Answer:
A.) The temperature of the junction between copper and sliver when they have come to equilibrium is 35 degree Celsius
B.) ice (in grams) melts per second = 0.078 kg/s
Explanation:
A.) Given that the two material are of the same area.
The Silver rod is 15.0 cm in length and the copper rod is 25.0 cm in length.
Silver temperature = 0 degree Celsius
Copper temperature = 100 degree Celsius
Thermal conductivity k of silver = 429 W/m•K
Thermal conductivity k of copper = 385W/m.k
Rate of energy transferred P in the two materials can be expressed as
P = k.A.dT/L
dT = change in temperature
Since the rate and the area are the same
429 ( T -0 )/0.15 = 385( 100 - T )/0.25
2860T = 1540(100 - T)
Open the bracket
2860T = 154000 - 1540T
Collect the like terms
2860T + 1540T = 154000
4400T = 154000
T = 154000/4400
T = 35 degree Celsius
The temperature of the junction between copper and sliver when they have come to equilibrium is 35 degree Celsius
B.) Rate of energy transferred P will be
P = 2860 × 35 = 100100
P = Q/t ..... (1)
Where Q = energy transferred
But Q = mcØ .....(2)
And specific heat capacity c of water = 4182J/k.kg
Substitutes Q into formula 1.
P = mcØ/t
Make m/t the subject of formula
m/t = P/cØ
m/t = 100100/ 4182( 35 + 273 )
m/t = 100100/1288056
m/t = 0.078 kg/s
It has been suggested that rotating cylinders about 12.5 mi long and 3.99 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth
Answer:
The correct answer to the following question will be "0.0562 rad/s".
Explanation:
[tex]r =\frac{3.9}{2}\times 1609.34[/tex]
[tex]=3138.213\ m[/tex]
As we know,
⇒ [tex]\omega^2 \ r=g[/tex]
On putting the values, we get
⇒ [tex]\omega^2\times 3138.213=9.8[/tex]
⇒ [tex]\omega = \sqrt{\frac{9.8}{3138.213}}[/tex]
⇒ [tex]\omega = 0.0562 \ rad /s[/tex]
Who first used the word atom to describe the smallest unit
Answer: It was Democritus, in fact, who first used the word atomos to describe the smallest possible particles of matter.
Explanation: hope this helped
Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field:________
Answer:
Del(ρ/ε₀) - (Del)²E = -dμ₀J/dt
Explanation:
From Maxwell's fourth equation
Curl B = μ₀J + μ₀ε₀dE/dt (1) where the second term is the displacement current.
If the oscillation conduction current in the conductor is much larger than the displacement current then, the displacement current goes to zero. So we have
Curl B = μ₀J (2)(since μ₀ε₀dE/dt = 0)
From maxwell's third equation
Curl E = -dB/dt (3)
taking curl of the above from the left
Curl(Curl E) = Curl(-dB/dt)
Curl(Curl E) = (-d(CurlB)/dt) (4)
Substituting for Curl B into (4), we have
Curl(Curl E) = -dμ₀J/dt
Del(DivE) - (Del)²E = -dμ₀J/dt (5)
From Maxwell's first equation,
DivE = ρ/ε₀
Substituting this into (5), we have
Del(ρ/ε₀) - (Del)²E = -dμ₀J/dt
There is a known potential difference between two charged plates of 12000 Volts. An object with a charge of 6.5 x 10-6 C charge and a mass of 0.02 kg is placed next to the positive plate. How fast will it be traveling when it gets to the negative plate
Answer:
1.97 m/s.
Explanation:
From the question,
Using the law of conservation of energy,
The energy stored in the charged plate = Kinetic energy of the mass
1/2(qV) = 1/2mv².......................... Equation 1
Where q = charge, V = voltage, m = mass, v = velocity.
make v the subject of the equation
v = √(qV/m)......................... Equation 2
Given: q = 6.5×10⁻⁶ C, V = 12000 Volts, m = 0.02 kg
Substitute these values into equation 2
v = √(6.5×10⁻⁶×12000 /0.02)
v = √3.9
v = 1.97 m/s.
g A 47.3 kg girl is standing on a 162 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.36 m/s relative to the plank. What is her velocity relative to the ice surface
Answer:
Explanation:
mass of the girl m₁ = 47.3 kg
mass of the plank m₂ = 162 kg
velocity of the girl with respect to surface = v₁
velocity of plank with respect to surface = v₂
v₁+ v₂ = 1.36
v₂ = 1.36 - v₁
applying conservation of momentum law to girl and plank.
m₁v₁ = m₂v₂
47.3 x v₁ = 162 x ( 1.36 - v₁ )
47.3 v₁ = 220.32 - 162v₁
209.3 v₁ = 220.32
v₁ = 1.05 m /s
If the frequency is 5 Hz, determine the speed of the wave in the spring?? Can someone pls help me??
Answer:
The speed of the wave is [tex]31.42 rad/s[/tex]Explanation:
yes, we can.Given data
frequency = 5 Hz
we know that the period T is expressed as
[tex]T= \frac{1}{f} \\[/tex]
Substituting we have
[tex]T= \frac{1}{5} \\T= 0.2s[/tex]
also the expression for angular velocity is
ω= [tex]\frac{2\pi}{T}[/tex]
Substituting we have
ω= [tex]\frac{2*3.142}{0.2}[/tex]
ω= [tex]\frac{6.284}{0.2} \\[/tex]
ω= [tex]31.42 rad/s[/tex]
For saving energy, bicycling and walking are far more efficient means of transportation than is travel by automobile. For example, when riding at 12.5 mi/h, a cyclist uses food energy at a rate of about 360 kcal/h above what he would use if merely sitting still. (In exercise physiology, power is often measured in kcal/h rather than in watts. Here 1 kcal = 1 nutritionist's Calorie = 4186 J). Walking at 3.20 mi/h requires about 220 kcal/h. It is interesting to compare these values with the energy consumption required for travel by car. Gasoline yields about 1.30.
A) Find the fuel economy in equivalent miles per gallon for a person walking.
B) Find the fuel economy in equivalent miles per gallon for a person bicycling.
Answer:
a. 451.72 mi/ga
b. 1078.33 mi/ga
Explanation:
The computation is shown below:
a. The fuel economy for a person walking is
Given that
Walking at 3.20 mi/h requires about 220 kcal/h so it is equal to
[tex]= 220 \times 4186[/tex]
= 920920 j/hr
Now
[tex]= \frac{mi}{j}[/tex]
[tex]= \frac{3.2}{920920}[/tex]
So,
[tex]= \frac{mi}{ga}[/tex]
[tex]= \frac{3.2}{920920}\times 1.3 \times 100000000[/tex]
= 451.72 mi/ga
b. Now
Bicycling 12.5 mi/h requires about 360 kcal/h energy so it is equal to
[tex]= 360 \times 4186[/tex]
= 1506960 j/hr
So,
[tex]= \frac{mi}{j}[/tex]
[tex]= \frac{12.5}{1506960}[/tex]
Now
[tex]= \frac{mi}{ga}[/tex]
[tex]= \frac{12.5}{1506960} \times 1.3 \times 100000000[/tex]
= 1078.33 mi/ga
We simply applied the above formula
An air track glider of mass m1 = 0.250 kg moving at 0.900 m/s to the right collides with a glider of mass m2 = 0.500 kg at rest. If m1 rebounds and moves to the left with a speed of 0.300 m/s, what is the speed and direction of m2 after the collision? kinetic energy
Answer:
The speed of m2 is 0.6 m/s and its direction is to the right.
Explanation:
This numerical can be solved easily by applying law of conservation of momentum to it. According to law of conservation of momentum:
Total Momentum Before Collision = Total Momentum After Collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where,
m₁ = Mass of 1st air glider = 0.25 kg
m₂ = Mass of 2nd air glider = 0.5 kg
u₁ = Speed of 1st air glider before collision = 0.9 m/s
u₂ = Speed of 2nd air glider before collision = 0 m/s (at rest)
v₁ = Speed of 1st air glider after collision = - 0.3 m/s (negative sign due to change in direction of velocity)
v₂ = Speed of 2nd air glider after collision = ?
Therefore,
(0.25 kg)(0.9 m/s) + (0.5 kg)(0 m/s) = (0.25 kg)(-0.3 m/s) + (0.5 kg)v₂
0.225 kg.m/s + 0.075 kg.m/s = (0.5 kg)v₂
v₂ = (0.3 kg.m/s)/(0.5 kg)
v₂ = 0.6 m/s
Positive sign indicates that v₂ is directed towards right
Two mirrors are at right angles to one another. A light ray is incident on the first at an angle of 30° with respect to the normal to the surface. What is the angle of reflection from the second surface?
Answer:
reflected angle - secod mirror = 60°
Explanation:
I attached an image with the solution to this problem below.
In the solution the reflection law, incident angle = reflected angle, is used. Furthermore some trigonometric relation is used.
You can notice in the image that the angle of reflection in the second mirror is 60°
For the circuit, suppose C=10µF, R1=1000Ω, R2=3000Ω, R3=4000Ω and ls=1mA. The switch closes at t=0s.1) What is the value of Vc (in volts) just prior to the switch closing? Assume that the switch had been open for a long time. 2) For the circuit above, what is the value of Vc after the switch has been closed for a long time?
3) What is the time constant of the circuit (in seconds)? Enter the answer below without units.
4) What is the value of Vc at t = 2msec (in volts).
Answer:
1.) Vc = 1V
2.) Vc = 2.7V
3.) Time constant = 0.03
4.) V = 2.53V
Explanation:
1.) The value of Vc (in volts) just prior to the switch closing
The starting current = 1mA
With resistance R1 = 1000 ohms
By using ohms law
V = IR
Vc = 1 × 10^-3 × 1000
Vc = 1 volt.
2.) The value of Vc after the switch has been closed for a long time.
R2 and R3 are in parallel to each other. Both will be in series with R1
The equivalent resistance R will be
R = (R2 × R3)/R2R3 + R1
Where
R1 = 1000Ω,
R2 = 3000Ω,
R3 = 4000Ω
R = (4000×3000)/(4000+3000) + 1000
R = 12000000/7000 + 1000
R = 1714.3 + 1000
R = 2714.3 ohms
By using ohms law again
V = IR
Vc = 1 × 10^-3 × 2714.3
Vc = 2.7 volts
3.) The time constant = CR
Time constant = 10 × 10^-6 × 2714.3
Time constant = 0.027
Time constant = 0.03 approximately
4.) The value of Vc at t = 2msec (in volts). Can be calculated by using the formula
V = Vce^-t/CR
Where
Vc = 2.7v
t = 2msec
CR = 0.03
Substitute all the parameters into the formula
V = 2.7 × e^-( 2×10^-3/0.03)
V = 2.7 × e^-(0.0667)
V = 2.7 × 0.935
V = 2.53 volts
A carnot heat engine has an efficiency of 0.800. if it operates between a deep lake with a constant temperature of 280.0 k and a hot reservoir, what is the temperature of the hot reservoir?
A and B Two vectors are in the xy plane. If 4= 3m , |B|= 4m , and |A– B= 2 m find
a) The angle between B and A
b) The unit vector in the direction of (4× B)
Answer:
Explanation:
a)
A = 3m , B = 4m .
(A-B)² = A² + B² - 2ABcosθ where θ is angle between A and B.
4 = 9 + 16 - 2. 3.4 cosθ
cosθ = .875
θ = 29° .
b ) Unit vector in the direction of A X B will be k vector because A and B are in X-Y plane and A X B lies perpendicular to both A and B .
At the equator, the earth's field is essentially horizontal; near the north pole, it is nearly vertical. In between, the angle varies. As you move farther north, the dip angle, the angle of the earth's field below horizontal, steadily increases. Green turtles seem to use this dip angle to determine their latitude. Suppose you are a researcher wanting to test this idea. You have gathered green turtle hatchlings from a beach where the magnetic field strength is 50 mu T and the dip angle is 56 degree. You then put the turtles in a 2.0 m diameter circular tank and monitor the direction in which they swim as you vary the magnetic field in the tank. You change the field by passing a current through a 50-tum horizontal coil wrapped around the tank. This creates a field that adds to that of the earth. In what direction should current pass through the coil, to produce a net field in the center of the tank that has a dip angle of 62 degree ? What current should you pass through the coil, to produce a net field in the center of the tank that has a dip angle of 62 degree ? Express your answer to two significant figures and include the appropriate units.
Answer:
Direction of current = clockwise
Magnitude of current, I = 0.36 A
Explanation:
The magnetic field strength, [tex]B_{E} = 50 \mu T[/tex]
The angle of dip, ∅ = 56°
The net magnetic field in the center of the tank is:
[tex]B_{net} = (B_{E} cos \phi ) (\hat{x} ) + ( B + B_{E} sin \phi)(-\hat{y})\\B_{net} = (50 cos 56 ) (\hat{x} ) + ( B +50 sin 56)(-\hat{y})\\B_{net} = (28 \mu T ) (\hat{x} ) + ( B +41.4 \mu T)(-\hat{y})\\[/tex]
The direction of the net magnetic field is:
[tex]\phi = tan^{-1} \frac{B + 41.4 }{28} \\tan \phi = \frac{B + 41.4 }{28}\\\phi = 62^0\\tan 62 = \frac{B + 41.4 }{28}\\28 tan 62 = B + 41.4\\52.66 = B + 41.4\\B = 11.26 \mu T[/tex]
The magnetic field due to the coil:
[tex]B = \frac{\mu_{0}NI }{2r} \\11.26 * 10^{-6} = \frac{4\pi * 10^{-7} * 50 *I }{2 *1}\\I = \frac{2 * 11.26 * 10^{-6}}{4\pi * 10^{-7} * 50} \\I = 0.36 A[/tex]
The current must be in clockwise direction to produce the field in downward direction
A "swing" ride at a carnival consists of chairs that are swung in a circle by 19.8 m cables attached to a vertical rotating pole, as the drawing shows. Suppose the total mass of a chair and its occupant is 137 kg. (a) Determine the tension in the cable attached to the chair. (b) Find the speed of the chair.
Answer:
a) T = 1342.6 cos θ, b) v = 13.93 √(sin θ tan θ)
Explanation:
We can solve this problem using Newton's second law
Let's fix a reference system with a horizontal axis and the other vertical, therefore the only force to decompose is the tension, in these problems the most common is to measure the angle with respect to the vertical. Let's use trigonometry to find the components of the dispute
cos θ = [tex]T_{y}[/tex] / T
T_{y} = T cos tea
sin θ = Tₓ / T
Tₓ = T sin θ
let's write Newton's second law
axis and vertical
T cos θ - W = 0
T = mg / cos θ
let's calculate
T = 137 9.8 cos θ
T = 1342.6 cos θ
unfortunately there is no drawing or indication of the angle
Axis x Horizontal
T sin θ = m a
acceleration is centripetal
a = v² / R
T sin θ = m v² / R
v² = (g / cos θ) R sin θ
v = √ (gR tan θ)
let's use trigonometry to find radius of gyration
sin θ = R / L
R = L sin θ
v = √ (g L sin θ tan θ)
let's calculate
v = √(9.8 19.8 sin θ tant θ)
v = 13.93 √(sin θ tan θ)
they do not give the angle for which the calculation cannot be finished
A duck flying horizontally due north at 10.7 m/s passes over East Lansing, where the vertical component of the Earth's magnetic field is 4.09×10-5 T (pointing down, towards the Earth). The duck has a positive charge of 6.47×10-8 C. What is the magnitude of the magnetic force acting on the duck?
Answer:
2.83×10⁻¹¹ N.
Explanation:
From the question,
Using
F = qvB....................... Equation 1
Where F = magnetic force acting on the duck, q = charge of the duck, v = velocity of the duck, B = magnetic field of the duck.
Given: q = 6.47×10⁻⁸ C, B = 4.09×10⁻⁵ T, v = 10.7 m/s.
Substitute these values into equation 1
F = 6.47×10⁻⁸×4.09×10⁻⁵×10.7
F = 2.83×10⁻¹¹ N.
Hence the magnetic force acting on the duck is 2.83×10⁻¹¹ N.
Two small plastic spheres each have a mass of 2.0 g and a charge of −50.0 nC. They are placed 2.0 cm apart (center to center). What is the magnitude of the electric force on each sphere? By what factor is the electric force on a sphere larger than its weight?
Answer:
a) F = 0.0561 N
b) F = 2.86*W
Explanation:
a) The magnitude of the electric force between the plastic spheres is given by the following formula:
[tex]F=k\frac{q_1q_2}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
q1 = q2: charge of the plastic spheres = -50.0nC = -50.0*10^-9 C
r: distance between the plastic spheres = 2.0 cm = 0.02 m
You replace the values of the parameters in the equation (1):
[tex]F=(8.98*10^9Nm^2/C^2)\frac{(-50.0*10^{-9}C)^2}{(0.02m)^2}\\\\F=0.0561N[/tex]
The electric force between the spheres is 0.0561 N
b) To calculate the relation between weight and electric force, you first calculate the weight of one of the spheres:
[tex]W=mg[/tex]
m: mass = 2.0g = 2.0*10^-3 kg
g: gravitational acceleration = 9.8 m/s^2
[tex]W=(2.0*10^{-3}kg)(9.8m/s^2)=0.0196N[/tex]
The ratio between W and F is:
[tex]\frac{F}{W}=\frac{0.0561N}{0.0196N}=2.86\\\\F=2.86W[/tex]
The electric force is 2.86 times the weight
(a) The magnitude of the electric force on each sphere is [tex]5.625 \times 10^{-2} \ N[/tex].
(b) The electric force on a sphere is larger than its weight by 2.87.
The given parameters:
mass of each sphere, m = 2.0 gcharge on each sphere, q = -50 nCdistance between the charges, d = 2.0 cmThe magnitude of the electric force on each sphere is calculated as follows;
[tex]F = \frac{kq^2}{r^2} \\\\F = \frac{9\times 10^9 \times (5 0 \times 10^{-9})^2}{(0.02)^2} \\\\F = 5.625 \times 10^{-2} \ N[/tex]
The weight of a sphere is calculated as follows;
[tex]W = mg\\\\W = 0.002 \times 9.8\\\\W = 0.0196 \ N[/tex]
Compare the electric force and the weight of a sphere;
[tex]= \frac{F}{W} = \frac{5.625 \times 10^{-2}}{0.0196} = 2.87[/tex]
Thus, the electric force on a sphere is larger than its weight by 2.87.
Learn more here:https://brainly.com/question/12533085
The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17 C and 10 C at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of 1.4 W/m K, what is the rate of heat loss through the slab
Answer:
Q = - 4312 W = - 4.312 KW
Explanation:
The rate of heat of the concrete slab can be calculated through Fourier's Law of heat conduction. The formula of the Fourier's Law of heat conduction is as follows:
Q = - kA dt/dx
Integrating from one side of the slab to other along the thickness dimension, we get:
Q = - kA(T₂ - T₁)/L
Q = kA(T₁ - T₂)/t
where,
Q = Rate of Heat Loss = ?
k = thermal conductivity = 1.4 W/m.k
A = Surface Area = (11 m)(8 m) = 88 m²
T₁ = Temperature of Bottom Surface = 10°C
T₂ = Temperature of Top Surface = 17° C
t = Thickness of Slab = 0.2 m
Therefore,
Q = (1.4 W/m.k)(88 m²)(10°C - 17°C)/0.2 m
Q = - 4312 W = - 4.312 KW
Here, negative sign shows the loss of heat.
You and a friend frequently play a trombone duet in a jazz band. During such performances it is critical that the two instruments be perfectly tuned. Since you take better care of your trombone, you decide to use your instrument as the standard. When you produce a tone that is known to be 470 Hz and your friend attempts to play the same note, you hear 4 beats every 3.00 seconds. Your ear is good enough to detect that your trombone is at a higher frequency. Determine the frequency of your friend's trombone. (Enter your answer to at least 1 decimal place.)
Answer:
f₂ = 468.67 Hz
Explanation:
A beat is a sudden increase and decrease of sound. The beats are produced through the interference of two sound waves of slightly different frequencies. Now we have the following data:
The higher frequency tone = f₁ = 470 Hz
No. of beats = n = 4 beats
Time period = t = 3 s
The lower frequency note = Frequency of Friend's Trombone = f₂ = ?
Beat Frequency = fb
So, the formula for beats per second or beat frequency is given as:
fb = n/t
fb = 4 beats/ 3 s
fb = 1.33 Hz
Another formula for beat frequency is:
fb = f₁ - f₂
f₂ = f₁ - fb
f₂ = 470 Hz - 1.33 Hz
f₂ = 468.67 Hz
A 60-turn coil has a diameter of 13 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.60 T so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf induced in the coil (in V) if the magnetic field is reduced to zero uniformly in the following times.
(a) 0.80 S 0.5973V
(b) 8.0 s 5.973 Xv
(c) 70 S 6.838- 3 v
Answer:
a) 0.5985 V
b) 0.05985 V
c) 0.00684 V
Explanation:
Given that
Number of turn in the coil, N = 60 turns
Magnetic field of the coil, B = 0.6 T
Diameter of the coil, d = 0.13 m
If area is given as, πd²/4, then
A = π * 0.13² * 1/4
A = 0.0133 m²
The induced emf, ε = -N(dΦ*m) /dt
Note, Φm = BA
Substituting for Φ, we have
ε = -NBA/t.
Now, we substitute for numbers in the equation
ε = -(60 * 0.6 * 0.0133)/0.8
ε = 0.4788/0.8
ε = 0.5985 V
at 8s,
ε = -(60 * 0.6 * 0.0132)/8
ε = 0.4788/8
ε = 0.05985 V
at 70s
ε = -(60 * 0.6 * 0.0132)/70
ε = 0.4788/70
ε = 0.00684 V
A 58.0 kg skier is moving at 6.00 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 3.65 m long. The coefficient of kinetic friction between this patch and her skis is 0.310. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 3.50 m high.
Required:
a. How fast is the skier moving when she gets to the bottom of the hill?
b. How much internal energy was generated in crossing the rough patch?
Answer:
a) v = 3.71m/s
b) U = 616.71 J
Explanation:
a) To find the speed of the skier you take into account that, the work done by the friction surface on the skier is equal to the change in the kinetic energy:
[tex]-W_f=\Delta K=\frac{1}{2}m(v^2-v_o^2)\\\\-F_fd=\frac{1}{2}m(v^2-v_o^2)[/tex]
(the minus sign is due to the work is against the motion of the skier)
m: mass of the skier = 58.0 kg
v: final speed = ?
vo: initial speed = 6.00 m/s
d: distance traveled by the skier in the rough patch = 3.65 m
Ff: friction force = Mgμ
g: gravitational acceleration = 9.8 m/s^2
μ: friction coefficient = 0.310
You solve the equation (1) for v:
[tex]v=\sqrt{\frac{2F_fd}{m}+v_o^2}=\sqrt{\frac{2mg\mu d}{m}+v_o^2}\\\\v=\sqrt{-2g\mu d+v_o^2}[/tex]
Next, you replace the values of all parameters:
[tex]v=\sqrt{-2(9.8m/s^2)(0.310)(3.65m)+(6.00m/s)^2}=3.71\frac{m}{s}[/tex]
The speed after the skier has crossed the roug path is 3.71m/s
b) The work done by the rough patch is the internal energy generated:
[tex]U=W_fd=F_fd=mg\mu d\\\\U=(58.0kg)(9.8m/s^2)(0.310)(3.50m)=616.71\ J[/tex]
The internal energy generated is 616.71J
A 1 900-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.00 m before coming into contact with the top of the beam, and it drives the beam 15.8 cm farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.
Answer:
471392.4 N
Explanation:
From the question,
Just before contact with the beam,
mgh = Fd.................... Equation 1
Where m = mass of the beam, g = acceleration due to gravity, h = height. F = average Force on the beam, d = distance.
make f the subject of the equation
F = mgh/d................ Equation 2
Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m
Constant: g = 9.8 m/s²
Substitute into equation 2
F = 1900(4)(9.8)/0.158
F = 471392.4 N
To increase the energy of an electromagnetic wave, which property should you decrease?
Shift,
Frequency
Speed
Wavelength
The increase in the energy of an electromagnetic wave can be achieved only by decreasing the wavelength. Hence, option (d) is correct.
The given problem is based on the fundamentals of electromagnetic wave and the energy stored in an electromagnetic wave.
The electromagnetic wave stores the energy in the form of radiations also known as the electromagnetic radiations. These radiations can take the several forms such as radio waves, microwaves, X-rays and gamma rays.The mathematical expression for the energy carried out by the electromagnetic waves is given as,
[tex]E = h \times \nu\\\\E = \dfrac{h \times c}{ \lambda}[/tex]
Here,
h is the Planck's constant.
[tex]\nu[/tex] is the frequency of the electromagnetic wave.
c is the speed of light.
[tex]\lambda[/tex] is the wavelength of wave.
Clearly, the energy of electromagnetic waves is directly proportional to the frequency of wave and inversely proportional to wavelength. So, decreasing the wavelength, we can easily increase the energy of electromagnetic wave.
Thus, we can conclude that the increase in the energy of an electromagnetic wave can be achieved only by decreasing the wavelength. Hence, option (d) is correct.
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Which electromagnetic wave transfers the least amount of energy?
Answer:
microwave
Explanation:
When a ball is dropped from a window, how much is the initial velocity in m/s2 ?
Explanation:
The initial velocity is 0 m/s.
The initial acceleration is -9.8 m/s².
Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^{-2CL}T≃e −2CL , where L is the width of the barrier and C is a term that includes the particle energy and barrier height. If the tunneling coefficient is found to be T = 0.050T=0.050 for a given value of LL, for what new value of L\text{'}L’ is the tunneling coefficient T\text{'} = 0.025T’=0.025 ? (All other parameters remain unchanged.) Express L\text{'}L’ in terms of the original LL.
Answer:
L' = 1.231L
Explanation:
The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:
[tex]T \approx e^{-2CL}[/tex]
L: width of the barrier
C: constant that includes particle energy and barrier height
You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.
To find the new value of the L' you can write down both situation for T and T', as in the following:
[tex]0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)[/tex]
Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):
[tex]ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)[/tex]
Next, you divide the equation (3) into (4), and finally, you solve for L':
[tex]\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L[/tex]
hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L
To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
When a star has fused most of its hydrogen and begins to collapse inward, it becomes a
A thin plastic rod of length 2.6 m is rubbed all over with wool, and acquires a charge of 98 nC, distributed uniformly over its surface. Calculate the magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod. Do the calculation two ways, first using the exact formula for a rod of any length, and second using the approximate formula for a long rod.
Answer:
By exact formula
5076.59N/C
And by approximation formula
5218.93N/C
Explanation:
We are given that
Length of rod,L=2.6 m
Charge,q=98nC=[tex]98\times 10^{-9} C[/tex]
[tex]1nC=10^{-9} C[/tex]
a=13 cm=0.13 m
1 m=100 cm
By exact formula
The magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{kq}{a}\times \frac{1}{\sqrt{a^2+\frac{L^2}{4}}}[/tex]
Where k=[tex]9\times 10^9[/tex]
Using the formula
The magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{9\times 10^9\times 98\times 10^{-9}}{0.13}\times \frac{1}{\sqrt{(0.13)^2+\frac{(2.6)^2}{4}}}=5076.59N/C[/tex]
In approximation formula
a<<L
[tex]a^2+(\frac{L}{2})^2=\frac{L^2}{4}[/tex]
Therefore,the magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{kq}{a}\times \frac{1}{\sqrt{\frac{L^2}{4}}}[/tex]
The magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod=[tex]\frac{9\times 10^9\times 98\times 10^{-9}}{0.13}\times \frac{1}{\sqrt{\frac{(2.6)^2}{4}}}=5218.93N/C[/tex]
Which scientist was the first to propose the heliocentric model of the universe
Answer:
Nicolaus Copernicus
Explanation:
Nevertheless, Copernicus began to work on astronomy on his own. Sometime between 1510 and 1514 he wrote an essay that has come to be known as the Commentariolus (MW 75–126) that introduced his new cosmological idea, the heliocentric universe, and he sent copies to various astronomers
An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s^{2} as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?
Answer:
a) [tex]t \approx 9.879\,s[/tex], b) [tex]v = 20.518\,\frac{m}{s}[/tex], c) [tex]t = 16.368\,s[/tex], d) [tex]v = 19.545\,\frac{m}{s}[/tex]
Explanation:
a) Since train is only translating in a straight line and experimenting a constant deceleration throughout the station, whose length is 210 meters. The time required for the nose of the train to reach the end of the station can be found with the help of the following motion formula:
[tex]210\,m = \left(22\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]
The following second-order polynomial needs to be solved:
[tex]-0.075\cdot t^{2} + 22\cdot t - 210 = 0[/tex]
Whose roots are presented herein:
[tex]t_{1}\approx 283.455\,s[/tex] and [tex]t_{2} \approx 9.879\,s[/tex]
Both solutions are physically reasonable, although second roots describes better the braking process of the train.
b) The speed of the nose leaving the station is given by this expression:
[tex]v = 22\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}}\right)\cdot (9.879\,s)[/tex]
[tex]v = 20.518\,\frac{m}{s}[/tex]
c) First, it is required to calculate the time when nose of the train reaches a distance of 130 meters.
[tex]130\,m = \left(22\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]
[tex]-0.075\cdot t^{2} + 22\cdot t - 130 = 0[/tex]
Roots of the second-order polynomial are:
[tex]t_{1} \approx 287.300\,s[/tex] and [tex]t_{2} \approx 6.033\,s[/tex]
Both solutions are physically reasonable, although second roots describes better the braking process of the train. Now, the speed experimented by the train at this instant is:
[tex]v = 22\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}}\right)\cdot (6.033\,s)[/tex]
[tex]v = 21.095\,\frac{m}{s}[/tex]
The distance traveled by the end of the train throughout station is modelled after the following equation:
[tex]210\,m = \left(21.095\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]
[tex]-0.075\cdot t^{2} + 21.095\cdot t - 210 = 0[/tex]
Roots of the second-order polynomial are:
[tex]t_{1} \approx 270.932\,s[/tex] and [tex]t_{2} \approx 10.335\,s[/tex]
Both solutions are physically reasonable, although second roots describes better the braking process of the train. The instant when the end of the train leaves the station is:
[tex]t = 6.033\,s + 10.335\,s[/tex]
[tex]t = 16.368\,s[/tex]
d) The velocity experimented by the end of the train is:
[tex]v = 21.095\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}} \right)\cdot (10.335\,s)[/tex]
[tex]v = 19.545\,\frac{m}{s}[/tex]
The negative sign indicates that the train comes to a stop before entering the station. Therefore, the nose of the train is not in the station at all. The negative sign indicates that the train is moving in the opposite direction of its initial velocity. The end of the train leaves the station after approximately 14.98 seconds. The velocity of the end of the train as it leaves the station is approximately 19.753 m/s.
(a) Using the equation :
v² = u² + 2as
0 = (22.0)² + 2 × (-0.150 ) × s
s = -(22.0 )^2 / (2 × -0.150)
s = 325.3 m
The negative sign indicates that the train comes to a stop before entering the station. Therefore, the nose of the train is not in the station at all.
(b) The velocity of the train when the nose leaves the station is given by:
v = u + at
v = 22.0 + (-0.150 ) × 210
v ≈ 22.0 - 31.5
v ≈ -9.5 m/s
The negative sign indicates that the train is moving in the opposite direction of its initial velocity.
(c)
s = ut + (1/2)at²
210= 22.0 × t + (1/2) × (-0.150) × t²
0.075 t² - 22.0 t + 210 = 0
we find two solutions for t: t ≈ 14.98 s and t ≈ 3.01 s.
The end of the train leaves the station after approximately 14.98 seconds.
(d) The velocity of the end of the train as it leaves the station is given by:
v = u + at
v = 22.0 + (-0.150) × 14.98
v = 22.0 - 2.247
v = 19.753 m/s
So, the velocity of the end of the train as it leaves the station is approximately 19.753 m/s.
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