Given:
initial velocity, u = 200 m/sFinal velocity, v = 300 m/s Time taken, t = 20 secTo be calculated:
Calculate the acceleration of given object ?
Formula used:
Acceleration = v - u / t
Solution:
We know that,
Acceleration = v - u / t
☆ Substituting the values in the above formula,we get
Acceleration ⇒ 300 - 200 / 20
⇒ 100/20
⇒ 5 m/s²
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
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?
How does the engine get the spacecraft to space?
Answer:
An electric power source is used to ionize fuel into plasma. Electric fields heat and accelerate the plasma while the magnetic fields direct the plasma in the proper direction as it is ejected from the engine, creating thrust for the spacecraft.
Explanation:
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.
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
Calcular el módulo del vector resultante de dos vectores fuerza de 9 [N] y 12 [N] concurrentes en un punto o, cuyas direcciones forman un ángulo de a) 30˚ b) 45˚ y c) 90˚
Answer:
a) 20.29N
b) 19.43N
c) 15N
Explanation:
To find the magnitude of the resultant vectors you first calculate the components of the vector for the angle in between them, next, you sum the x and y component, and finally, you calculate the magnitude.
In all these calculations you can asume that one of the vectors coincides with the x-axis.
a)
[tex]F_R=(9cos(30\°)+12)\hat{i}+(9sin(30\°))\hat{j}\\\\F_R=(19.79N)\hat{i}+(4.5N)\hat{j}\\\\|F_R|=\sqrt{(19.79N)^2+(4.5N)^2}=20.29N[/tex]
b)
[tex]F_R=(9cos(45\°)+12)\hat{i}+(9sin(45\°))\hat{j}\\\\F_R=(18.36N)\hat{i}+(6.36N)\hat{j}\\\\|F_R|=\sqrt{(18.36N)^2+(6.36N)^2}=19.43N[/tex]
c)
[tex]F_R=(9cos(90\°)+12)\hat{i}+(9sin(90\°))\hat{j}\\\\F_R=(12N)\hat{i}+(9N)\hat{j}\\\\|F_R|=\sqrt{(12N)^2+(9N)^2}=15N[/tex]
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.
A wire loop is suspended from a string that is attached to point P in the drawing. When released, the loop swings downward, from left to right, through a uniform magnetic field, with the plane of the loop remaining perpendicular to the plane of the paper at all times. Determine the direction of the current induced in the loop as it swings past the locations labeled (a) I and (b) II. Specify the direction of the current in terms of the points x, y, and z on the loop (e.g., x→y→z or z→y→x). The points x, y, and z lie behind the plane of the paper. What is the direction of the induced current at the locations (c) I and (d) II when the loop swings back, from right to left?
Complete Question
The complete question iws shown on the first uploaded image
Answer:
a
[tex]y \to z \to x[/tex]
b
[tex]x \to z \to y[/tex]
Explanation:
Now looking at the diagram let take that the magnetic field is moving in the x-axis
Now the magnetic force is mathematically represented as
[tex]F = I L[/tex] x B
Note (The x is showing cross product )
Note the force(y-axis) is perpendicular to the field direction (x-axis)
Now when the loop is swinging forward
The motion of the loop is from y to z to to x to y
Now since the force is perpendicular to the motion(velocity) of the loop
Hence the force would be from z to y and back to z
and from lenze law the induce current opposes the force so the direction will be from y to z to x
Now when the loop is swinging backward
The motion of the induced current will now be x to z to y
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.
Learn more about the electromagnetic wave here:
https://brainly.com/question/3101711
There are seven ___ included in the periodic table.
we have seven groups in the periodic table
A coin is placed 17.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 26.0 rpm (revolutions per minute) is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?
Answer: The coefficient of static friction between the coin and the turntable is 0.13.
Explanation:
As we know that,
Centripetal force = static frictional force
[tex]\frac{mv^{2}}{r} = F_{s}[/tex]
or, [tex]\frac{mv^{2}}{r} = \mu_{s} \times m \times g[/tex]
v = [tex]\sqrt{\mu_{s} \times r \times g}[/tex]
or, [tex]\mu_{s} = \frac{v^{2}}{rg}[/tex] ......... (1)
Here, it is given that
r = 17 cm, [tex]\omega[/tex] = 26 rpm,
and v = [tex]r \omega[/tex] ..........(2)
Putting equation (2) in equation (1) we get the following.
[tex]\mu_{s} = \frac{r^{2}\omega^{2}}{rg}[/tex]
= [tex]\frac{17 \times 10^{-2} \times (26 \times [\frac{2 \times \pi}{60}]^{2})}{9.8}[/tex]
= 0.128
= 0.13 (approx)
Thus, we can conclude that the coefficient of static friction between the coin and the turntable is 0.13.
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 charge q1 of -5.00 × 10‐⁹ C and a charge q2 of -2.00 × 10‐⁹ C are separated by a distance of 40.0 cm. Find the equilibrium position for a third charge of +15.0 × 10‐⁹ C.
P.S. 10-⁹ is 10^-9
Answer:
Explanation:
The equilibrium position will be in between q₁ and q₂ .
Let this position of third charge be x distance from q₁
q₁ will pull the third charge towards it with force F₁
F₁ =9 x 10⁹x 5 x 10⁻⁹x15 x 10⁻⁹ / x²
= 675 x 10⁻⁹ / x²
q₂ will pull the third charge towards it with force F₂
F₂ =9 x 10⁹x 2 x 10⁻⁹x15 x 10⁻⁹ /( .40-x )²
= 270 x 10⁻⁹ / ( .40-x )²
For equilibrium
675 x 10⁻⁹ / x² = 270 x 10⁻⁹ / ( .40-x )²
5 / x² = 2 / ( .40-x )²
( .40-x )² / x² = 2/5 = .4
.4 - x / x = .632
.4 - x = .632x
.4 = 1.632 x
x = .245 .
24.5 cm
so third charge must be placed at 24.5 cm away from q₁ charge.
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]
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°
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
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]
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
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
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
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
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
Mass Center Determine the coordinates (x, y) of the center of mass of the area in blue in the figure below. Answers: x=(3)/(8)a and y=(2)/(5)h
Explanation:
The x and y coordinates of the center of mass are:
xcm = ∫ x dm / m = ∫ x ρ dA / ∫ ρ dA
ycm = ∫ y dm / m = ∫ y ρ dA / ∫ ρ dA
Assuming uniform density, the center of mass is also the center of area.
xcm = ∫ x dA / ∫ dA = ∫ x y dx / A
ycm = ∫ y dA / ∫ dA = ∫ ½ y² dx / A
First, let's find the area:
A = ∫ y dx
A = ∫₀ᵃ (-h/a² x² + h) dx
A = -⅓ h/a² x³ + hx |₀ᵃ
A = -⅓ h/a² (a)³ + h(a)
A = ⅔ ha
Now, let's find the x coordinate of the center of mass:
xcm = ∫ x y dx / A
xcm = ∫₀ᵃ x (-h/a² x² + h) dx / (⅔ ha)
xcm = ∫₀ᵃ (-h/a² x³ + hx) dx / (⅔ ha)
xcm = (-¼ h/a² x⁴ + ½ hx²) |₀ᵃ / (⅔ ha)
xcm = (-¼ h/a² (a)⁴ + ½ h(a)²) / (⅔ ha)
xcm = (¼ ha²) / (⅔ ha)
xcm = ⅜ a
Next, we find the y coordinate of the center of mass:
ycm = ∫ y² dx / A
ycm = ∫₀ᵃ ½ (-h/a² x² + h)² dx / (⅔ ha)
ycm = ∫₀ᵃ ½ (h²/a⁴ x⁴ − 2h²/a² x² + h²) dx / (⅔ ha)
ycm = ½ (⅕ h²/a⁴ x⁵ − ⅔ h²/a² x³ + h² x) |₀ᵃ / (⅔ ha)
ycm = ½ (⅕ h²/a⁴ (a)⁵ − ⅔ h²/a² (a)³ + h² (a)) / (⅔ ha)
ycm = ½ (⁸/₁₅ h²a) / (⅔ ha)
ycm = ⅖ h
A lens of focal length 12cm forms an
three times the size of the
to the object. The distance between the object and the image is what
b) 16 cm
Magnification, m = v/u
3 = v/u
⇒ v = 3u
Lens formula : 1/v – 1/u = 1/f
1/3u = 1/u = 1/12
-2/3u = 1/12
⇒ u = -8 cm
V = 3 × (-8) = -24
Distance between object and image = u – v = -8 – (-24) = -8 + 24 = 16 cm
6. A capacitor of charge 3 x 10 coulomb has a potential of 50volts. What is the capacitance of the capacitor?
Answer:
Explanation:
Sry
Answer:
C = Q/V
where C is capacitance, Q is charge and V is voltage
C = (3×10)/50
C = 30/50
= 0.6F where F is in Farads
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
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
Calculate the Reynold's number using a viscosity of air as 1.81E-05 kilograms/(meters-seconds), the density of air (see above), the diameter as 0.15 m, and, from the data, 0.89 m/s.
Answer:
8924.6
Explanation:
We are given that
Viscosity of air,[tex]\eta=1.81\times 10^{-5}kg/m-s[/tex]
Density of air,[tex]\rho=1.21kg/m^3[/tex]
Diameter,d=0.15 m
v=0.89m/s
We have to find the Reynold's number.
Reynold's number,R=[tex]\frac{\rho vd}{\eta}[/tex]
Substitute the values then we get
[tex]R=\frac{1.21\times 0.89\times 0.15}{1.81\times 10^{-5}}[/tex]
R=[tex]8924.6[/tex]
Hence, the value of Reynold's number=8924.6
Which electromagnetic wave transfers the least amount of energy?
Answer:
microwave
Explanation:
Upon impact, bicycle helmets compress, thus lowering the potentially dangerous acceleration experienced by the head. A new kind of helmet uses an airbag that deploys from a pouch worn around the rider's neck. In tests, a headform wearing the inflated airbag is dropped onto a rigid platform; the speed just before impact is 6.0 m/s. Upon impact, the bag compresses its full 12.0 cm thickness, slowing the headform to rest.What is the acceleration, in g's, experienced by the headform? (An acceleration greater than 60g is considered especially dangerous.)
Answer:
This is approximately 16 g's.
Explanation:
For the person’s head to stop falling, the rigid platform must exert a force that is equal to the sum of weight and force that caused the velocity to decrease from 6 m/s to 0 m/s.
Weight = m * -9.8
Let’s use the following equation to determine the acceleration.
vf^2 = vi^2 + 2 * a * d
0 = 36 + 2 * a * 0.12
a = -36 ÷ 0.24 = -150 m/s^2
The acceleration is negative, because it caused the velocity to decrease.
Total acceleration = -159.8 m/s^2
To determine the number of g, divide this by -9.8.
N g’s = -159.8 ÷- 9.8
This is approximately 16 g's.