uppose that white light strikes a flat piece of flint glass in air, coming in at an angle of 60 degrees to the surface (30 degrees from the normal or perpendicular). The index of refraction of this dense glass for red light is 1.710, for green light is 1.723, and for blue light is 1.735. What is the order of colors you would see in the refracted light inside the glass as they leave the surface

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

          Wₐ = 1500 lb

          W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

             W = mg

             m = W / g

             mₐ = Wₐ / g

             m_b = W_b / g

             mₐ = 1500/32 = 46.875 slug

             m_b = 125/32 = 35,156 slug

Let's reduce to the english system

             v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

           p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

           p_f = (mₐ + m_b) v

the moment is preserved

           p₀ = p_f

           mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

           v = [tex]\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}[/tex]

we substitute the values

           v = [tex]\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6[/tex]

           v = 0.559 v₀ₐ - 26.40                  (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

              M = mₐ + m_b

              M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

              K₀ = ½ M v²

final point. When they stop

             K_f = 0

The work is

             W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

           N-W = 0

           N = W

the friction force has the expression

            fr = μ N

we substitute

            -μ W x = Kf - Ko

            -μ W x = 0 - ½ (W / g) v²

            v² = 2 μ g x  

            v = [tex]\sqrt{ 2 \ 0.750 \ 32 \ 17.5}[/tex]Ra (2 0.750 32 17.5  

            v = 28.98 ft / s

Answer:

It is actually Albert Bandura! Hope this helps<3

Explanation:

In contrast to Skinner's idea that the environment alone determines behavior, Bandura (1990) proposed the concept of reciprocal determinism, in which cognitive processes, behavior, and context all interact, each factor simultaneously influencing and being influenced by the others.

The theorist who expanded the behavioristic perspective to include cognitive influences on personality was Albert Bandura. So, the correct option is B.

Personality is defined as a distinctive set of behavior, cognition, and emotional patterns that are formed by biological and environmental factors, and that change over time. This explained as the enduring characteristics and behavior which comprise a person's unique adjustment to life, including major traits, interests, drives, values, self-concept, abilities, and emotional patterns.

The four theory is also called as the proto-psychological theory which suggests that there are four fundamental personality types: sanguine, choleric, melancholic, and phlegmatic. Albert Bandura (1990) proposed the concept of reciprocal determinism in which cognitive processes, behavior, and context all interact, with each factor simultaneously affecting and being affected by the others.

Thus, the theorist who expanded the behavioristic perspective to include cognitive influences on personality was Albert Bandura. So, the correct option is B.

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Answer:

B). half-ton car

Explanation:

We can find Kinectic using below expresion.

m= mass of the object

v= velocity

mass (1) 1 ton= 1000kg

Mass(2)0.5= 500kg

K.E= 1/2 mv^2

For 1 ton car

Substitute the values

Kinectic energy= 1/2( 1000) 30^2

= 450,000J

For 0.5 ton car

Kinectic energy= 1/2(500)60^2

= 900,000J

Hence, half-ton car has greater Kinectic energy

Answer: A Punnett square can be used to predict genotype and phenotypes of offspring from genetic crosses. ... In the P generation, one parent has a dominant yellow phenotype and the genotype YY, and the other parent has the recessive green phenotype and the genotype yy.

Explanation:

Answer:

A)  a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

               a_c = v² / r

the angular and linear variables are related

              v = w r

we substitute

               a_c = w² r

let's reduce the magnitudes to the SI system

              w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

               r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

              a_c = 418.88² 0.1

               a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

                v² = v₀² - 2g (y- y₀)

                v² = 0 - 2 9.8 (0 + 1)

                v =√19.6

                v = -4.427 m / s

now let's look for the applied steel to stop the test tube

                v_f = v + a t

                0 = v + at

                a = -v / t

                a = 4.427 / 0.001

                a = 4.43 10³ m / s²

Answer:

a) The frequency of the waves are = 2 Hz (since you are making two pulses every second)

b)The period of the waves is 0.19 m (as the consecutive crests are separated by 0.19 m)

c) V = L x f

v = 0.19 m x 2 S-1 (Hz is actually per second)

v = 0.38 ms-1

pe nis

Answer:

a. 34.45 V

b. 89.80  pJ

c. 355.57  pJ

d. -265.67 pJ

Explanation:

(a) the potential difference between the plates

Since charge is conserved, charge before separation, Q = charge after separation, Q'

Q = C₁V₁ and C₁ = capacitance before separation = ε₀A/d₁ where A = area of plates = 6.70 cm² = 6.70 × 10⁻⁴ m²and d₁ = initial separation of plates = 2.50 mm = 2.50 × 10⁻³ m, V₁ = initial potential difference across plates = 8.70 V

Q' = C₂V₂ and C₂ = capacitance before separation = ε₀A/d₂ where A = area of plates = 6.70 cm² = 6.70 × 10⁻⁴ m² and d₁ = final separation of plates = 9.90 mm = 9.90 × 10⁻³ m. V₂ = final potential difference across plates = unknown

So, Q = Q'

C₁V₁ = C₂V₂

ε₀AV₁/d₁ = ε₀AV₂/d₂

So, V₁/d₁ = V₂/d₂

V₂ = V₁d₂/d₁

Substituting the values of the variables into the equation, we have

V₂ = 8.70 V × 9.90 mm/2.50 mm

V₂ =  8.70 V × 3.96

V₂ = 34.452 V

V₂ ≅ 34.45 V

(b) the initial stored energy

The energy stored in a capacitor U = 1/2CV².

So, the initial energy stored U₁ = 1/2C₁V₁² = ε₀AV₁²/2d₁

Substituting the values of the variables into the equation, we have

U₁ = ε₀AV₁²/2d₁

U₁ = 8.854 × 10⁻¹² F/m × 6.70 × 10⁻⁴ m² × (8.70V)²/(2 × 2.50 × 10⁻³ m)

U₁ = 4490.07 × 10⁻¹⁶ FmV²/5 × 10⁻³ m

U₁ = 898.01 × 10⁻¹³ J

U₁ = 89.801 × 10⁻¹² J

U₁ = 89.801  pJ

U₁ ≅ 89.80  pJ

(c) the final stored energy

The energy stored in a capacitor U = 1/2CV².

So, the initial energy stored U₂ = 1/2C₂V₂² = ε₀AV₂²/2d₂

Substituting the values of the variables into the equation, we have

U₂ = ε₀AV₂²/2d₂

U₂ = 8.854 × 10⁻¹² F/m × 6.70 × 10⁻⁴ m² × (34.45 V)²/(2 × 9.90 × 10⁻³ m)

U₂ = 70403.26 × 10⁻¹⁶ FmV²/19.8 × 10⁻³ m

U₂ = 3555.72 × 10⁻¹³ J

U₂ = 355.572 × 10⁻¹² J

U₂ = 355.572  pJ

U₂ ≅ 355.57  pJ

(d) the work required to separate the plates.

The work required to separate the plates W = -ΔU is the difference between the energy stored at the separation of 9.90 mm and 2.50 mm.

So, W = -(U₂ - U₁) = -(355.57  pJ - 89.80  pJ) = -265.67 pJ.

The disadvantage of sending information over long distances wirelessly using digital signals is "the signals become weaker the farther the receiver is from the source."

Thus , The disadvantage of sending information over long distances wirelessly using digital signals is "the signals become weaker the farther the receiver is from the source."

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Answer:

0.6

Explanation:

Given that :

Radius, R = 7m

Period, T = 6.9s

The Coefficient of static friction, μs can be obtained using the relation :

μs = v² / 2gR

Recall, v = 2πR/T

μs becomes ;

μs = (2πR/T)² / 2gR

μs = (4π²R² / T²) ÷ 2gR

μs = (4π²R² / T²) * 1/ 2gR

μs = 4π²R / T²g

μs = 4π²*7 / 6.9^2 * 9.8

μs = 28π² / 466.578

μs = 276.34892 / 466.578

μs = 0.5922887

μs = 0.6

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

       [tex]a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)[/tex]

       [tex]v_{f} ^{2} -v_{o} ^{2} = 2* a* \Delta x (2)[/tex]

       [tex]\Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)[/tex]

b)

       [tex]v_{f} = v_{o} + a*\Delta t (4)[/tex]

       [tex]\Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s (5)[/tex]

Answer:

Angular momentum

Explanation:

Stars are formed as a result of a collapse of a low-temperature cloud of gas and dust. During the colapse conservation of angular momentum causes any small net rotation of the cloud to increase thus forcing the material into rotating

Answer:

Same as before.

Explanation:

we can use momentum conservation to solve the problem.

Let mass of child be m and of the sled be M. Also let both have initial velocity u.

since, child fells off on the sled it will still have same velocity of ''u''.

By momentum conservation we have,

(M + m) x u = m x u + M x v

⇒ Mu=Mv

⇒ u=v

Hence, velocity of the sled remains the same.

Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2               ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1                  ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Answer:

The answer is C. time. Please mark as brainliest.

Answer:

lane 3    Ф   = 450 vehicles,  ρ = 0.1 vehicle / s

lane 2    Ф_{average} = 300 vehicles,  ρ _{average} = 6.66 10⁻² vehicles/s

lane 1      Ф_{average} = 300 vehicles,  ρ_{average} = 2.66 10⁻² vehicle/s

Explanation:

Before solving this exercise we must clarify the concepts the flow is defined as the occurrence of an event in a time interval, in this case the passage of a car through time

Flux Density is the flux between unit area or unit time

Let's start by calculating the calculation for lane 3

the flow.

Let's use a direct rule of proportions (rule three) if the number of vehicles per unit of time (t₀ = 10s), for the observation time how many vehicles passed in the observation time (t_total = 75 * 60 = 4500 s)

            Ф = 4500 s (1 vehicle / 10 s)

            Ф   = 450 vehicles

The flux density is the flux per unit area, in this case the area is not indicated, so we can define the flux density as the flux per unit of time.

           ρ = 450/4500

           ρ = 0.1 vehicle / s

Lane 2

we look for the flow

we can have separates the interval into two parts

* for the first t₁1 = 30 * 60 = 1800 s

            Ф₁ = 1800 s (1 vehicle / 6s)

            Ф₁ = 300 vehicles during t₁

* for the rest of the time t₂ = 4500-1800 = 2700 s

           Ф₂ = 0

the average density is the total number of vehicles between the total time

           #_ {vehicle} = 300 +0

            Ф_{average) = # _vehicle

            Ф_{average} = 300 vehicles in all time

The density is

            ρ 1 = fi1 / t1

            ρ1 = 300/1800

            ρ1 = 1.66 10-1 vehicles / s

the average density is

            ρ_{average} = [tex]\frac{\phi_1 + \phi_2}{ t_{total}}[/tex]

            ρ _{average} = (300 +0) / 4500

            ρ _{average} = 6.66 10⁻² vehicles / s

Lane 1

flow

* first time interval t₁ = 10 * 60 = 600 s

             Ф₁ = 600 s (1 vehicle / 5s)

             Ф₁ = 120 vehicles in interval t₁

* second interval t₂ = 4500-600 = 3900 s

              Ф2 = 0

 average flow

             Ф = Ф1 + Ф2

             Ф = 120 vehicles at all time

Density

* first interval

          ρ₁ = 120/600

          ρ₁ = 0.2 vehicles / s

* second interval

          ρ₂ = 0

average density

         ρ+{average} = 120/4500

         ρ_{average} = 2.66 10⁻² vehicle/s

Answer:

4.4pF

Explanation:

the capacitance of the circuit required for reception is given:

wL =  [tex]\frac{1}{wC}[/tex]

w = 2π [tex]f[/tex]

Using both equation

Capacitance is given

C =   1 - 4π2  f2 L

1- 4×9.8969×144×10 10 ×0.004

​=4.4pF

​

​

Answer:

T =  30.42°C

Explanation:

According to the conservation of energy principle:

[tex]Energy\ Given\ by\ Resistor = Heat\ Gain\ by\ Copper + Heat\ Gain\ by\ Water\\E = m_{c}C_{c}(T_{2c} - T_{1c}) + m_{w}C_{w}(T_{2w} - T_{1w})[/tex]

E = 120 KJ

mc = mass of copper = 13 kg

Cc = specific heat capacity of copper = 0.385 KJ/kg.°C

T2c = T2w = Final Equilibrium Temperature = T = ?

T1c = Initial Temperature of Copper = 27°C

T1w = Initial Temperature of Water = 50°C

mw = mass of water = 4 kg

Cw = specific heat capacity of water = 4.2 KJ/kg.°C

Therefore,

[tex]120\ KJ = (13\ kg)(0.385\ KJ/kg^oC)(T-27^oC) + (4\ kg)(4.2\ KJ/kg^oC)(T-50^oC)\\120\ KJ - 135.135\ KJ - 840\ KJ = (- 5.005T - 16.8 T)\ KJ/^oC\\T = \frac{-855.135\ KJ}{-28.105\ KJ/^oC}\\[/tex]

T =  30.42°C

Answer:

Tie aluminium foil on each end of the battery using rubber band, then use copper wire to attach the led to the aluminium foil...... probably would work

Answer:

Tetherball is an interesting game in which two players tries to hit the ball hard so that it goes around the

pole.Each time the player hits the ball, it's orbit rises higher off the ground.Let's understand the physics

behind this.The motion of a tetherball is governed by two forces.These two forces combine to generate a

net force, i.e. centripetal force.If the ball is moving more quickly, it requires a greater centripetal force,

which in turn requires a greater tension force.Since the ball's weight hasn't changed, the angle of the

tension force changes until the ball is in vertical equilibrium.

To access this physics simulation visit: http://goo.gl/xVdwgO Page 02Exploration Series www.ck12.org

Ball Mass : This slider controls the mass of the ball. A ball with more mass will have more inertia, requiring

a greater net force to accelerate it. A ball with more mass will ALSO have a greater gravitational force

acting on it. Watch both of these effects occur when you manipulate this slider.

Cable Length : This slider controls the length of the cable. A longer cable is capable of allowing a greater

circular radius of motion for the ball. It is important to remember that the radius of the circular motion is

NOT equal to the length of the cable. Instead, if you want to understand the size of the circle of the ball's

motion, ignore the cable and just imagine the path of the ball.

Ball Speed : This slider controls the speed of the ball - imagine a kid just hit the ball and it sped up. A ball

moving more quickly is also accelerating more quickly because its velocity is changing as it moves in a

circle (remember that changes in DIRECTION of velocity 'count' as changes to velocity).

Force Diagram : This allows you to turn on or off the diagram of the forces acting on the ball. Look for the

ball to be in vertical force balance, which means the vertical component of tension is canceled by the

gravitational force. The ball should NOT be in horizontal force balance - it is accelerating towards the center

of the circle! It is important to note that this free body diagram should really be moving with the ball so that

To access this physics simulation visit: http://goo.gl/xVdwgO Page 03Exploration Series www.ck12.org

the tension force always points along the cord - we are just showing the forces at the moment the ball is at

the furthest-right on this screen.

Centripetal force vs Tetherball speed : This is a plot of centripetal force required to keep the ball in

circular orbit about the pole as a function of its speed. As expected, a more quickly-moving ball is changing

in velocity more often in a given amount of time, and so is accelerating more. This greater (centripetal, or

center-pointing) acceleration requires a greater net force.

Explanation:

Answer:

true

Explanation:

Gravity (or the acceleration due to gravity) is 9.81 meters per second squared, on the surface of Earth, because of the size of Earth and the distance we are on its surface from its center.

Answer:

[tex]\boxed{\text{\sf \Large 3.0 s}}[/tex]

Explanation:

Use distance formula

[tex]\displaystyle d=ut+\frac{1}{2} at^2[/tex]

[tex]u= \text{\sf initial velocity}\\d= \text{\sf distance}\\a= \text{\sf acceleration}\\t= \text{\sf time taken}[/tex]

[tex]\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2[/tex]

[tex]t=3[/tex]

A block starting from rest slides down the length of an 18 m plank with an acceleration of 4.0 meters per second square. Time taken by the block  to reach at the bottom is 3 sec.

The rate at which an object changes its velocity is known as acceleration. Acceleration is a vector quantity. If an object's velocity is changing then it is accelerating and an object with a constant velocity is not accelerating.

The speed at which something moves in a specific direction is known as its velocity. As an illustration, think of the speed of a car travelling north on a highway or the speed at which a rocket takes off.

Given that in the question block slides down an 18 m plank length with an acceleration of 4.0 meters per second square when it begins at rest.

Using equation of motion,

S = ut + (1/2)at²

s is distance, s = 18 m

u is initial velocity, u = 0

a is acceleration, a = 4 m/sec²

t is time

18 = 0*t + (1/2)*4*t

solving we get t = 3 sec.

Time taken by the block  to reach at the bottom is 3 sec.

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Answer:A)u =4.295m/s  , B)a = 29.746m/s²   C) F=3,153N

Explanation:

Using the kinematic expression  

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

Given that he reaches a height of 0.940 m above the floor,

the final velocity  = 0

Here, acceleration due to gravity is acting in  opposite the initial direction of motion. So, a=-9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.940

- u² = 2 × - 9.81 × 0.920

- u² = -18.4428

cancelling the minus in both sides , we have that  

u² = 18.4428

u = √18.4428

u =4.295m/s

(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2

Using v² = u² + 2as

where u = initial speed of basketball player before lengthening = 0 m/s,

v = final speed of basketball player after lengthening =  4.295m/s,

a = acceleration while  straightening his legs

s = distance moved during lengthening = 0.310m

v² = u² + 2as  

 a = (v² - u²)/2s

a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)

a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)

a = (18.4428 m²/s²/(0.62 m)

a = 29.746m/s²

c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N

Force= mass x acceleration.

F = 106 kg X 29.746m/s²

 F = 3,153.076 rounded to  3,153N

Answer:

All planets including the outer larger planets were formed at the same time somewhere around 4.5 Billion years ago.

Explanation:

the young sun drove away most of the gas from the inner solar system, leaving behind the rocky cores also known as the terrestrial planets.

Answer:

A) a_max = 4.9 m/s²

B) the metal cabinet will not slip.

C) a_max = 9.8 m/s². The cabinet will slip.

Explanation:

A) We are given;

Coefficient of static friction; μ_s = 1

Coefficient of kinetic friction; μ_k = 0.7

Formula for maximum static friction;

F_s = μ_s•N

We are dealing with half of the weight of a flatbed truck. Thus;

N = mg/2

Thus;

F_s(max) = ½μ_s•mg

Now, the maximum acceleration it can achieve on dry concrete will be when the Maximum static friction is reached.

Because after maximum static friction, the cabinet will slip.

Thus; F_s(max) = ma

Therefore,

ma = ½μ_s•mg

m will cancel out to give;

a = ½μ_s•g

a = ½ × 1 × 9.8

a_max = 4.9 m/s²

B) we are told that the coefficient of static friction is now 0.55

Thus, the friction between the metal and the wood is;

F = μ_s•g = 0.55 × 9.8

F = 5.39 N

The acceleration gotten in the first part is less than μ_s•g = 5.39 N, then the metal cabinet will not slip.

C) Now, if we are considering a 4 wheel drive, then we will not divide the mass by 2 and so; N = mg

Like we did in A above;

ma = μ_s•mg

a = μ_s•g

a = 1 × 9.8

a = 9.8 m/s²

Now, this value of a_max is greater than μ_s•g in answer A above.

Thus, a_max > μ_s•g and thus, the cabinet will slip.

Answer:

x and y, or 0 and 1. These two symbols form hundreds of patterns that represent data like numbers, letters, even pictures, and videos. Digital images are made up of pixels, and each pixel contains binary code. When all that code comes together and the pixels are formed we can see things like family photo albums or we can watch a movie on our phones. We can even see a digital picture of a kitten I guess. k, bye.

Explanation:

Answer:

[tex]\boxed{\text{\sf \Large 42 m}}[/tex]

Explanation:

Use height formula

[tex]\displaystyle \sf H=\frac{u^2 sin(\theta)^2}{2g}[/tex]

u is initial velocity

θ = 90° (fired vertically upward)

g is acceleration of gravity

[tex]\displaystyle \sf H=\frac{29^2 \times sin(90 )^2}{2 \times 10}=42.05[/tex]

Answer:

i don't know but i hope you get it right

Explanation

Answer:

Tectonic Plate Movement

Explanation:

Each continent and ocean sits on its own tectonic plate which floats on the Earths upper mantle. They move very little over time.

Answer:

tectonic plates movement