1}{-26.2}[/tex]
=> [tex]v=- \frac{1}{0.0232}[/tex]
=> [tex]v=- 43 \ cm[/tex]
The near point is evaluated as
[tex]n = -v + k[/tex]
=> [tex]n =-(-43) + 1.8[/tex]
=> [tex]n =44.8 \ cm[/tex]
A long bar slides on two contact points and is in motion with velocity ν. A steady, uniform, magnetic field B is present. The induced current through resistor R is:
Answer:
The induced current in the resistor is I = BLv/R
Explanation:
The induced emf ε in the long bar of length, L in a magnetic field of strength, B moving with a velocity, v is given by
ε = BLv.
Now, the current I in the resistor is given by
I = ε/R where ε = induced emf in circuit and R = resistance of resistor.
So, the current I = ε/R.
substituting the value of ε the induced emf, we have
I = ε/R
I = BLv/R
So, the induced current through the resistor is given by I = BLv/R
a ball is kicked on level ground with a speed of 30 m/s at angle of 40 degrees above horizontal g Find the minimum velocity of the ball during its flight
Answer:
The minimum velocity of the ball during its flight is 22.98 m/s.
Explanation:
The velocity of the ball v = 30 m/s
The angle it makes with the horizontal ∅ = 40°
The minimum velocity of the ball during flight will be the horizontal axis component of the velocity, as acceleration is zero on this axis.
[tex]V_{x}[/tex] = v cos ∅
[tex]V_{x}[/tex] = 30 cos 40°
[tex]V_{x}[/tex] = 30 x 0.766 = 22.98 m/s
When a magnet moves above a conducting ladder, the currents induced in the ladder produces a magnetic field. This field interacts with the magnetic field of the magnet to produce a force that A) pulls the ladder in the direction opposite to the direction of the moving magnet. B) pulls the ladder in the same direction as the moving magnet. C) brakes the ladder. D) None of the above.
Answer:
) pulls the ladder in the direction opposite
Explanation:
This is in line with lenz law that states that the magnetic field induced in a conductor act to oppose the magnetic field that produced it
If the
refractive index of benzere is 2.419,
what is the speed of light in benzene?
Answer:
[tex]v=1.24\times 10^8\ m/s[/tex]
Explanation:
Given that,
The refractive index of benzene is 2.419
We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,
[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]
So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].
When light travels from one medium to another with a different index of refraction, how is the light's frequency and wavelength affected
Answer:
The frequency does not change, but the wavelength does
Explanation:
Here are the options
A. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, the frequency changes and the wavelength does not.
B. The frequency does change, but the wavelength remains unchanged.
C. Both the frequency and wavelength change.
D. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, neither the wavelength nor the frequency changes.
E. The frequency does not change, but its wavelength does.
When light goes through one medium to the next, the frequency doesn't really change seeing as frequency is dependent on wavelength and light wave velocity. And when the wavelength shifts from one medium to the next.
[tex]n= \frac{C}{V} \ and\ \frac{\lambda_o}{\lambda_m}[/tex]
where [tex]\lambda_o[/tex] indicates wavelength in vacuum
[tex]\lambda_m[/tex] indicates wavelength in medium
n indicates refractive index
v indicates velocity of light wave
c indicates velocity of light
And wavelength is medium-dependent. Frequency Here = v[tex]\lambda[/tex] and shift in wavelength and velocity, not shifts in overall frequency.
Therefore the correct option is E
To get an idea of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 1500 loops of wire wound on a rod 13 cm long with radius 2 cm
Answer:
The self-inductance in henries for the solenoid is 0.0274 H.
Explanation:
Given;
number of turns, N = 1500 turns
length of the solenoid, L = 13 cm = 0.13 m
radius of the wire, r = 2 cm = 0.02 m
The self-inductance in henries for a solenoid is given by;
[tex]L = \frac{\mu_oN^2A}{l}[/tex]
where;
[tex]\mu_o[/tex] is permeability of free space = [tex]4\pi*10^{-7} \ H/m[/tex]
A is the area of the solenoid = πr² = π(0.02)² = 0.00126 m²
[tex]L = \frac{4\pi *10^{-7}(1500)^2*(0.00126)}{0.13} \\\\L = 0.0274 \ H[/tex]
Therefore, the self-inductance in henries for the solenoid is 0.0274 H.
The sun generates both mechanical and electromagnetic waves. Which statement about those waves is true?
OA. The mechanical waves reach Earth, while the electromagnetic waves do not.
OB. The electromagnetic waves reach Earth, while the mechanical waves do not.
OC. Both the mechanical waves and the electromagnetic waves reach Earth.
OD. Neither the mechanical waves nor the electromagnetic waves reach Earth.
Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.
Hope this helps!
Answer:
The electromagnetic waves reach Earth, while the mechanical waves do not
Sunlight reflected from a smooth ice surface is completely polarized. Determine the angle of incidence. (nice = 1.31.)
Answer:
Explanation:
We shall apply Brewster's law to solve the problem . Let i be the angle of reflection required. According to this law ,
tan i = μ where i is angle of incidence , μ is refractive index of the medium .
Here i = ? , μ = 1.31
tani = 1.31
i = 53° approx .
At the angle of 53° incidence , the reflected ray will be completely polarised .
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by:_____
a. a whole number of half-wavelengths.
b. a whole number of wavelengths.
c. an odd number of half-wavelengths.
Answer:
(B) a whole number of wavelengths.
Explanation:
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by a whole number of wavelengths.
When the resultant effect of the combination of two identical waves result in their annihilation or complete cancellation of the effect of each other, destructive interference takes place. Hence to have two wave sources producing waves that have the same frequency wavelength and amplitude and which are always in phase with each other or have a constant phase difference are said to be Coherent source
The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature in degrees Celsius when 1.50 kcal of heat enters 1.50 kg of the following, originally at 15.0°C.(a) water
(b) concrete
(c) steel
(d) mercury
Answer:
Final temperature Water = 20.99-degree celsius
Final temperature Concrete = 24.98 degree celsius
Final temperature Steel = 50.1 degree celsius
Final temperature Mercury = 29.26 degree celsius
Explanation:
Given the mass of each substance = 1.50 kg
Ti = 15
Q = 1.5 kcal = 6276 joule
We have to use the heat capacity of each object so find the heat capacity from the table.
Heat capacity of water = 4186 J/kg degree celsius.
Heat capacity of concrete = 840 J/kg degree celsius.
Heat capacity of steel = 452 J/kg degree celsius.
Heat capacity of mercury = 139 J/kg degree celsius.
Use the below formula to find the final temperature.
[tex]T_f = T_i + \frac{Q}{mc_w} \\[/tex]
[tex]\text{Temperature in the case of water.} \\= 20 + \frac{6276}{1.5 \times 4186 } \\= 20.99 \ degree \ celsius \\\text{Temperature in the case of concrete.} \\= 20 + \frac{6276}{1.5 \times 840 } \\= 24.98 \ degree \ celsius \\\text{Temperature in the case of steel.} \\= 20 + \frac{6276}{1.5 \times 452 } \\= 29.26 \ degree \ celsius \\\text{Temperature in the case of mercury.} \\= 20 + \frac{6276}{1.5 \times 139 } \\= 50.1 \ degree \ celsius \\[/tex]
A high school physics student claims her muscle car can achieve a constant acceleration of 10 ft/s/s. Her friend develops an accelerometer to confirm the feat. The accelerometer consists of a 1 ft long rod (mass=4 kg) with one end attached to the ceiling of the car, but free to rotate. During acceleration, the rod rotates. What will be the angle of rotation of the rod during this acceleration? Assume the road is flat and straight.
Answer: Ф = 17.2657 ≈ 17°
Explanation:
we simply apply ET =0 about the ending of the rod
so In.g.L/2sinФ - In.a.L/2cosФ = 0
g.sinФ - a.cosФ = 0
g.sinФ = a.cosФ
∴ tanФ = a/g
Ф = tan⁻¹ a / g
Ф = tan⁻¹ ( 10 / 32.17405)
Ф = tan⁻¹ 0.31080948777
Ф = 17.2657 ≈ 17°
Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°
Matter's resistance to a change in motion is called _____ and is directly proportional to the mass of an object
Answer:
Matter's resistance to a change in motion is called INERTIA and is directly proportional to the mass of an object.
Explanation:
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that carries a 4.5-A current
Answer:
Explanation:
From the question we are told that
The radius is [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]
The current is [tex]I = 4.5 \ A[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{J}{\sigma }[/tex]
Where [tex]\sigma[/tex] is the conductivity of aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]
J is the current density which mathematically represented as
[tex]J = \frac{I}{A}[/tex]
Here A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (1.4*10^{-3})^2[/tex]
[tex]A = 6.158*10^{-6} \ m^2[/tex]
So
[tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]
[tex]J = 730757 A/m^2[/tex]
So
[tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]
[tex]E = 0.021 \ N/C[/tex]
Water flows at speed v in a pipe of radius R. At what speed does the water flow through a constriction in which the radius of the pipe is R/3
Answer:
v₂ = 9 v
Explanation:
For this exercise in fluid mechanics, let's use the continuity equation
v₁ A₁ = v₂ A₂
where v is the velocity of the fluid, A the area of the pipe and the subscripts correspond to two places of interest.
The area of a circle is
A = π R²
let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint
In this case v₁ = v and the area is
A₁ = π R²
in the second point
A₂= π (R / 3)²
we substitute in the continuity equation
v π R² = v₂ π R² / 9
v = v₂ / 9
v₂ = 9 v
Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.630 cm thick and heat conduction occurs through an area of 1.45 ✕ 10−2 m2 at a rate of 500 J/s, what is the temperature difference across it (in °C)? Ceramic has the same thermal conductivity as glass and concrete brick.
Answer:
The temperature difference [tex]\Delta T = 258.6 \ ^ o\ C[/tex]
Explanation:
From the question we are told that
The thickness is [tex]\Delta x = 0.630 cm = 0.0063 m[/tex]
The area is [tex]A = 1.45 *10^{-2 } \ m^2[/tex]
The rate is [tex]P = 500 J/s[/tex]
The thermal conductivity is [tex]\sigma = 0.84J[\cdot s \cdot m \cdot ^oC ][/tex]
Generally the rate heat conduction mathematically represented as
[tex]P = \sigma * A * \frac{\Delta T}{\Delta x }[/tex]
=> [tex]\Delta T = \frac{P * \Delta x }{\sigma * A }[/tex]
=> [tex]\Delta T = \frac{ 500 * 0.00630 }{ 0.84 * 1.45 *10^{-2} }[/tex]
=> [tex]\Delta T = 258.6 \ ^ o\ C[/tex]
A 590-turn solenoid is 12 cm long. The current in it is 36 A . A straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?
Complete Question
A 590-turn solenoid is 12 cm long. The current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?
Answer:
The force is [tex]F = 0.1602 \ N[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 590 \ turns[/tex]
The length of the solenoid is [tex]L = 12 \ cm = 0.12 \ m[/tex]
The current is [tex]I = 36 \ A[/tex]
The diameter is [tex]D = 4.5 \ cm = 0.045 \ m[/tex]
The current carried by the wire is [tex]I = 27 \ A[/tex]
The length of the wire is [tex]l = 2 cm = 0.02 \ m[/tex]
Generally the magnitude of the force on this wire assuming the solenoid's field points due east is mathematically represented as
[tex]F = B * I * l[/tex]
Here B is the magnetic field which is mathematically represented as
[tex]B = \frac{\mu_o * N * I }{L}[/tex]
Here [tex]\mu _o[/tex] is permeability of free space with value [tex]\mu_ o = 4\pi *10^{-7} \ N/A^2[/tex]
substituting values
[tex]B = \frac{4 \pi *10^{-7} * 590 * 36 }{ 0.12}[/tex]
[tex]B = 0.2225 \ T[/tex]
So
[tex]F = 0.2225 * 36 * 0.02[/tex]
[tex]F = 0.1602 \ N[/tex]
