Which of the units of the following physical quantities are derived

A concave mirror is an example of curved mirrors. So that the appropriate answer to the given question is option D. The rays will cross at the focal point.

A concave mirror is a type of mirror in which its inner part is the reflecting surface, while its outer part is the back of the mirror.  This mirror reflects all parallel rays close to the principal axis to a point of convergence. It can also be referred to as the converging mirror.

In this type of mirror, all rays of light parallel to the principal axis of the mirror after reflection will cross at the focal point.

Therefore, the required answer to the given question is option D. i.e The rays will cross at the focal point.

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Answer:

This causes the amplitude of the oscillation to decay over time. The damped oscillation frequency does not equal the natural frequency. Damping causes the frequency of the damped oscillation to be slightly less than the natural frequency

Answer:

9 Brainly hahaha ............huh

Answer:

One way to look at this is to consider the forces acting on any point in a string.

For a very small portion of string F = M a must still hold. As M approaches zero the small portion of string would have to approach infinite acceleration if the net force on that portion of string were not zero.

One generally considers the net force acting on the center of mass of an object not  the individual forces acting on each infinitesimal mass composing

the object.

Answer:

26 km

Explanation:

Let's say our "cable" has a cross section of 1 m²

Then each meter of cable would weight 7900(9.8) = 77420 N

A Pascal is a Newton per square meter

2 x 10⁹ / 77420 = 25840 m or about 26 km or about 16 miles

Answer:

you are going to expend energy to give a lot of velocity (and momentum) to your foot in order to transfer it the stone air drag this time the kicking speed is for superior to walking speed.

                                                  Thank You

Answer:

ans is c

Explanation:

chk photo

The axial magnetic field 15 cm from the center of the loop is approximately 21 μT. Hence, option (c) is correct.

In the vicinity of a magnet, an electric current, or a shifting electric field, there is a vector field called a magnetic field where magnetic forces can be seen. Electric charges in motion and the intrinsic magnetic moments of elementary particles connected to the fundamental quantum characteristic known as spin create a magnetic field.

Given parameters:

Radius of the circular loop: r = 10 cm = 0.10 m.

Current passing through the loop: I = 20 A.

Axial distance of the point: z = 15 cm = 0.15 m.

Hence, the axial magnetic field at that point:

B = (μ₀/4π) 2πR²I/(z²+R²)^(3/2)

By putting these values, we get B = 21 × 10⁻⁶ T = 21 μT.

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I assume friction is the only force acting on the book as it slides.

(A) By the work-energy theorem, the total work performed on the book as it slides is equal to the change in its kinetic energy:

W = ∆K

W = 1/2 (1.50 kg) (1.25 m/s)² - 1/2 (1.50 kg) (3.21 m/s)²

W ≈ -6.56 J

(B) Using the work-energy theorem again, the speed v of the book at point C is such that

-0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²

==>   v = 0.750 m/s

(C) Take the left side to be positive, then solve again for v.

0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²

==>   v ≈ 1.60 m/s

Answer:

a = 4.9(1 - sinθ - 0.4cosθ)

Explanation:

Really not possible without a complete setup.

I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g

                                     F = ma

mg - mgsinθ - μmgcosθ = (m + m)a

      mg(1 - sinθ - μcosθ) = 2ma

      ½g(1 - sinθ - μcosθ) = a

maximum acceleration is about 2.94 m/s² when θ = 0

acceleration will be zero when θ is greater than about 46.4°

Answer:

The number of neutrons or electrons in an atom can change without changing the identity of the element.

Answer:

Their is a direct relationship between the number of batteries and the increase in power. The voltage is the product of the number of batteries and the voltage which is 9 volts. As the batteries touch ends the voltages of all three combines.

Explanation:

We have that the total distance covered by the runner is

[tex]d_t=20.723miles[/tex]

The total distance covered by the runner is a sum of all miles covered by the runner

Therefore

With

[tex]d_t[/tex]=Total distance

[tex]d_t=d_1+d_2+d_3\\\\d_t=9.5+8.89+2.333[/tex]

[tex]d_t=20.723miles[/tex]

in conclusion

The total distance covered by the runner is

[tex]d_t=20.723miles[/tex]

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The SI unit of pressure is the pascal: 1Pa=1N/m2 1 Pa = 1 N/m 2 . Pressure due to the weight of a liquid of constant density is given by p=ρgh p = ρ g h , where p is the pressure, h is the depth of the liquid, ρ is the density of the liquid, and g is the acceleration due to gravity.

Answer:

Explanation:

Unknown fork frequency is either

335 + 5.3 = 340.3 Hz

or

335 - 5.3 = 329.7 Hz

After we modify the known fork, the unknown fork frequency equation becomes either

(335 - x) + 8 = 340.3

(335 - x)  = 332.3

x = 2.7 Hz

or

(335 - x) + 8 = 329.7

(335 - x) = 321.7

x = 13.3 Hz

IF the unknown fork frequency was 340.3 Hz,

THEN the 335 Hz fork was detuned to 335 - 2.7 = 332.3 Hz

IF the unknown fork frequency was 329.7 Hz,

THEN the 335 Hz fork was detuned to 335 - 13.3 = 321.7 Hz

Answer:

record is information created, received and maintained as evidence and information by an organization or person.in simpler terms it's a collection of of fields probably of different data types.

sound is however something loud or soft.which can be defined as vibrations that travel through the air or another medium.

I hope this helps

The acceleration of the block over the rough surface is 1.22625 m/s²

The process through which the acceleration is obtained is presented as follows of approach to

The given parameters are;

Mass of block, m = 2 kg

Nature of the surface of the quarter circle = Frictionless

The length of the horizontal, d = 8 m

The coefficient of friction of the horizontal surface, μ = 0.4

The unknown parameter;

The acceleration of the block over the rough surface

Method;

Find the work done by friction to stop the block and divide the result by the mass of the block

The work done by friction, [tex]W_f[/tex] = (Force of friction) × (Distance the block moves on the rough surface before coming to rest)

[tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d

[tex]F_f[/tex] = Normal reaction of surface on block, [tex]N_r[/tex] × μ

Normal reaction on block, [tex]\mathbf{N_r}[/tex] = Weight of block

[tex]\mathbf{N_r}[/tex] ≈ 2 kg × 9.81 m/s² = 19.62 N

Therefore;

The work done by friction [tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d = [tex]\mathbf{N_r}[/tex] × μ × d

[tex]\mathbf{W_f}[/tex] = 19.62 N × 0.4 × 8 m = 62.784 J

The work done by the block, W = Force, F × d

Force, F = m × a

Where;

a = The acceleration of the block

According to the principle of conservation of energy, we have;

[tex]\mathbf{W_f}[/tex]  = W

∴ 19.62 J = 2 kg × a × 8 m

a = 19.62/(2 kg × 8 m) = 1.22625 m/s²

The acceleration of the block over the rough surface, a = 1.22625 m/s²

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Answer:

5 kilometers per hour

Explanation:

Speed = distance / time

Distance: 50km

Time: 10 hours

Speed = 50/10 = 5kph

Answer:

5kmph

Explanation:

if the bus traveled 50 km in 10 hours, we have to divide 50 by 10 to see how fast it traveled per hour.

50/10 = 5

therefore, the bus was traveling 5 km per hour

hope this helps :)

OK.  

Good luck on your experimental demonstration.

It's a nice exercise.

Answer:

don't know what class are you you are using which mobile or laptop

Answer:

Counterclockwise

explanation in attachment

We're given

[tex]\displaystyle y_1(x) = 1 + \frac x3 + \frac{x^2}{24} + \frac{x^3}{360} + \cdots = \sum_{n=0}^\infty a_nx^n[/tex]

so let's see if we can find a closed form for the n-th term's coefficient.

Notice that

[tex]\displaystyle a_0 = 1 \\\\ a_1 = \frac13 = \frac1{1\times3} \\\\ a_2 = \frac1{24} = \frac1{(1\times3) \times (2\times4)} \\\\ a_3 = \frac1{360} = \frac1{(1\times3) \times (2\times4) \times (3\times5)}[/tex]

If the pattern continues, the next few terms are likely

[tex]\displaystyle a_4 = \frac1{8640} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6)} \\\\ a_5 = \frac1{302400} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6) \times (5\times7)} \\\\ a_6 = \frac1{14515200} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6) \times (5\times7) \times (6\times8)}[/tex]

which leads up to the n-th term,

[tex]\displaystyle a_n = \frac1{(1\times3) \times (2\times4) \times \cdots \times (n\times(n+2))} = \frac2{n!(n+2)!}[/tex]

where the numerator is multiplied by 2 in order to "complete" the factorial pattern in (n + 2)!.

So we have

[tex]\displaystyle y_1(x) = \sum_{n=0}^\infty \frac2{n!(n+2)!} x^n[/tex]

Now we use reduction of order to find a linearly independent solution of the form [tex]y_2(x) = v(x)y_1(x)[/tex], with derivatives

[tex]\displaystyle \frac{\mathrm dy_2}{\mathrm dx} = v(x) \frac{\mathrm dy_1}{\mathrm dx} + y_1(x) \frac{\mathrm dv}{\mathrm dx} \\\\ \frac{\mathrm d^2y_2}{\mathrm dx^2} = v(x) \frac{\mathrm d^2y_1}{\mathrm dx} + 2 \frac{\mathrm dv}{\mathrm dx} \frac{\mathrm dy_1}{\mathrm dx} + y_1(x) \frac{\mathrm d^2v}{\mathrm dx^2}[/tex]

Substitute [tex]y_2[/tex] and its derivatives into the DE, and simplify the resulting expression to get a DE in terms of v(x) :

[tex]\displaystyle x y_1 \frac{\mathrm d^2v}{\mathrm dx^2} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)\frac{\mathrm dv}{\mathrm dx} + \left(x\frac{\mathrm d^2y_1}{\mathrm dx^2}+3\frac{\mathrm dy_1}{\mathrm dx}-y_1\right)v = 0[/tex]

but since we know [tex]y_1(x)[/tex] satisfies the original DE, the last term vanishes and we're left with

[tex]\displaystyle x y_1 \frac{\mathrm d^2v}{\mathrm dx^2} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)\frac{\mathrm dv}{\mathrm dx} = 0[/tex]

Reduce the order by substituting [tex]w(x)=\dfrac{\mathrm dv}{\mathrm dx}[/tex] to get yet another DE in w(x) :

[tex]\displaystyle x y_1 \frac{\mathrm dw}{\mathrm dx} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)w = 0[/tex]

This equation is separable:

[tex]\displaystyle \frac{\mathrm dw}w = - \frac{2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1}{xy_1}\,\mathrm dx \\\\ \frac{\mathrm dw}w = -\left(\frac2{y_1}\frac{\mathrm dy_1}{\mathrm dx} + \frac3x\right)\,\mathrm dx[/tex]

From here you would integrate to solve for w(x), then integrate again to solve for v(x), and finally for [tex]y_2(x)[/tex] by multiplying [tex]y_1(x)[/tex] by v(x). Using the fundamental theorem of calculus, you would find

[tex]\displaystyle \ln|w| = -2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi - 3\ln|x| + C_1 \\\\ w = \frac{C_1}{x^3} \exp\left(-2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)} \,\mathrm d\xi\right)\right) \\\\ v = C_1 \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega + C_2[/tex]

so that you end up with

[tex]\displaystyle y_2(x) = C_1 y_1(x) \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega + C_2y_1(x)[/tex]

But the second term is already accounted for by [tex]y_1(x)[/tex] itself, so the second solution is

[tex]\displaystyle y_2(x) = \boxed{y_1(x) \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega}[/tex]

You could go the extra mile and try to find a power series expression for this solution, but that's a lot of work for little payoff IMO.

Answer:

THIS IS YOUR ANSWER:

☺✍️HOPE IT HELPS YOU ✍️☺

Answer:

5 m/s I hope it will help you

Explanation:

mark me as a brainlist answer

Answer:

first order date and most recent order date

Explanation:

it was switched. column 5 should be most recent order date because it's 2020 while column 6 should be first order date because it was in 2019

Explanation:

The heart rate of the astronaut is 78.5 beats per minute, which means that the time between heart beats is 0.0127 min. This will be the time t measured by the moving observer. The time t' measured by the stationary Earth-based observer is given by

[tex]t' = \dfrac{t}{\sqrt{1 - \left(\dfrac{v^2}{c^2}\right)}}[/tex]

a) If the astronaut is moving at 0.480c, the time t' is

[tex]t' = \dfrac{0.0127\:\text{min}}{\sqrt{1 - \left(\dfrac{0.2304c^2}{c^2}\right)}}[/tex]

[tex]\:\:\:\:=0.0145\:\text{min}[/tex]

This means that time between his heart beats as measured by Earth-based observer is 0.0145 min, which is equivalent to 69.1 beats per minute.

b) At v = 0.940c, the time t' is

[tex]t' = \dfrac{0.0127\:\text{min}}{\sqrt{1 - \left(\dfrac{0.8836c^2}{c^2}\right)}}[/tex]

[tex]\:\:\:\:=0.0372\:\text{min}[/tex]

So at this speed, the astronaut's heart rate is 1/(0.0372 min) or 26.9 beats per minute.

1. Based on Scenario A, multiple frames will minimize re-transmission overhead and should be preferred in the encapsulation of packets.

2. Based on Scenario B, the encapsulation of packets should be in a single frame because of the high level of network reliability and accuracy.

Justification:

There will not be further need to re-transmit the packets in a highly reliable and accurate network environment, unlike in an environment that is very prone to errors.  The reliable and accurate network environment makes a single frame economically better.

Encapsulation involves the process of wrapping code and data together within a class so that data is protected and access to code is restricted.

With encapsulation, each layer:

Thus, in a reliable and accurate network environment, single frames should be used to enhance transmission and minimize re-transmission overhead.  This is unlike in an environment that is very prone to errors, where multiple frames should rather be used to minimize re-transmission overhead.

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