What are regular and irregular reflection of light? plz help its
urgent..​

Answer:

Im 99.99999% sure its c

Explanation:

i cant see the pictures too well

Answer:

D

Explanation:

The net force is the combination of all forces acting on an object.

Answer:

5.05225 moles

Explanation:

The computation of the number of moles of gas in the tank is shown below:

Given that

Volume = V = 50 L = 50.0 × 10^-3m^3

Pressure = P = 2.45 atm = 2.45 × 101325

Temperature = T = 22.5°C  = (22.5 + 273)k = 295.5 K

As we know thta the value of gas constant R is 8.314 J/mol.K

Now

PV = nRT

n = PV ÷ RT

= ((2.45 × 101325) (50.0 × 10^-3)) ÷ ((8.314) (295.5))

= 5.05225 moles

Answer:

a

Explanation:

Answer:

a proton because it has a positive charge

Answer:

The answer is

B)  

Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

[tex]a_{c}[/tex] = rω²

ω² = [tex]a_{c}[/tex] / r

ω = √( [tex]a_{c}[/tex] / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a [tex]a_{c}[/tex]  = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²  

ω = 0.0500647 ≈ 0.05 rad/s  

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM  

1 rpm = 60 rph  

so  

ω = 0.478082 × 60

ω = 28.6849  revolutions per hour  

Therefore, the required revolution per hour is 28.6849

The required revolution per hour is 28.6849

The expression for the linear acceleration with respect to the angular velocity is

= rω²

So,

ω² =  / r

ω = √(  / r )

Here r represent the radius of the cylinder

ω represent the angular velocity

Now

diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

And,

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

So,

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²  

ω = 0.0500647 ≈ 0.05 rad/s  

Now

1 rad/s = 9.5493 revolution per minute

So,

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM  

Now

1 rpm = 60 rph  

so  

ω = 0.478082 × 60

ω = 28.6849  revolutions per hour  

Learn more about the cylinder here: https://brainly.com/question/17262276

Answer:

Velocity time graph

Explanation:

Draw on graph paper two straight lines originating at the same point and perpendicular to each other. This is the x-y axis. The x-axis is the horizontal line and the y-axis is the vertical line.

Mark appropriate equally-spaced time intervals on the x-axis so that you can easily graph the time values from the table.

Mark appropriate velocity increments on the y-axis so that you can easily graph the velocity values from the table. If you have negative velocity values, extend the y-axis downward.

Find the first time value from the table and locate it on the x-axis. Look at the corresponding velocity value and find it on the y-axis.

Put a dot where a straight line vertically drawn up through the x-axis value and a straight line horizontally drawn through the y-axis value intersect.

Plot in similar fashion for all other velocity-time pairs in your table.

Draw a straight line with a pencil, connecting each dot you have put down on the graph paper, going from left to right

Answer:

In my textbook's words-

Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path.

Explanation:

The correct definition of amplitude is that it is a maximum displacement

that occurs on a vibrating body from one point to the other.

The initial point of the wave is regarded as its equilibrium position which is

equal to one-half the length of the vibration path.

Amplitude helps to calculate the peak value of different types of waves

such as water waves and in electrical appliances so as to know the peak

current suitable for it.

Read more on https://brainly.com/question/21632362

Answer:

it’s 2.5 m/s

Explanation:

i’m too lazy but trust

This question can be solved by using the law of conservation of momentum.

The final speed of the lighter 2 kg train is " 2.5 ".

When two moving objects collide with each other, the law of conservation of momentum can be applied to them as follows:

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

where,

m₁ = mass of heavier train = 3 kg

m₂ = mass of lighter train = 2 kg

u₁ = initial speed of heavier train = 2.8

u₂ = initial speed of lighter train = 1.6

v₁ = final speed of heavier train = 2.2

v₂ = final speed of lighter train = ?

Therefore,

(3 kg)(2.8) + (2 kg)(1.6) = (3 kg)(2.2) + (2 kg)(v₂)

[tex]v_2 = \frac{5 kg}{2 kg}[/tex]

v₂ = 2.5

Learn more about the law of conservation of momentum here:

https://brainly.com/question/1113396?referrer=searchResults

The attached picture illustrates the law of conservation of momentum.

Answer:

3 is the GCF for all these numbers if thats what you're asking

Answer:

The person who did the comment is correct, it B - Dew will form on leaves

Explanation:

Answer:

B

Explanation:

Valeriie.07 tap in g

Answer:

9

Explanation:

Answer:

C= 50km/hr west

Explanation:

150/3= 50

Because it asks for velocity, make sure to include the direction as well.

Answer:

D. Weight

Explanation:

Hope that helps:)

Answer:

0.0119502868 kilocalorie

Explanation:

Answer:

increases by 50

Explanation:

Answer:

true

Explanation:

Answer:

true

Explanation:

please give brainlest need 1

Answer:

6. A

7. C

8. B

9. The average speed is the distance (a scalar quantity) per time ratio. Speed is ignorant of direction. The average velocity is the displacement or position change (a vector quantity) per time ratio.

Answer:

Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and electric field intensity.

Answer:

They become oppositely charged, which causes them to be attracted to each other.

Answer: number A

Explanation:

If a positive charge and a negative charge interact, their forces act in the same direction, from the positive to the negative charge. As a result opposite charges attract each other: The electric field and resulting forces produced by two electrical charges of opposite polarity. Have a nice day <3

Answer: I think it C and B but I am really confident in C

Explanation:

Answer:

The minimum coefficient of friction required is 0.35.  

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

[tex] -F_{f} + F = 0 [/tex]      

[tex] -F_{f} + ma = 0 [/tex]      

[tex] \mu mg = ma [/tex]

[tex] \mu = \frac{a}{g} [/tex]

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]

[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} [/tex]

Where:  

d: is the distance traveled = 46.1 m

[tex] v_{f}[/tex]: is the final speed of the truck = 0 (it stops)      

[tex]v_{0}[/tex]: is the initial speed of the truck = 17.9 m/s

[tex] a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2} [/tex]        

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.  

[tex] \mu = \frac{a}{g} [/tex]  

[tex] \mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}} [/tex]

[tex] \mu = 0.35 [/tex]

Therefore, the minimum coefficient of friction required is 0.35.  

I hope it helps you!

Answer:

b)  k Δx - W cos θ - μ mg cos θ = m a ,  c)  θ = 86.6º, d)  Δx = 1.18 m

Explanation:

a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.

F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring

b) Let's apply Newton's second law for when the spring is compressed

let's use trigonometry to break down the weight

            sin θ = Wₓ / W

            cos θ = W_y / W

             Wₓ = W sin θ

             W_y = W cos θ

Y axis  

               N - W_y = 0

               N = W_y

               N = W cos θ

X axis

           F -Wₓ -fr = ma

the force applied by the spring is given by hooke's law

           F = k Δx

friction force has the expression

           fr = μ N

           fr = μ W cos θ

we substitute

            k Δx - W cos θ - μ mg cos θ = m a           ( 1)

c) If the plane has no friction, what is the angle so that Δx = 0.1m

We write the equation 1, with fr = 0 and since the system is still a = 0

            k Δx - W cos θ -0 = 0

            cos θ = [tex]\frac{k \Delta x}{ m g}[/tex]

            cos θ = [tex]\frac{880 \ 0.1}{ 150 \ 9.8}[/tex]

            cos θ = 0.0598

            θ = cos⁻¹ 0.0598

            θ = 86.6º

d) In this part they give the angle θ = 45º and there is no friction, they ask the compression

the acceleration is zero, we substitute in 1

            k Δx - W cos θ - 0 = 0

            Δx = [tex]\frac{mg \ cos \ \theta}{k}[/tex]

            Δx = [tex]\frac{ 150 \ 9.8 \ cos45}{880}[/tex]

            Δx = 1.18 m

Answer:

a) attractiva, b) dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex], c)  F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex], d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

           dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex]

where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

         ∫ dF = [tex]k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2[/tex]

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

          λ = dQ₂ / dx

          DQ₂ = λ dx

we substitute

         F = [tex]k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}[/tex]

         F = k Q1 λ ([tex]-\frac{1}{x}[/tex])  

we evaluate the integral

        F = k Q₁ λ [tex](- \frac{1}{d+L} + \frac{1}{d} )[/tex]

        F = k Q₁ λ  [tex]( \frac{L}{d \ (d+L)})[/tex]

we change the linear density by its value

      λ = Q2 / L

       F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex]

d) we calculate the magnitude of F

       F =9 10⁹ (-4.2 10⁻⁶)   [tex]\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}[/tex]

       F = -1.09 N

the sign indicates that the force is attractive

Answer:

a)Toward the rod

b)|dF| = k|Q1|Q2(dx/L)/x^2

c)|F| = k|Q1|Q2/(d(d+L))

d)Plug in for answer c and solve

Explanation:

A)

Q1 is negative and Q2 is positive so it is an attractive force to  where the rod is located.

B)

The formula for Force due to electric charges is F=kQ1Q2/r^2

In this case, Q2 is distrusted through the length of the rod as opposed to a single point charge. As such Q2 is actually Q2*dx/L as dx is a small portion of the full length, L.

The radius between Q1 and Q2 depends on the section of the rod taken so r will be the variable x distance from Q1.

The force is only from a small portion of the rod so more accurately, we are finding |dF| as opposed to the full force, F, caused by the whole rod.

The final formula is |dF| = k|Q1|Q2(dx/L)/x^2

C)

Integrating with respect to the only changing variable, x, which spans the length of the rod, from radius = d to d+L we get this:

F = integral from d to d+L of k|Q1|Q2(dx/L)/x^2

factor out constants

F = kQ1Q2/L * integral d to d+L(1/x^2)dx

F = kQ1Q2/L * (-1/x)| from d to d+L

F = kQ1Q2/L * (-1/d+L - -1/d)

F = kQ1Q2/L * (-d/(d(d+L)) + (d+L)/(d(d+L))

F = kQ1Q2/L * (L)/(d(d+L))

F = kQ1Q2/(d(d+L))

D)

Plug in the given values into c and you have your answer.

Answer:

The electric potential at the center of the meter stick is 54 KV.

Explanation:

Electric potential (V) is given as:

i.e V = [tex]\frac{kq}{r}[/tex]

Where: k is the Coulomb constant, q is the charge and r is the distance.

Given: q = 3.0 μC = 3.0 x [tex]10^{-6}[/tex] C, r = 0.5 m

So that,

V = [tex]\frac{9*10^{9}*3.0*10^{-6} }{0.5}[/tex]

   = [tex]\frac{2.7*10^{4} }{0.5}[/tex]

V = 54000

  = 54 000 volts

The electric potential at the center of the meter stick is 54 KV.

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

[tex] v = \frac{I}{nqA} [/tex]

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

[tex] n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3} [/tex]                  

Now, we can find the drift speed:

[tex]v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s[/tex]              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

Answer:

i think 692m/s2 is the correct answer

Answer:

A. pushing a car that isn't moving