Answer:
Explanation:
I catch the ruler 5.7 cm from lower end that means my reaction time is equal to time of fall of ruler as free fall under gravity .
h = 1/2 gt²
t = [tex]\sqrt{\frac{2h}{g} }[/tex]
= [tex]\sqrt{\frac{2\times 5.7}{9.8} }[/tex]
= 1.078 s
= 1.1 s .
The drawing shows four sheets of polarizing material, each with its transmission axis oriented differently. Light that is polarized in the vertical direction is incident from the left and has an average intensity of 32 W/m2. Determine the average intensity of the light that emerges on the right in the drawing (a) when sheet A alone is removed, (b) when sheet B alone is removed, (c) when sheet C alone is removed, and (d) when sheet D alone is removed.
Answer:
Explanation:
When sheet A is removed
[tex]I_B=32\cos^230=24W/m^2\\\\I_C=24 \cos^260=6W/m^2\\\\I_D=6\cos^230=4.5W/m^2[/tex]
When sheet B is removed
[tex]I_A=32\cos^20=32W/m^2\\\\I_C=24 \cos^290=0W/m^2\\\\I_D=0\cos^230=0W/m^2[/tex]
When sheet C is removed
[tex]I_A=32\cos^20=32W/m^2\\\\I_D=32 \cos^230=24W/m^2\\\\I_B=24\cos^290=0W/m^2[/tex]
When sheet D is removed
[tex]I_A=32\cos^20=32W/m^2\\\\I_B=32\cos^230=24W/m^2\\\\I_C=24\cos^260=6W/m^2[/tex]
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
The motion of a free falling body is an example of __________ motion
Answer:
accelerated
Explanation:
The motion of the body where the acceleration is constant is known as uniformly accelerated motion. The value of the acceleration does not change with the function of time.
Newton's second law of motion
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 girl throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball
Answer:
F = 800N
the magnitude of the average force exerted on the wall by the ball is 800N
Explanation:
Applying the impulse-momentum equation;
Impulse = change in momentum
Ft = m∆v
F = (m∆v)/t
Where;
F = force
t = time
m = mass
∆v = v2 - v1 = change in velocity
Given;
m = 0.80 kg
t = 0.050 s
The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.
v2 = 25 m/s
v1 = -25 m/s
∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s
Substituting the values;
F = (m∆v)/t
F = (0.80×50)/0.05
F = 800N
the magnitude of the average force exerted on the wall by the ball is 800N
In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0580 s, during which time it experiences an acceleration of 376 m/s2. The ball is launched at an angle of 59.9° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.
Answer:
V₀ₓ = 10.94 m/s
V₀y = 18.87 m/s
Explanation:
To find the launch velocity, we use 1st equation of motion.
Vf = Vi + at
where,
Vf = Final Velocity of Ball = Launch Speed = V₀ = ?
Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)
a = acceleration = 376 m/s²
t = time = 0.058 s
Therefore,
V₀ = 0 m/s + (376 m/s²)(0.058 s)
V₀ = 21.81 m/s
Now, for x-component:
V₀ₓ = V₀ Cos θ
where,
V₀ₓ = x-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀ₓ = (21.81 m/s)(Cos 59.9°)
V₀ₓ = 10.94 m/s
for y-component:
V₀ₓ = V₀ Sin θ
where,
V₀y = y-component of launch velocity = ?
θ = Angle with horizontal = 59.9⁰
V₀y = (21.81 m/s)(Sin 59.9°)
V₀y = 18.87 m/s
If a 2,000-kg car hits a tree with 500 N of force over a time of 0.5 seconds,
what is the magnitude of its impulse?
O A. 14,700 kg-m/s
B. 2,500 kg-m/s
C. 250 kg-m/s
D. 10,000 kg-m/s
Answer:
c 250 kg-m/s
Explanation:
happy to help!!
Answer: 250 kg
Explanation:
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
A 1,150-kg car moving east at 80 km/h collides head-on with a 1,900-kg car moving west at 40 km/h , and the two cars stick together. It’s initial speed is 5km/h. How much KE is lost in the collision?
Answer:
Ask a pro,bro
Explanation:
I just ryhmed
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]
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.
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 cylindrical capacitor is made of two concentric cylinders. The inner cylinder has radius r1 = 4 mm, and the outer one a radius r2= 8 mm. The common length of the cylinders is L = 150 m. What is the potential energy stored in this capacitor when a potential difference V = 4 V is applied between the inner and outer cylinder?
Answer:
E = 9.62*10^-8 J
Explanation:
The energy stored in a capacitor is given by the following formula:
[tex]E=\frac{1}{2}CV^2[/tex] (1)
E: energy stored
C: capacitance
V: potential difference of the capacitor = 4 V
The capacitance for a concentric cylindrical capacitor is:
[tex]C=\frac{2\pi \epsilon_o L}{ln(\frac{r_2}{r_1})}[/tex] (2)
L: length of the capacitor = 150m
r2: radius of the outer cylinder = 8mm = 8*10^-3m
r1: radius of the inner cylinder = 4mm = 4*10^-3m
εo: dielectric permittivity of vacuum = 8.85*10^-12C^2/Nm^2
You replace the expression (2) into the equation (1) and replace the values of all parameters:
[tex]E=\frac{1}{2}(\frac{2\pi \epsilon_o L}{ln(\frac{r_2}{r_1})})V^2\\\\E=\frac{\pi \epsilon_o L}{ln(\frac{r_2}{r_1})}V^2\\\\E=\frac{\pi (8.85*10^{-12}C^2/Nm^2)(150m)}{ln(\frac{8*10^{-3}m}{4*10^{-3}m})}(4V)^2\\\\E=9.62*10^{-8}J[/tex]
The energy stored in the cylindrical capacitor is 9.62*10-8 J
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
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]
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:
Imagine that the ball on the left is given a nonzero initial velocity in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed vx must the ball on the left start with so that it hits the ground at the same position as the ball on the right? Express your answer in meters per second to two significant figures.
Answer:
vₓ = xg/2y
Explanation:
In this question, let us find the time it takes for the ball on the right that has zero initial velocity to reach the ground.
By newton equation of motion we know that
y = v₀ t - ½ g t²
t = 2y / g
This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance
vₓ = x/t
vₓ = xg/2y
vₓ = xg/2y
Where we assume that x and y are known.
Answer:
The answer is 3.0
Explanation:
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]
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.
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
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
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
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
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.
In which of the following is negative work performed by an individual?
A) A mover picks up a packing box up off the floor.
B) A person leans against a wall.
C) A shopper lowers a can of beans from a store shelf into the grocery cart.
Answer:
c
Explanation:
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
A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of 5.90 m: (a) the initially stationary spelunker is accelerated to a speed of 2.30 m/s; (b) he is then lifted at the constant speed of 2.30 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 69.0 kg rescue by the force lifting him during each stage
Answer:
W₃ = 3310.49 J , W3 = 3310.49 J
Explanation:
We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections
We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics
v2 = v₀² + 2 a₁ y
as they rest part of the rest the ricial speed is zero
v² = 2 a₁ y
a₁ = v² / 2y
a₁ = 2.3² / (2 5.90)
a₁ = 0.448 m / s²
with this acceleration we can calculate the applied force, using Newton's second law
F -W = m a₁
F = m a₁ + m g
F = m (a₁ + g)
F = 69 (0.448 + 9.8)
F = 707.1 N
Work is defined by
W₁ = F.y = F and cos tea
As the force lifts the man, this and the displacement are parallel, therefore the angle is zero
W₁ = 707.1 5.9
W₁ = 4171.89 J W3 = 3310.49 J
Let's calculate for the second part
the speed is constant, therefore they relate it to zero
F - W = 0
F = W
F = m g
F = 60 9.8
F = 588 A
the job is
W² = 588 5.9
W2 = 3469.2 J
finally the third part
in this case the initial speed is 2.3 m / s and the final speed is zero
v² = v₀² + 2 a₂ y
0 = vo2₀² + 2 a₂ y
a₂ = -v₀² / 2 y
a₂ = - 2.3²/2 5.9
a2 = - 0.448 m / s²
we calculate the force
F - W = m a₂
F = m (g + a₂)
F = 60 (9.8 - 0.448)
F = 561.1 N
we calculate the work
W3 = F and
W3 = 561.1 5.9
W3 = 3310.49 J
total work
W_total = W1 + W2 + W3
W_total = 4171.89 +3469.2 + 3310.49
w_total = 10951.58 J
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.
There are seven ___ included in the periodic table.
we have seven groups in the periodic table
Two wires carry current I1 = 47 A and I2 = 29 A in the opposite directions parallel to the x-axis at y1 = 9 cm and y2 = 15 cm. Where on the y-axis (in cm) is the magnetic field zero?
Answer:
Approximately [tex]25\; \rm cm[/tex], assuming that the permeability between the two wires is constant, and that the two wires are of infinite lengths.
Explanation:
Note that [tex]y_1 < y_2[/tex], which means that the first wire (with current [tex]I_1[/tex]) is underneath the second wire (with current [tex]I_2[/tex].) Let [tex]y[/tex] denote the [tex]y[/tex]-coordinate (in [tex]\rm cm[/tex]) of a point of interest. The two wires partition this region into three parts:
The region under the first wire, where [tex]y < y_1[/tex];The region between the first and the second wire, where [tex]y_1 < y < y_2[/tex]; andThe region above the second wire, where [tex]y > y_2[/tex].Apply the right-hand rule to find the direction of the magnetic field in each of the three regions. Assume that [tex]I_1[/tex] points to the left while [tex]I_2[/tex] points to the right. Let [tex]B_1[/tex] and [tex]B_2[/tex] denote the magnetic field due to [tex]I_1[/tex] and [tex]I_2[/tex], respectively.
In the region below the first wire, [tex]B_1[/tex] points out of the [tex]x y[/tex]-plane while [tex]B_2[/tex] points into the [tex]x y[/tex]-plane.In the region between the two wires, both [tex]B_1[/tex] and [tex]B_2[/tex] point into the [tex]x y[/tex]-plane.In the region above the second wire, [tex]B_1[/tex] points into the [tex]x y[/tex]-plane while [tex]B_2[/tex] points out of the [tex]x y[/tex]-plane.The (net) magnetic field on this plane would be zero only in regions where [tex]B_1[/tex] and[tex]B_2[/tex] points in opposite directions. That rules out the region between the two wires.
At a distance of [tex]R[/tex] away from a wire with current [tex]I[/tex] and infinite length, the formula for the magnitude of the magnetic field [tex]B[/tex] due to that wire is:
[tex]\displaystyle B = \frac{\mu\, I}{2\pi\, R}[/tex],
where [tex]\mu[/tex] is the permeability of the space between the wire and the point of interest (should be constant.) The exact value of [tex]\mu[/tex] does not affect the answer to this question, as long as it is constant throughout this region.
Note that the value of [tex]R[/tex] in this formula is supposed to be positive. Let [tex]R_1[/tex] and [tex]R_2[/tex] denote the distance between the point of interest and the two wires, respectively.
In the region under the first wire, [tex]R_1 = y_1 - y = 9 - y[/tex], while [tex]R_2 = y_2 - y = 15 - y[/tex]. In the region above the second wire, [tex]R_1 = y - 9[/tex], while [tex]R_2 = y - 15[/tex].Make sure that given the corresponding range of [tex]y[/tex], these distances are all positive.
Strength of the magnetic field due to the first wire: [tex]\displaystyle B_1 = \frac{\mu\, I_1}{4\pi\, R_1}[/tex].Strength of the magnetic field due to the second wire: [tex]\displaystyle B_1 = \frac{\mu\, I_2}{4\pi\, R_2}[/tex].For the net magnetic field to be zero at a certain [tex]y[/tex]-value, the strength of the two magnetic fields at that point should match. That is:
[tex]B_1 = B_2[/tex].
[tex]\displaystyle \frac{\mu\, I_1}{4\pi\, R_1} = \frac{\mu\, I_2}{4\pi\, R_2}[/tex].
Simplify this equation:
[tex]\displaystyle \frac{I_1}{R_1} = \frac{I_2}{R_2}[/tex].
In the region under the first wire (where [tex]y < 9[/tex],) this equation becomes [tex]\displaystyle \frac{47}{9 - y} = \frac{29}{15 - y}[/tex].In the region above the second wire (where [tex]y > 15[/tex],) this equation becomes [tex]\displaystyle \frac{47}{y - 9} = \frac{29}{y - 15}[/tex].These two equations give the same result: [tex]y = 25[/tex]. However, based on the respective assumptions on the value of [tex]y[/tex], this value corresponds to the region above the second wire.