A beam of light, incident on a flat water surface, reflects from the mirror-like surface so that the angle of incidence equals the angle of reflection. The water has waves. Would individual light beams obey the law of reflection in this case?

Answer:b

Explanation:

Guess

Answer:

The speed of this particle is constantly [tex]c[/tex].

Explanation:

Position vector of this particle at time [tex]t[/tex]:

[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].

Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:

[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].

Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].

Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:

[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].

The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :

[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].

Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)

Answer:

The Big Bang theory predicts that the early universe was a very hot place and that as it expands, the gas within it cools. Thus the universe should be filled with radiation that is literally the remnant heat left over from the Big Bang, called the “cosmic microwave background", or CMB.

Explanation:

The Big Bang theory suggest  that the universe in early stage was at very hot place and which can be expanded, the gas within it cools. It is in an infinite universe and it has no edge.

The Big Bang theory is a cosmological model which explain the existence of the observable universe from the earliest periods to the large-scale evolution.

The model describes the mechanism behind the universe expansion from an initial state of high density and temperature, it is very important concept as a lot of research is going on in this field to find out exactly how the universe began billions of years ago.

The universe began to cool down in order to allow the formation of particles become atoms after its initial phase of expansion,  Primordial elements  such as Hydrogen, Helium, and Lithium are condensed through gravity are formed early stars and galaxies.

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Answer:

Numerous studies have shown that the economic costs of divorce fall more heavily on women. After separation, women experience a sharper decline in household income and a greater poverty risk (Smock 1994; Smock and Manning

Answer:

sadness and stress...................

Answer:

1. Temperature= 869.35 K

2. Pressure of combustion = 12994.043 kpa

3. Thrust = 127x10⁶N

Explanation:

this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.

1.

The temperature = (273+2400k) - (3800)²/2(4003)

= 2673 - 14440000/8006

= 2673 - 1803.65

= 869.35 K

Approximately 869.4K

2. We first get mach number

= 3800/√1.3(923.8)(869.35)

= 3800/1021.78

= 3.719

Pressure = 100kpa[1+2.07464415]^1.3/0.3

= 12995.043kpa

C. Thrust

Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)

= 12678.621

= 126.781 kN

Thrust is approximately 127kN = 127x10⁶N

Answer: Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force.

Answer:

roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge

Explanation:

Answer:

b)  a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N

Explanation:

a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle

b) Let's start at point A

Let's use that the acceleration is centripetal

           a = v² / r

let's calculate

            a = 28² / 15.0

            a = 52.26 m / s²

as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards

Point B

           a ’= 142/15

           a ’= 13.06 m / s²

in this case the acceleration is vertical downwards

c) The values ​​of the normal force

point A

let's use Newton's second law

           ∑ F = m a

           N- W = m a

           N = mg + ma

           N = m (g + a)

           N = 450.0 (9.8 + 52.25)

           N = 2.79 10⁴ N

d) Point B

            -N -W = m (-a)

             N = ma -m g

             N = m (a-g)

             N = 450.0 (14.0 - 9.8)

             N = 1.89 10³ N

Answer:

A) t = 1.249 10⁻² s, B)  f = 80 Hz, C) f = 20 Hz,

D)  slowly advancing an angle of approximately    Δθ = 0.05 rad each flash

E) In each flash it seems to go backward an angle of Δθ = -0.05 rad

Explanation:

A) To make it appear that the blades are immobile, it implies that every time the light turns on, a blade should be in the same position, therefore, as we have 4 blades, they must rotate an angle of 2π/4,  

         θ = π / 2  

         θ = 1.57 rad  

taking the angle let's use the endowment kinematics relations  

          θ = w₀ t + ½ α t²  

in general the fans rotate at constant speed α= 0  

         θ = w₀ t  

         t = θ / w₀  

let's reduce the magnitudes to the SI system  

        w₀ = 20 rev / s (2π rad / 1rev) = 125.66 rad / s  

let's calculate  

        t = 1.57 / 125.66  

        t = 1.249 10⁻² s  

B) the fastest speed for the blades to rotate is when one blade of a complete turn , we use the relationship between the fecuance and the period  

        f = 1 / T  

        f = 1 / 1.25 10⁻²  

       f = 80 Hz

C) we have two possibilities:  

* a yellow dot is placed on each sheet  

In this case the angular velocity of the blade is the same at all points, therefore the results obtained should not change

* a yellow dot is placed on a single sheet.  

Here for the point to remain fixed the angle of rotation must be

       θ= 2π rad  

the time is  

       t = 2π / 125.66  

       t = 5 10⁻² s  

the maximum frequency is  

      f = 1/5 10⁻²  

      f = 20 Hz

D) The copy strobe rotates at f = 19 Hz, the time between each flash is  

      t = 1/19  

      t = 5.26 10⁻² s  

this time is higher, so the angle turned is large  

       θ = w t  

       θ = 125.66 5.26 10⁻²  

       θ = 6.61 rad  

the relationship between this angle and the angle of a circle is  

θ = 1,052

We can see that it is this time the blade rotates 1 complete turns, for this the position of the blade changes us, for the other 0.052 rad the blade rotates a little more than the circumference therefore it seems that it is slowly advancing an angle of approximately  

         Δθ = 0.05 rad each flash  

E) in this case changes the flash speed  

       t = 1/21  

       t = 4.76 10⁻² s  

the angle rotated is  

      θ = 125.66 4.76 10⁻²  

      θ = 5.984 rad  

      θ / 2π = 0.95  

in that case, the blade did not complete the turn, therefore in each flash it seems to go backward an angle

Δθ = -0.05 rad

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

a) The total electric potential is 2282000 V

b) The total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to a specific point in an electric field.

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

Electric potential at p in diagram 1 below is;

[tex]V_P=V_1+V_2[/tex]

[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]

we know that; the Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)The total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

[tex]r_1^2=0.015^2+0.0125^2[/tex]

[tex]r_1 = \sqrt{[ 0.015^2 + 0.0125^2 ][/tex]

[tex]r_1 = \sqrt{0.00038125}[/tex]

[tex]r_1 = 0.0195[/tex]

Also

[tex]r_2^2 = 0.015^2 + 0.018^2[/tex]

[tex]r_2 = \sqrt{0.015^2 + 0.018^2}[/tex]

[tex]r_2 = \sqrt{0.000549[/tex]

[tex]r_2 = 0.0234[/tex]

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

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Answer:

5.51 m/s^2

Explanation:

Initial scale reading = 50 kg  

assume the greatest scale reading = 78.09 kg

Determine the maximum acceleration for these elevators

At rest the weight is = 50 kg

Weight ( F ) = mg = 50 * 9.81 = 490.5 N

At the 10th floor weight = 78.09 kg

Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N

F = change in weight

Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)

50 * a = 275.61

Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2

Answer:

5.7375 seconds

Explanation:

The computation of the time required to reach that maximum velocity is shown below:

Given that

Mass = m =  5100 kg

Net upward force F = 8 × times 10^5 N

Initial speed = V_i = 0

Maximum speed = V = 900 m.s

Based on the above information

Impluse J = m(V - V_i)

= 5100 (900 - 0)

= 459 × 10^4 kg m.s

As we know that

J = FT

So

T = J ÷ F

= (459 × 10^4) ÷ (8 × 10^5)

= 5.7375 seconds

Answer:

[tex]E=30.78\ J[/tex]

Explanation:

The force constant of the spring, k = 76 N/m

The extension in the spring, x = 0.9 m

We need to find the energy is stored in the extended spring. The energy stored in the spring is given by :

[tex]E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 76\times (0.9)^2\\\\E=30.78\ J[/tex]

So, 30.78 J of energy is stored in the spring.

Answer:

B

Explanation:

The crest of all three waves are of equal height

Answer:

Kinetic energy increases and potential energy decrease when velocity of an object increase.

Answer:

7,812 J

Explanation:

Using the relation:

Q = mcΔθ

Q = quantity of heat

C = specific heat capacity of lead

Δθ = temperature change (T2 - T1)

M = mass of substance

Q = mass * specific heat * Δθ

Q = 0.125kg * 128 * (327 – 20)

Q = 0.125 * 128 * 307

Q = 4912 J

For melting:

Q = mass * Hf

0.125 * (2.32 * 10^4)

= 2,900 J

Total = 4,912 J + 2,900 J = 7,812 J

Answer:

answer = opposite

they are not changing being in a place as being they-self

the same would not be true because its different then the Styrofoam and it could only be different and its make the most sense aswell.

Both ball and rod has opposite charges on their bodies because of presence of different charges on it.

In the Styrofoam ball investigation, it is likely that the charges on the ball and rod are opposite because Styrofoam ball is negatively charged due to the presence of electrons while on the other hand, the rod is positively charged because of lining of positive charges on the rod.

So we can conclude that both ball and rod has opposite charges on their bodies because of presence of different charges on it.

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Answer:

a

Explanation:

ajjkiikkkkkhglutfgkitfgghiiij

Answer:

They are the resultant vector.

Answer:

the impulse must be the same in these two cases    F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])

Explanation:

For this exercise we use the relationship between momentum and momentum

         I = Δp

         F t = m v_f - m v₀

To know the speed we use the conservation of energy

starting point. Highest point

       Em₀ = U = m g h

fincla point. Just before the crash

      Em_f = K = ½ m v²

energy is conserved

        Em₀ = Em_f

        m g h = ½ m v²

         v = [tex]\sqrt{2gh}[/tex]

we substitute in the impulse relation

     F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])

therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases

Answer:

A. v = √2gh

B. No! The final velocity does not depend on the mass of the car.

C. Yes! the final velocity depends on the steepness of the hill

D. 3.28 m/s

Explanation:

A. Determination of the final velocity.

½mv² = mgh

Cancel out m

½v² = gh

Cross multiply

v² = 2gh

Take the square root of both side

v = √2gh

B. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is no mass (m) in the formula.

Thus, the final velocity does not depend on the mass of the car.

C. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is height (h) in the formula.

Thus, the final velocity depends on the steepness of the hill

D. Determination of the final velocity.

Height (h) = 0.55 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = √2gh

v = √(2 × 9.8 × 0.55)

v = √10.78

v = 3.28 m/s

Answer:

-the ratio of the speed of light

in air to the speed of light in the substance.

-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.

-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5

Explanation:

Answer:

yes. that is how a baby is conceived.

Answer:

Change in momentum is zero.

Explanation:

The following data were obtained from the question:

Mass (m) = 5000 kg

Time (t) = 1 h

Net force (F) = 0

Change in momentum =?

Force = Rate of change of momentum

0 = change in momentum

Change in momentum = 0

We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.

Answer:

75N

Explanation:

a = v/t = 3/2

F = ma = 50(3/2) = 75

Answer:34.6 m/s

Explanation: It is asking how long meaning the answer is in time

Answer:

Are you sure it was soccer ball? Or meine hearts

Explanation:

Answer:

C

Explanation:

Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.  

A person's self-efficacy relates to their confidence in their ability to carry out the behaviors required to achieve particular performance goals (Bandura, 1977, 1986, 1997).

The belief in one's capacity to exercise control over one's own motivation, behavior, and social environment is known as self-efficacy. The goals for which people strive, the amount of effort put out to obtain goals, and the possibility of achieving particular levels of behavioral performance are all influenced by these cognitive self-evaluations.

Self-efficacy beliefs, unlike conventional psychological notions, are anticipated to change according to the operating domain and the environment in which an action occurs.

Therefore, Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.  

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