Answer:
Explanation:
From the information given:
The motional emf can be computed by using the formula:
[tex]E = L^{\to}*(V^\to*\beta^{\to})[/tex]
[tex]E = L^{\to}*((x+y+z)*\beta^{\to})[/tex]
[tex]E = 0.50*((18\hat i+24 \hat j +72 \hat k )*0.0800)[/tex]
[tex]E = 0.50*((18*0.800)\hat k +0j+(72*0.080) \hat -i ))[/tex]
[tex]E = 0.50*((18*0.800)[/tex]
E = 0.72 volts
According to the question, suppose the wire segment was parallel, there will no be any emf induced since the magnetic field is present along the y-axis.
As such, for any motional emf should be induced, the magnetic field, length, and velocity are required to be perpendicular to one another .
Then the motional emf will be:
[tex]E = 0.50 \hat j *((18*0.800)\hat k -(72*0.080) \hat i ))[/tex]
E = 0 (zero)
Many collisions, like the collision of a bat with a baseball, appear to be instantaneous. Most people also would not imagine the bat and ball as bending or being compressed during the collision. Consider the following possibilities: The collision is instantaneous. The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact. The collision takes a finite amount of time, during which the ball and bat are bending or being compressed. How can two of these be ruled out based on energy or momentum considerations?
The collision is instantaneous.
The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact.
The collision takes a finite amount of time, during which the ball and bat are bending or being compressed.
How can two of these be ruled out based on energy or momentum considerations?
Answer:
The collision takes a finite amount of time, during which the ball and bat are bending or being compressed
Explanation:
These two conditions can be ruled out on the fact that :The collision takes a finite amount of time, during which the ball and bat are bending or being compressed
The rule of energy is been broken here because during the collision of objects energy and momentum is conserved. i.e. the change in shape of the ball when hit by the bat should not be noticed because the compression and returning to normal shape happens instantaneously
What has a wind speed of 240 kph or greater?
Answer:
SUPER TYPHOON (STY), a tropical cyclone with maximum wind speed exceeding 220 kph or more than 120 knots.
3
Select the correct answer.
What is a substance?
Answer:
physical material from which something is made or which has discrete existence
Explanation:
Why is the melting of ice a physical change?
A. It changes the chemical composition of water.
B. It does not change the chemical composition of water.
C. It creates new chemical bonds.
D. It forms new products.
E. It is an irreversible change that forms new products.
It does not change the chemical composition of water.
The force an ideal spring exerts on an object is given by , where measures the displacement of the object from its equilibrium position. If , how much work is done by this force as the object moves from to
Answer:
The correct answer is "1.2 J".
Explanation:
Seems that the given question is incomplete. Find the attachment of the complete query.
According to the question,
x₁ = -0.20 mx₂ = 0 mk = 60 N/mNow,
⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]
⇒ [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]
⇒ [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]
By putting the values, we get
⇒ [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]
⇒ [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]
⇒ [tex]=1.2 \ J[/tex]
An ice skater with a mass of 50 kg is gliding acrossthe ice at a speed of 8 m/s when herfriend comes up from behind and gives her a push,causing her speed to increase to 12m/s. How much work did the friend do on the skater
Answer:
[tex]W=2KJ[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=50kg[/tex]
Initial Velocity [tex]v_1=8m/s[/tex]
Final Velocity [tex]v_2=12m/s[/tex]
Generally the equation for Work-done is mathematically given by
W=\triangle K.E
Therefore
[tex]W=0.5M(v_2^2-v_1^2)[/tex]
[tex]W=0.5*50(12^2-8^2)[/tex]
[tex]W=2KJ[/tex]
An aircraft has a glide ratio of 12 to 1. (Glide ratio means that the plane drops 1 m in each 12 m it travels horizontally.) A building 45 m high lies directly in the glide path to the runway. If the aircraft dears the building by 12 m, how far from the building does the aircraft touch down on the runway
The aircraft is 12 meters higher than the building so it is at 45 + 12 = 57 meters high.
For every 12 meters it travels it drops 1 m.
Divide the height by 12 to find the distance it travels:
57 / 12 = 4.75
It touches down 4.75 meters from the building.
The building is 684 meters away from the aircraft touching down on the runway.
What are trigonometric functions?A right-angled triangle's side ratios are the easiest way to express a function of an arc or angle, such as the sine, cosine, tangent, cotangent, secant, or cosecant. These functions are known as trigonometric functions.
As given in the problem an aircraft has a glide ratio of 12 to 1. (Glide ratio means that the plane drops 1 m in each 12 m it travels horizontally.) A building 45 m high lies directly in the glide path to the runway. If the aircraft clears the building by 12 m,
the total height of the aircraft when it clears the building = 45 +12
the total height of the aircraft when it clears the building is 57 meters
It is given that the Glide ratio is 12:1,
The distance of the building from touch down on the runway = 12 ×57
The distance of the building from the touch-down on the runway is 684 meters.
Thus, the building is 684 meters away from the aircraft touching down on the runway.
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At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 mls2, and friction is negligible. The coupling bars have negligible mass. By how much would the tension in each of the coupling bars A, B, and C change if 39 kg of luggage were removed from car 2 and placed in (a) car I and (b) car 3
Answer:
a) ΔT₁ = -4.68 N, ΔT₂ = 4.68 N, b) ΔT₂ = 4.68 N, ΔT₁ = 4.68 N
Explanation:
In this exercise we will use Newton's second law.
∑F = m a
Let's start with the set of three cars
F_total = M a
F_total = M 0.12
where the total mass is the sum of the mass of each charge
M = m₁ + m₂ + m₃
This is the force with which the three cars are pulled.
Now let's write this law for each vehicle
car 1
F_total - T₁ = m₁ a
T₁ = F_total - m₁ a
car 2
T₁ - T₂ = m₂ a
T₂ = T₁ - m₂ a
car 3
T₂ = m₃ a
note that tensions are forces of action and reaction
a) They tell us that 39 kg is removed from car 2 and placed on car 1
m₂’= m₂ - 39
m₁'= m₁ + 39
m₃ ’= m₃
they ask how much each tension varies, let's rewrite Newton's equations
The total force does not change since the mass of the set is the same F_total ’= F_total
car 1
F_total ’- T₁ ’= m₁’ a
T₁ ’= F_total - m₁’ a
T₁ ’= (F_total - m₁ a) - 39 a
T₁ '= T₁ - 39 0.12
ΔT₁ = -4.68 N
car 2
T₁’- T₂ ’= m₂’ a
T₂ ’= T₁’- m₂’ a
T₂ '= (T₁'- m₂ a) + 39 a
T₂ '= T₂ + 39 0.12
ΔT₂ = 4.68 N
b) in this case the masses remain
m₁ '= m₁
m₂ ’= m₂ - 39
m₃ ’= m₃ + 39
we write Newton's equations
car 3
T₂ '= m₃' a
T₂ ’= (m₃ + 39) a
T₂ '= m₃ a + 39 a
T₂ '= T₂ + 39 0.12
ΔT₂ = 4.68 N
car 1
F_total - T₁ ’= m₁’ a
T₁ ’= F_total - m₁ a
car 2
T₁' -T₂ '= m₂' a
T₁ ’= T₂’- m₂’ a
T₁ '= (T₂'- m₂ a) + 39 a
T₁ '= T₁ + 39 0.12
ΔT₁ = 4.68 N
The tension in each of the coupling bars A, B, and C of the luggage carrier changes as,
When luggage were removed from car 2 and placed in car 1, the tension is A and C does not change and the tension in B is decreased by 4.68 N.When luggage were removed from car 2 and placed in car 3, the tension is A and B does not change and the tension in C is increased by 4.68 N.What is tension force?Tension is the pulling force carried by the flexible mediums like ropes, cables and string.
Tension in a body due to the weight of the hanging body is the net force acting on the body.
At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 m/s², and friction is negligible.
The acceleration is the same, Tension due to the horizontal component of the forces for car 1, 2 and 3 can be given as,
[tex]\sum F_{1h}=T_A-T_B=m_1a\\\sum F_{2h}=T_B-T_C=m_2a\\\sum F_{3h}=T_C=m_3a[/tex]
On solving the above 3 equation, we get the values of tension in each bar as,
[tex]T_A=(m_1+m_2+m_3)a\\T_B=(m_3+m_2)a\\T_C=m_3a[/tex]
Case 1- When 39 kg of luggage were removed from car 2 and placed in car IThe tension is A and C does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,
[tex]\Delta T_B=39\times0.12\\\Delta T_B=4.68\rm \;N[/tex]
Case 2- When 39 kg of luggage were removed from car 2 and placed in car IIIThe tension is A and B does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,
[tex]\Delta T_C=39\times0.12\\\Delta T_C=4.68\rm \;N[/tex]
Hence, the tension in each of the coupling bars A, B, and C of the luggage carrier changes as,
When luggage were removed from car 2 and placed in car 1, the tension is A and C does not change and the tension in B is decreased by 4.68 N.When luggage were removed from car 2 and placed in car 3, the tension is A and B does not change and the tension in C is increased by 4.68 N.
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The bulk modulus of water is B = 2.2 x 109 N/m2. What change in pressure ΔP (in atmospheres) is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C?
Answer:
A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.
Explanation:
The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:
[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)
Where:
[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.
[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.
[tex]\Delta P[/tex] - Pressure change, in pascals.
If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:
[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]
[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]
[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]
A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.
Stationary waves are
A) transverse waves
B) longitudinal waves
C) mechanical waves
Answer:
stationary waves are transverse waves
Electrical resistance is a measure of resistance to the flow of _?____
Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.
Hope this helps!!!!
Answer:
electric current
Explanation:
The answer is electric current
A metre rule is used to measure the length of a piece of string in a certain experiment. It is found to be 20 cm long to the nearest millimeter. How should thisresult be recorded in a table of results? a. 0.2000m b. 0.200m c. 0.20m d. 0.2m
Answer:
C
Explanation:
20 cm = 0.2m
since uncertainty is 0.1 cm (0.001 m), should be recorded to same number of decimal place as uncertainty
therefore it's 0.200m
What's the speed of a sound wave through water at 25 Celsius?
A. 1,000 m/s
B. 1,500 m/s
C. 1,250 m/s
D. 750 m/s
Answer:
B) 1500m/s
Explanation:
Ans is 1500m/s
A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for 0.1 , what force did the wall impart onto the ball?
Answer:
5N
Explanation:
We have a simple problem of momentum here.
ΔMomentum= mΔv= FΔt
Solve for F
mΔv/Δt=F
Plug in givens
1*(2-1.5)/0.1=F
F=5N
The amount of force that the wall imparts on the ball is 5.0N
According to Newton's second law, the formula for calculating the force applied is expressed as:
[tex]F=ma[/tex]
m is the mass of the object
a is the acceleration of the object
Since acceleration is the change in velocity of an object, hence [tex]a=\frac{\triangle v}{t}[/tex]
The applied force formula becomes [tex]F=\frac{m\triangle v}{t}[/tex]
Given the following parameters
m = 1.0kg
[tex]\triangle v=2.0-1.5\\\triangle v=0.5m/s[/tex]
t = 0.1sec
Substitute the given parameter into the formula
[tex]F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N[/tex]
Hence the amount of force that the wall imparts on the ball is 5.0N
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In an exciting game, a baseball player manages to safely slide into second base. The mass of the baseball player is 88.9 kg and the coefficient of kinetic friction between the ground and the player is 0.53. (a) Find the magnitude of the frictional force in newtons. N (b) It takes the player 1.7 s to come to rest. What was his initial velocity (in m/s)
Answer:
Look at explanation
Explanation:
a) Kinetic Friction= μmg
μmg=0.53*88.9*9.8=461.75N
b) -461.75N=ma
a= -5.19m/s^2
v=v0+at
5.19*1.7=v0
v0=8.81m/s^2
(a) The magnitude of the frictional force will be 461.75N
(b)The initial velocity will be 8.81 m/s.
What is kinetic friction?A force that acts among sliding parts is referred to as kinetic friction. A body moving on the surface is subjected to a force that opposes its progressive motion
The size of the force will be determined by the kinetic friction coefficient between the two materials.
The given data in the problem is;
μ is the coefficient of kinetic friction= 0.53.
m is the mass = 88.9 kg
g is the acceleration due to gravity= 9.81 m/s²
v is the speed =?
The formula for friction force is;
[tex]\rm F= \mu R \\\\ R=mg \\\\ F= \mu mg \\\\\ F=0.53 \times 88.9 \times 9.81 \\\\ F= 461.75 \ N[/tex]
Mechanical force is found as;
F=ma
-461.75=(88.9)a
(-ve shows the -ve work done)
a=-5.19 m/s
From the Newton's first equation of motion;
v=u+at
0=u+at
u=-at
u=(- (-5.19)(1.7)
u=8.81 m/s²
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If a bale of hay behind the target exerts a constant friction force, how much farther will your arrow burry itself into the hay than the arrow from the younger shooter
Answer:
The arrow will bury itself farther by 3S₁
Explanation:
lets assume; the Arrow shot by me has a speed twice the speed of the arrow fired by the younger shooter
Given that ; acceleration is constant , Frictional force is constant
A₂ = A₁
Vf²₂ - Vi²₂ / 2s₂ = Vf₁² - Vi₁² / 2s₁ ---- ( 1 )
final velocities = 0
Initial velocities : Vi₂ = 2(Vi₁ )
Back to equation 1
0 - (2Vi₁ )² / 2s₂ = 0 - Vi₁² / 2s₁
hence :
s₂ = 4s₁
hence the Arrow shot by me will burry itself farther by :
s₂ - s₁ = 3s₁
Note : S1 = distance travelled by the arrow shot by the younger shooter
if a body covers 100m in 5 second from rest find the acceleration produced by a body in 10 second
Answer:
a=10m/s^2
Explanation:
acceleration= final velocity+ initial velocity/time taken
v-u/t=a
100-0/5=a
100/5=a
a=20m/s^2
case2
100-0/10=a
100/10=a
a=10m/s^2
Don't forget to write the units.
Hope this helps
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convert 2.4 milimetre into metre
Answer is 0.0024
Explanation
divide the length value by 1000.
Three spheres (water, iron and ice) of the exact same volume are submerged in a tub of water. After the spheres are lined up, they are released. The spheres are made of plastic with the same density as water, ice, and iron.
Required:
a. Compare the weights of the three spheres.
b. Compare the buoyant forces on the three spheres.
c. What direction does the net force push on each of the spheres?
d. What happens to each sphere after it is released?
Answer:
(a) Iron > plastic > ice
(b) Same on all
(c) Iron downwards, plastic net force zero, ice upwards.
(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.
Explanation:
Three spheres have same volume , plastic, ice and iron.
(a) The weight is given by
Weight = mass x gravity = volume x density x gravity
As the density of iron is maximum and the density of ice is least so the order of the weight is
Weight of iron > weight of plastic > weight of ice
(b) Buoyant force is given by
Buoyant force = Volume immersed x density of fluid x g
As they have same volume, density of fluid is same so the buoyant force is same on all the spheres.
(c) Net force is
F = weight - buoyant force
So, the net force on the iron sphere is downwards
On plastic sphere is zero as the density of plastic sphere is same as water. On ice sphere it is upwards.
(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.
Question 7 of 10
A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s
Answer:
a
Explanation:
you take 32,000kg ÷2.0m
93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures
Answer:
828 kg/m³ or 0.828 g/cm³
Explanation:
Applying,
D = m/V............. Equation 1
Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.
From the question,
Given: m = 77 g , V = 93 cm³
Substitute these values into equation 1
D = 77/93
D = 0.828 g/cm³
Converting to kg/m³
D = 828 kg/m³
find the distance travelled by a moving body if it attained acceleration of 2m/s2 after starting from rest in 5min
Answer:
300 meters
Explanation:
a= 2m/s^2
t= 5 min
Convert into seconds, 5*60= 300seconds
v0= 0
x0=0
use x-x0= v0t + 1/2at^2
plug in values
x= 1/2*2*(300)
Solve
x= 300 meters
What is the value of the charge that experiences a force of 2.4×10^-3N in an electric field of 6.8×10^-5N/C
Hi there!
[tex]\large\boxed{\approx 35.29 C}[/tex]
Use the following formula:
E = F / C, where:
E = electric field (N/C)
F = force (N)
C = Charge (C)
Thus:
6.8 × 10⁻⁵ = 2.4 × 10⁻³ / C
Isolate for C:
C = 2.4 × 10⁻³ / 6.8 × 10⁻⁵
Solve:
≈ 35.29 C
A block of mass M is connected by a string and pulley to a hanging mass m.
The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg.
b. Find the acceleration of the system and tensions on the string.
c. How far will block m drop in the first seconds after the system is released?
d. How long will block M move during the above time?
e. At the time, calculate the velocity of block M
f. Find out the deceleration of block M if the connection string is removal by cutting after the first second. Then, calculate the time taken to contact block M and pulley
How far will block m drop in the first seconds after the system is released?
(b) Use Newton's second law. The net forces on block M are
• ∑ F (horizontal) = T - f = Ma … … … [1]
• ∑ F (vertical) = n - Mg = 0 … … … [2]
where T is the magnitude of the tension, f is the mag. of kinetic friction between block M and the table, a is the acceleration of block M (but since both blocks are moving together, the smaller block m also shares this acceleration), and n is the mag. of the normal force between the block and the table.
Right away, we see n = Mg, and so f = µn = 0.2Mg.
The net force on block m is
• ∑ F = mg - T = ma … … … [3]
You can eliminate T and solve for a by adding [1] to [3] :
(T - 0.2Mg) + (mg - T ) = Ma + ma
(m - 0.2M) g = (M + m) a
a = (10 kg - 0.2 (20 kg)) (9.8 m/s²) / (10 kg + 20 kg)
a = 1.96 m/s²
We can get the tension from [3] :
T = m (g - a)
T = (10 kg) (9.8 m/s² - 1.96 m/s²)
T = 78.4 N
(c/d) No time duration seems to be specified, so I'll just assume some time t before block M reaches the edge of the table (whatever that time might be), after which either block would move the same distance of
1/2 (1.96 m/s²) t
(e) Assuming block M starts from rest, its velocity at time t is
(1.96 m/s²) t
(f) After t = 1 s, block M reaches a speed of 1.96 m/s. When the string is cut, the tension force vanishes and the block slows down due to friction. By Newton's second law, we have
∑ F = -f = Ma
The effect of friction is constant, so that f = 0.2Mg as before, and
-0.2Mg = Ma
a = -0.2g
a = -1.96 m/s²
Then block M slides a distance x such that
0² - (1.96 m/s²) = 2 (-1.96 m/s²) x
x = (1.96 m/s²) / (2 (1.96 m/s²))
x = 0.5 m
(I don't quite understand what is being asked by the part that says "calculate the time taken to contact block M and pulley" …)
Meanwhile, block m would be in free fall, so after 1 s it would fall a distance
x = 1/2 (-9.8 m/s²) (1 s)
x = 4.9 m
a. What do you mean by chromatic aberration in lenses?
You have three identical metallic spheres, A, B and C, fixed to isolating pedestals. They all start off uncharged. You then charge sphere A to +32.0 uC. You use rubber gloves to move sphere A so that it briefly touches sphere B, and then is separated. Next, sphere A briefly touches sphere C, and again is separated. Finally, sphere A touches sphere B a second time, and is again separated. What will be the final charge of sphere B?
Answer:
Charge on B is 12 uC.
Explanation:
Initial charge on A = 32 uC
Initial charge on B and C = 0
Now A touches to B, so the charge on A and B both is
q = (32 + 0) / 2 = 16 uC
Now A touches to C, so the charge on A and C both is
q' = (16 + 0) / 2 = 8 uC
Now again A touches to B so the charge on B is
q''= (8 + 16) / 2 = 12 uC
Traveling waves propagate with a fixed speed usually denoted as v (but sometimes c). The waves are called __________ if their waveform repeats every time interval T.
a. transverse
b. longitudinal
c. periodic
d. sinusoidal
Answer:
periodic
Explanation:
Una cuerda horizontal tiene una longitud de 5 m y masa de 0,00145 kg. Si sobre esta cuerda se da un pulso generando una longitud de onda de 0,6 m y una frecuencia de 120 Hz. La tensión a la cual está sometida la cuerda es:
a. 1,5 N
b. 15,0 N
c. 3,1 N
d. 5,2 N
Answer:
Option (A) is correct.
Explanation:
A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is
mass of string, m = 0.00145 kg
Frequency, f = 120 Hz
wavelength = 0.6 m
Speed = frequency x wavelength
speed = 120 x 0.6 = 72 m/s
Let the tension is T.
Use the formula
[tex]v =\sqrt\frac{T L}{m}\\\\72 = \sqrt\frac{T\times 5}{0.00145}\\\\T = 1.5 N[/tex]
Option (A) is correct.
Light of intensity I0 and polarized horizontally passes through three polarizes. The first and third polarizing axes are horizontal, but the second one is oriented 20.0� to the horizontal. In terms of I0, what is the intensity of the light that passes through the set of polarizers?
A) 0.442 I0
B) 0.180 I0
C) 0.780 I0
D) 0.883 I0
Answer:
Option C.
Explanation:
Suppose that we have light polarized in some given direction with an intensity I0, and it passes through a polarizer that has an angle θ with respect to the polarization of the light, the intensity that comes out of the polarizer will be:
I(θ) = I0*cos^2(θ)
Ok, we know that the light is polarized horizontally and comes with an intensity I0
The first polarizer axis is horizontal, then the intensity after this polarizer is:
then θ = 0°
I(0°) = I0*cos^2(0°) = I0
The intensity does not change. The axis of polarization does not change.
The second polarizer is oriented at 20° from the horizontal, then the intensity that comes out of this polarizer is:
I(20°) = I0*cos^2(20°) = I0*0.88
And the axis of polarization of the light that comes out is now 20° from the horizontal
Now the light passes through the last polarizer, which has an axis oriented horizontally, so the final intensity of the light will be:
note that here the initial polarization is I0*0.88
and the angle between the axis is 20° again.
Then the final intensity is:
I(20°) = I0*0.88*cos^2(20°) = I0*0.78
Then the correct option is C.
A 64-ka base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
Required:
a. How much mechanical energy is tout due to friction acting on the runner?
b, How far does he slide?
Answer:
Explanation:
From the given information:
mass = 64 kg
speed = 3.2 m/s
coefficient of friction [tex]\mu =[/tex] 0.70
The mechanical energy touted relates to the loss of energy in the system as a result of friction and this can be computed as:
[tex]W = \Delta K.E[/tex]
[tex]\implies \dfrac{1}{2}m(v^2 -u^2)[/tex]
[tex]= \dfrac{1}{2}(64.0 \kg) (0 - (3.2 \ m/s^2))[/tex]
Thus, the mechanical energy touted = 327.68 J
According to the formula used in calculating the frictional force
[tex]F_r = \mu mg[/tex]
= 0.70 × 64 kg× 9.8 m/s²
= 439.04 N
The distance covered now can be determined as follows:
d = W/F
d = 327.68 J/ 439.04 N
d = 0.746 m