Answer:
The velocity just before hitting the ground is [tex]v_f = 30 m/s[/tex]
Explanation:
From the question we are told that
The initial speed is [tex]u = 10 m/s[/tex]
The final speed is [tex]v = 30 \ m/s[/tex]
From the equations of motion we have that
[tex]v^2 =u^2 + 2as[/tex]
Where s is the distance travelled which is the height of the cliff
So making it the subject of the the formula we have that
[tex]s = \frac{v^2 - u^2 }{2a}[/tex]
Where a is the acceleration due to gravity with a value [tex]a = 9.8m/s^2[/tex]
So
[tex]s = \frac{30^2 - 10^2 }{2 * 9.8 }[/tex]
[tex]s = 40.8 \ m[/tex]
Now we are told that was through horizontally with a speed of
[tex]v_x =10 m/s[/tex]
Which implies that this would be its velocity horizontally through out the motion
Now it final velocity vertically can be mathematically evaluated as
[tex]v_y = \sqrt{2as}[/tex]
Substituting values
[tex]v_y = \sqrt{(2 * 9.8 * 40.8)}[/tex]
[tex]v_y = 28.3 \ m/s[/tex]
The resultant final velocity is mathematically evaluated as
[tex]v_f = \sqrt{v_x^2 + v_y^2}[/tex]
Substituting values
[tex]v_f = \sqrt{10^2 + 28.3^2}[/tex]
[tex]v_f = 30 m/s[/tex]
The shaft of a motor has an angular displacement θ that is a function of time given by the equation: θ(t) = 4.40 t 3 rad/s3 + 1.40 t2 rad/s2 . At time t = 0.00 s the wheel is at rest and is oriented at θ = 0.00 rad. a) Derive the equation that specifies the angular velocity of the shaft as a function of time. b) Derive the equation that specifies the angular acceleration as a function of time.
Answer:
a) [tex]\omega = 13.2t^2\frac{rad}{s^3}+2.80t\frac{rad}{s^2}[/tex]
b) [tex]\alpha=26.4t\frac{rad}{s^3}+2.80\frac{rad}{s^2}[/tex]
Explanation:
You have that the angular displacement is given by:
[tex]\theta=4.40t^3\frac{rad}{s^3}+1.40t^2\frac{rad}{s^2}[/tex]
a) the angular velocity is given by the derivative in time, of the angular displacement, that is:
[tex]\omega=\frac{d\theta}{dt}=\frac{d}{dt}[4.40 t^3 rad/s^3 + 1.40 t^2 rad/s^2]\\\\\omega=\frac{d\theta}{dt}=13.2t^2\frac{rad}{s^3}+2.80t\frac{rad}{s^2}[/tex]
b) the angular acceleration is the derivative, in time, of the angular velocity:
[tex]\alpha=\frac{d\omega}{dt}=\frac{d}{dt}[13.2t^2\frac{rad}{s^3}+2.80t\frac{rad}{s^2}]\\\\\alpha=26.4t\frac{rad}{s^3}+2.80\frac{rad}{s^2}[/tex]
Which is the correct representation of the right-hand rule for a current flowing to the right?
Answer:
The third image
Explanation:
The one with the thumb pointing to the right
Answer:
3, correct on Edge 2020
A. A PH202 student lives next to a construction site and sees a crane with a wrecking ball demolish the building next door. The wrecking ball swings along the wall between her house and the neighbor’s house. In an effort to determine the length of the cable on the wrecking ball the student builds a pendulum using a random rock and a string. Her pendulum turns out to be 0.500m long. While she plays with her pendulum she realizes that the wrecking ball swings back and forth in the same amount of time that it takes the rock to complete 5 full oscillations. What is the length of the cable on the wrecking ball?
Answer:
The length of cable is 12.5 m
Explanation:
Since, the wrecking ball completes 1 oscillation, in the same time, as it takes for the rock to complete 5 oscillations.
Therefore,
Time Period of Wrecking Ball = 5 (Time Period of Rock)
Since,
Time Period of Pendulum = 2π√(L/g)
Therefore,
2π√(L₁/g) = 5[2π√(L₂/g)]
√L₁ = 5√L₂
Squaring on both sides:
L₁ = 25 L₂
where,
L₁ = Length of Cable = ?
L₂ = Length of string = 0.5 m
Therefore,
L₁ = 25 (0.5 m)
L₁ = 12.5 m
water is pumped from a stream at the rate of 90kg every 30s and sprayed into a farm at a velocity of 15m/s. Calculate the power of the pump.
Answer:
340 W
Explanation:
Power = change in energy / change in time
P = ΔKE / Δt
P = ½ mv² / Δt
P = ½ (90 kg) (15 m/s)² / (30 s)
P = 337.5 W
Rounded to 2 significant figures, the power is 340 W.
A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500.0 N/m (assume that the spring is massless). The block is held in position such that the spring is compressed 4.00 cm shorter than its undisturbed length. The block is suddenly released and allowed to slide away on the frictionless surface. Find the speed the block will be traveling when it leaves the spring.
Answer:
6 m/s
Explanation:
Given that :
mass of the block m = 200.0 g = 200 × 10⁻³ kg
the horizontal spring constant k = 4500.0 N/m
position of the block (distance x) = 4.00 cm = 0.04 m
To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.
i.e [tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2[/tex]
[tex]kx^2 = mv^2[/tex]
[tex]4500* 0.04^2 = 200*10^{-3} *v^2[/tex]
[tex]7.2 =200*10^{-3}*v^{2}[/tex]
[tex]v^{2} =\frac{7.2}{200*10^{-3}}[/tex]
[tex]v =\sqrt{\frac{7.2}{200*10^{-3}}}[/tex]
v = 6 m/s
Hence,the speed the block will be traveling when it leaves the spring is 6 m/s
Which of the following actions would decrease the energy stored in a parallel plate capacitor when a constant potential difference is applied across the plates? (Choose from: Increasing the area of the plates; Decreasing the area of the plates; Increasing the separation between the plates; Decreasing the separation between the plates; Inserting a material with a higher dielectric constant between the plates
Answer:
increasing the separation between the plates
Explanation:
The increase in the vacuum/separation between the plates in a parallel plate capacitor connected to a constant potential difference decreases the energy stored in the capacitor. the increase in the separation of the plates of a parallel plate capacitor reduces the capacitance of the capacitor because
Q(charge) = CV V = VOLTAGE , c = capacitance
E = 1/2 eAV^2/ D ( ENERGY STORED )
where D = distance between plates, e = dielectric, A = area of capacitor , V = potential difference
A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2300 kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.75, and the cars slide to a halt at a point 5.54 m west and 6.19 m south of the impact point. How fast was sedan traveling just before the collision? How fast was SUV traveling just before the collision?
Answer:
Explanation:
momentum of sedan of 1600 kg = 1600x v , where v is its velocity
momentum of suv of 2300 kg = 2300 x u where u is its velocity .
force of friction = ( 1600 + 2300 ) x 9.8 x .75 ( fiction = μ mg )
= 28665 N
distance by which friction acted = √ (5.54² + 6.19²)
= 8.3 m
work done by friction
= 28665 x 8.3
= 237919.5 J
Total kinetic energy of cars = work done by friction
1/2 x 1600 x v² + 1/2 x 2300 u² = 237919.5
16 v² + 23 u² = 4758.4
1600 x v / 2300 u = 6.19 / 5.54
v / u = 1.6
v = 1.6 u
putting this equation in fist equation
40.96 u² + 23 u² = 4758.4
= 63.96 u² = 4758.4
u² = 74.4
u = 8.62 m /s
v = 13.8 m /s
A Texas cockroach of mass 0.157 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has a radius 14.9 cm, rotational inertia 5.92 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 2.92 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 3.89 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
Answer:
-7.23 rad/s
Explanation:
Given that
Mass of the cockroach, m = 0.157 kg
Radius of the disk, r = 14.9 cm = 0.149 m
Rotational Inertia, I = 5.92*10^-3 kgm²
Speed of the cockroach, v = 2.92 m/s
Angular velocity of the rim, w = 3.89 rad/s
The initial angular momentum of rim is
Iw = 5.92*10^-3 * 3.89
Iw = 2.3*10^-2 kgm²/s
The initial angular momentum of cockroach about the axle of the disk is
L = -mvr
L = -0.157 * 2.92 * 0.149
L = -0.068 kgm²/s
This means that we can get the initial angular momentum of the system by summing both together
2.3*10^-2 + -0.068
L' = -0.045 kgm²/s
After the cockroach stops, the total inertia of the spinning disk is
I(f) = I + mr²
I(f) = 5.92*10^-3 + 0.157 * 0.149²
I(f) = 5.92*10^-3 + 3.49*10^-3
I(f) = 9.41*10^-3 kgm²
Final angular momentum of the disk is
L'' = I(f).w(f)
L''= 9.41*10^-3w(f)
Using the conservation of total angular momentum, we have
-0.068 = 9.41*10^-3w(f) + 0
w(f) = -0.068 / 9.41*10^-3
w(f) = -7.23 rad/s
Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed
b)
The mechanical energy of the cockroach is not converted as it stops
A particle is projected at an angle 60 degrees to the horizontal with a speed of 20m/s. (i) calculate total time of flight of the particle. (i) speed of the particle at its maximum height
Answer:
Time of flight=3.5 seconds
Speed at maximum height is 0
Explanation:
Φ=60°
initial velocity=u=20m/s
Acceleration due to gravity=g=9.8 m/s^2
Total time of flight=T
Final speed=v
question 1:
T=(2 x u x sinΦ)/g
T=(2 x 20 x sin60)/9.8
T=(2 x 20 x 0.8660)/9.8
T=34.64/9.8
T=3.5 seconds
Question 2
Speed at maximum height is 0
A"boat"is"moving"to"the"right"at"5"m/s"with"respect"to"the"water."A"wave"moving"to"the"left,"opposite"the"motion"of"the"boat."The"waves"have"2.0"m"between"the"top"of"the"crests"and"the"bottom"of"the"troughs."The"period"of"the"wave"is"8.3"s"and"their"wavelength"is"110"m."At"one"instant"the"boat"sits"on"a"crest"of"the"wave,"20"seconds"later,"what"is"the"vertical"displacement"of"the"boat
Answer:
0.99m
Explanation:
Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:
[tex]v=\lambda f=\lambda\frac{1}{T}=(110m)\frac{1}{8.3s}=13.25\frac{m}{s}[/tex]
the relative velocity is:
[tex]v'=13.25m/s-5m/s=8.25\frac{m}{s}[/tex]
This velocity is used to know which is the distance traveled by the boat after 20 seconds:
[tex]x'=v't=(8.25m/s)(20s)=165m[/tex]
Next, you use the general for of a wave:
[tex]f(x,t)=Acos(kx-\omega t)=Acos(\frac{2\pi}{\lambda}x-\omega t)[/tex]
you take the amplitude as 2.0/2 = 1.0m.
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{8.3s}=0.75\frac{rad}{s}[/tex]
by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:
[tex]f(165,20)=1.0m\ cos(\frac{2\pi}{110m}(165)-(0.75\frac{rad}{s})(20s))\\\\f(165,20)=0.99m[/tex]
A bicycle coasting downhill reaches its maximum speed at the bottom of the
hill.
This speed would be even greater if some of the bike's
energy had
not been transformed into
energy
A) kinetic; heat
OB) heat; potential
C) kinetic; potential
OD) potential; kinetic
OB
mmnjnjlkdhfutydjfyiudtkcgvyftdcgvjyiluftgyiuyu ( had to do that cuz it wouldn't let through)
6. The two ends of an iron rod are maintained at different temperatures. The amount of heat thatflows through the rod by conduction during a given time interval does notdepend uponA) the length of the iron rod.B) the thermal conductivity of iron.C) the temperature difference between the ends of the rod.D) the mass of the iron rod.E) the duration of the time interval.Ans: DDifficulty: MediumSectionDef: Section 13-27. The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twiceas long and twice the diameter conduct heat between the same two temperatures
Answer:
20cal/s
Explanation:
Question:
There are two questions. The first one has been answered:
From the formular, Power = Q/t = (kA∆T)/l
the amount heat depends on the duration of time interval, length of the iron rod, the thermal conductivity of iron and the temperature difference between the ends of the rod.
The amount of heat that flows through the rod by conduction during a given time interval does not depend upon the mass of the iron rod (D).
Second question:
The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?
Solution:
Power = 10cal/s
Power = energy per unit time = Q/t
Where Q = energy
Power = (kA∆T)/l
k = thermal conductivity of iron
A = area
Area = πr^2
r = radius
Diameter = d = 2r
r = d/2
Area = (πd^2)/4
Length = l
∆T = change in temperature
10 = (kA∆T)/l
For a steel rod with length doubled and diameter doubled:
Let Length (L) = 2l
Diameter (D)= 2d
Area = π [(2d)^2]/4 = (π4d^2)/4
Area = 4(πd^2)/4
Using the formula Power = (kA∆T)/l, insert the new values for A and l
Power = [k × 4(πd^2)/4 × ∆T]/2l
Power = [4k((πd^2)/4) ∆T]/2l
Power = [(4/2)×k((πd^2)/4) ∆T]/l
Power = [2k(A) ×∆T]/l = 2(kA∆T)/l
Power of a steel that has its length doubled and diameter doubled = 2(kA∆T)/l
Recall initial Power = (kA∆T)/l = 10cal/s
And ∆T is the same
2[(kA∆T)/l] = 2 × 10
Power of a steel that has its length doubled and diameter doubled = 20cal/s
Part F A system experiences a change in internal energy of 14 kJkJ in a process that involves a transfer of 36 kJkJ of heat into the system. Simultaneously, which of the following is true? A system experiences a change in internal energy of 14 in a process that involves a transfer of 36 of heat into the system. Simultaneously, which of the following is true? 22 kJkJ of work is done by the system. 22 kJkJ of work is done on the system. 50 kJkJ of work is done by the system. 50 kJkJ of work is done on the system
Answer:
Explanation:
According to first law of thermodynamics :
Q = ΔE + W
Q is heat added , ΔE is increase in the internal energy of the system and W is work done by the system .
Here Q = 36 KJ
ΔE = 14 kJ
Putting the values in the equation
36 = 14 + W
W = 36 - 14
= 22 kJ .
Work done by gas or system = 22 kJ.
Electric fields are MOST associated with ________.
Yellow light with wavelength 600 nm is travelling to the left (in the negative x direction) in vacuum. The light is polarized along the z direction. (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave. (c) If the wave were to represent blue light instead of yellow light, how would your pictures in parts a and b change? If there is no change, say so explicitly.
Answer: (a) and (b) => check attached file.
(c). Picture (a) and (b) will both remain the same.
Explanation:
IMPORTANT: The solution to the question (a) and (b) that is (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave is there in the ATTACHED FILE/PICTURE.
It is also worthy of note to know that in anything Electromagnetic wave, the magnetic field, the Electric Field and their direction of propagation are perpendicular to each other.
Therefore, knowing the fact above we can say that in yellow light, the magnetic field is in the y-direction and the Electric Field is in the z-direction.
Hence, the solution to option C is given below;
(C).If the wave were to represent blue light instead of yellow light, picture (a) will remain the same because both light are Electromagnetic wave, although the wavelength will have to change. Picture (b) will also remain the same because they are both Electromagnetic waves and possess similar properties.
How are the elements in the same row similar
Answer:
All elements in a row have the same number of electron shells. Each next element in a period has one more proton and is less metallic than its predecessor. Arranged this way, groups of elements in the same column have similar chemical and physical properties, reflecting the periodic law.
What do you think will be different about cars in the future? Describe a change that is already being developed or that you think should be invented.
Answer:
Flying cars.
Explanation:
By which process does the heat from the Sun reach the Earth? (AKS 4b DOK 1) *
A piston with stops containing water goes through an expansion process through the addition of heat. State 1 the pressure is 200 kPa and the volume is 2 m3. After half of the heat has been delivered the piston hits the stops corresponding to a volume of 5 m3. After all the heat has been delivered, state 2, the pressure is 1000 kPa with the piston resting on the stops. What is the work?
Answer:
The work will be "600 kJ/kg".
Explanation:
(1-a) ⇒ Constant Pressure
(a-2) ⇒ Constant Volume
The given values are:
In state 1,
Pressure, P₁ = 200 kPa
Volume, V₁ = 2m³
In state 2,
Pressure, P₂ = 1000 kPa
Volume, V₁ = 5m³
Now,
In process (1-a), work will be:
⇒ W₁₋ₐ = P₁(Vₐ - V₁)
On putting the values, we get
⇒ W₁₋ₐ = 200(5-2)
⇒ = 200(3)
⇒ = 600 kJ/kg
In process (a-2), work will be:
⇒ Wₐ₋₂ = 0
∴ (The change in the volume will be zero.)
So,
⇒ Total work = (W₁₋ₐ) + (Wₐ₋₂)
⇒ = 600 + 0
⇒ = 600 kJ/kg
Help ill give you brainliest !!!
Answer:
1. B
2. A
3. C
4. B
5. A
6. Muscular strength is different than muscular endurance because of the fact that muscular strength is the amount of force that can be exerted in one instance. Muscular endurance is how long that you can exert that force without being completely exhausted.
7. Some benefits to strength training is the increase in muscular endurance. There is also the benefit of better muscular strength.
Explanation:
A student performs an experiment that involves the motion of a pendulum. The student attaches one end of a string to an object of mass M and secures the other end of the string so that the object is at rest as it hangs from the string. When the student raises the object to a height above its lowest point and releases it from rest, the object undergoes simple harmonic motion. As the student collects data about the time it takes for the pendulum to undergo one oscillation, the student observes that the time for one swing significantly changes after each oscillation. The student wants to conduct the experiment a second time. Which two of the following procedures should the student consider when conducting the second experiment?
a) Make sure that the length of the string is not too long.
b) Make sure that the mass of the pendulum is not too large.
c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.
d) Make sure that the experiment is conducted in an environment that has minimal wind resistance.
Answer:
the answers the correct one is cη
Explanation:
In this simple pendulum experiment the student observes a significant change in time between each period. This occurs since an approximation used is that the sine of the angle is small, so
sin θ = θ
with this approach the equation will be surveyed
d² θ / dt² = - g / L sin θ
It is reduced to
d² θ / dt² = - g / L θ
in which the time for each oscillation is constant, for this approximation the angle must be less than 10º so that the difference between the sine and the angles is less than 1%
The angle is related to the height of the pendulum
sin θ = h / L
h = L sin θ.
Therefore the student must be careful that the height is small.
When reviewing the answers the correct one is cη
Considering the approximation of simple harmonic motion, the correct option is:
(c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.
Simple Harmonic MotionAccording to Newton's second law in case of rotational motion, we have;
[tex]\tau = I \alpha[/tex]
Applying this, in the case of a simple pendulum, we get;
[tex]-mg\,sin\,\theta =mL^2 \,\frac{d^2 \theta}{dt^2}[/tex]
On, rearranging the above equation, we get;
[tex]mL^2 \,\frac{d^2 \theta}{dt^2} + mg\,sin\,\theta=0\\\\\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} sin \,\theta=0[/tex]
Now, if angular displacement is very small, i.e.; the bob of the pendulum is only raised slightly.
Then, [tex]sin\, \theta \approx \theta[/tex]
[tex]\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} \,\theta=0[/tex]
This is now in the form of the equation of a simple harmonic motion.
[tex]\frac{d^2 \theta}{dt^2} +\omega^2 \,\theta=0[/tex]
Comparing both these equations, we can say that;
[tex]\omega = \sqrt{\frac{g}{L}}[/tex]
[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]
This relation for the time period can only be obtained if the angular displacement is very less.
So, the correct option is;
Option (c): Make sure that the difference in height between the pendulum's release position and rest position is not too large.
Learn more about simple harmonic motion here:
https://brainly.com/question/26114128
How the musculoskeletal and nervous system develop as a human grows
Answer:
Explanation:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an efficiency of 30%, and taken in 50 kJ from the hot steam per cycle. If a Carnot engine takes in the same amount of heat per cycle and operates at these temperatures, the work it can turn into is most likely to be:a) 15 kJ. b) 20 kJ. c) 10 kJ. d) 50 kJ.
Answer:
b) 20 kJ
Explanation:
Efficiency of carnot engine = (T₁ - T₂ ) / T₁ Where T₁ is temperature of hot source and T₂ is temperature of sink .
T₁ = 270 + 273 = 543K
T₂ = 50 + 273 = 323 K
Putting the given values of temperatures
efficiency = (543 - 323) / 543
= .405
heat input = 50 KJ
efficiency = output work / input heat energy
.405 = output work / 50
output work = 20.25 KJ.
= 20 KJ .
A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. Long after contact is made with the battery (a) the voltage across the capacitor is A) equal to the battery's terminal voltage. B) less than the battery's terminal voltage, but greater than zero. C) zero. (b) the voltage across the resistor is A) equal to the battery's terminal voltage. B) less than the battery's terminal voltage, but greater than zero. C) zero.
Answer:
A) equal to the battery's terminal voltage.
Explanation:
When the capacitor is fully charged after long hours of charging , its potential becomes equal to the emf of the battery and its polarity is opposite to that of battery . Hence net emf becomes equal . The capacitor itself becomes a battery which is connected in the circuit with opposite polarity . This results in the net emf and current becoming zero . There is no charging current when the capacitor is fully charged .
8. At temperature 15°C, aluminum rivets have a diameter of 0.501 cm, and holes drilled in a titanium sheet have a diameter of 0.500 cm. If both the aluminum rivets and the titanium sheet are cooled together, at what temperature will the rivets just fit into the appropriate holes in the titanium sheet? Use 25x10-6 (°C)-1 for the coefficient of linear expansion for aluminum, and 8.5x10-6 (°C)-1 for titanium
Answer:
The temperature is [tex]T = -106 ^oC[/tex]
Explanation:
From the question we are told that
The temperature is [tex]T_1 = T_t= T_a=15^oC[/tex]
The diameter is [tex]d_1 = 0.5001 cm[/tex]
The diameter of the hole [tex]d_2 = 0.500 \ cm[/tex]
The coefficient of linear expansion for aluminum is [tex]\alpha _1 = 25 *10^{-6} \ ^oC^{-1}[/tex]
The coefficient of linear expansion for titanium is [tex]\alpha _2 = 8.5 *10^{-6} \ ^o C^{-1}[/tex]
According to the law of linear expansion
[tex]d = d_o (1 + \alpha \Delta T )[/tex]
Where [tex]d_o[/tex] represents the original diameter
So for aluminum
[tex]d_a = d_1 (1 + \alpha_1 (T- T_a) )[/tex]
Where [tex]d_a[/tex] is the new diameter of aluminum
[tex]T_a[/tex] is the new temperature of the aluminum
So for titanium
[tex]d_t = d_2 (1 + \alpha_1 (T- T_t) )[/tex]
Where [tex]d_t[/tex] is the new diameter of titanium
[tex]T_t[/tex] is the new temperature of the aluminum
So for the aluminum rivets to fit into the holes
[tex]d_a = d_t[/tex]
=> [tex]d_1 (1 + \alpha_1 (T- T_a) ) = d_2 (1 + \alpha_2 (T- T_t) )[/tex]
Making T the subject of the formula
[tex]T = \frac{(d_1 - d_2 ) + (d_2 *\alpha_2 T_t) - d_1 \alpha_1 * T_a }{d_2 \alpha_2 - d_1 \alpha_1 }[/tex]
Substituting values
[tex]T = \frac{(0.501 - 0.500 ) + (0.500 *(8.5*10^{-6}) * 15) - 0.500* (25*10^{-6}) * 15 }{0.500 * (8.5 *10^{-6}) - 0.501 * (25 *10^{-6}) }[/tex]
[tex]T = -106 ^oC[/tex]
A long solid conducting cylinder with radius a = 12 cm carries current I1 = 5 A going into the page. This current is distributed uniformly over the cross section of the cylinder. A cylindrical shell with radius b = 21 cm is concentric with the solid cylinder and carries a current I2 = 3 A coming out of the page. 1)Calculate the y component of the magnetic field By at point P, which lies on the x axis a distance r = 41 cm from the center of the cylinders.
Answer:
Explanation:
We shall use Ampere's circuital law to find magnetic field at required point.
The point is outside the circumference of two given wires so whole current will be accounted for .
Ampere's circuital law
B = ∫ Bdl = μ₀ I
line integral will be over circular path of radius r = 41 cm .
Total current I = 5A -3A = 2A .
∫ Bdl = μ₀ I
2π r B = μ₀ I
2π x .41 B = 4π x 10⁻⁷ x 2
B = 2 x 10⁻⁷ x 2 / .41
= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.
You could use an analytical or triple beam balance to determine a ___ called ____
A)
physical property; mass.
B)
chemical property, mass.
C)
physical property; weight.
D)
physical property; density.
Answer:
a and b are the correct answers
Explanation:
Answer:
A) physical property; mass.
Explanation:
took the test
A mutation causes a dog to be born with a tail that is shorter than normal.
Which best describes this mutation?
Answer:
A mutation causes a dog to be born with a tail that is shorter than normal. Which best describes this mutation? It is harmful because it obviously affects the dog’s survival. It is harmful because it affects the dog’s physical appearance. It is neutral because it does not obviously affect the dog’s survival. It is beneficial because it affects the dog’s physical appearance.
Explanation:
Answer:
C
Explanation:
:)))
To understand thermal linear expansion in solid materials. Most materials expand when their temperatures increase. Such thermal expansion, which is explained by the increase in the average distance between the constituent molecules, plays an important role in engineering. In fact, as the temperature increases or decreases, the changes in the dimensions of various parts of bridges, machines, etc., may be significant enough to cause trouble if not taken into account. That is why power lines are always sagging and parts of metal bridges fit loosely together, allowing for some movement. It turns out that for relatively small changes in temperature, the linear dimensions change in direct proportion to the temperature.
For instance, if a rod has length L0 at a certain temperature T0 and length L at a higher temperature T, then the change in length of the rod is proportional to the change in temperature and to the initial length of the rod: L - L0 = αL0(T - T0),
or
ΔL = αL0ΔT.
Here, α is a constant called the coefficient of linear expansion; its value depends on the material. A large value of α means that the material expands substantially as the temperature increases; smaller values of α indicate that the material tends to retain its dimensions. For instance, quartz does not expand much; aluminum expands a lot. The value of α for aluminum is about 60 times that of quartz!
Questions:
A) Compared to its length in the spring, by what amount ΔLwinter does the length of the bridge decrease during the Teharian winter when the temperature hovers around -150°C?
B) Compared to its length in the spring, by what amount ΔLsummer does the length of the bridge increase during the Teharian summer when the temperature hovers around 700°C?
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
What types of mediums are involved in the energy transfer
Answer:
In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.
