Computer use ___code to transmit information

570k or 1050k

270k+290k= 570k

270k•290k = 1050k

Answer:

a.49 n

b. 63 n

c. 112 n

Explanation:

a.10 times 9.8 from gravity/2 = 49 n

b. 49n times 4.5/8-4.5 = 63 n

c 49n + 63 n = 112 n

Answer:

Answer: Heat energy transfer by conduction, convection and radiation

Heat energy is a very difficult energy to store as it can transfer in three different ways from warm surroundings to cooler surroundings. The three processes are conduction, convection or radiation.

Understanding energy, how it is transferred and how the amount of energy that is usefully transferred can be used to measure the efficiency is very important to physics and to the world.

Answer:

Solid is the state in which Matter maintains a fixed volume

Answer:

The state of matter that has a fixed volume is Solid.

Explanation:

Solid substances will maintain a fixed volume and shape.

Answer:

The answer is "Option D".

Explanation:

Please find the complete question in the attached file.

As plate separation increased to 2d the capacitance get halred but the change remain same

[tex]\therefore V=\frac{Q}{C}[/tex]

The voltage doubles are now electric field remain same because both the distance and voltage get doubled.

[tex]\to E=\frac{v}{d}\ = \frac{2v}{2d}\\\[/tex]

So,

[tex]energy=\frac{1}{2}\ \frac{Q^2}{C}\\\\c'=\frac{C}{2}\\\\E'=2E[/tex]

Answer:

Option (a).

Explanation:

Let the angular velocity is w.

The centripetal acceleration is given by

[tex]a = r w^2[/tex]

where, r is the distance between the axle and the spoke.

So, more is the distance more is the centripetal acceleration.

(a) For the points on the spoke closer to the hub than for points closer to the rim is larger distance, so the centripetal force is more.

The statement is true.  

(b) The direction of centripetal acceleration is always towards the center, so the statement is false.

(c) It is false.

(d) It is false.

Option (a) is correct.

Answer:

The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

Explanation:

Given;

magnitude of the attractive force, F = 17 mN = 0.017 N

distance between the two objects, r = 24 cm = 0.24 m

The attractive force is given by Coulomb's law;

[tex]F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C[/tex]

The charge of 1 electron = 1.602 x 10⁻¹⁹ C

n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷

[tex]n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons[/tex]

Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

Answer:

C. 6.9 watts

Explanation:

Power = work/time

if work = force×distance...

Then... power= (force×distance)/time

Power = (20×12)/35

= 6.9 watts

Answer:

C the same as pressure on top

Answer:

The required wavelength is 1.19 μm

Explanation:

In the double-slit study, the formula below determines the position of light fringes [tex]y_m[/tex] on-screen.

[tex]y_m = \dfrac{m \lambda D}{d}[/tex]

where;

m = fringe order

d = slit separation

λ = wavelength

D = distance between screen to the source

For the first bright fringe, m = 1, and we make (d) the subject, we have:

[tex]d = \dfrac{(1) \lambda D}{y_1}[/tex]

[tex]d = \dfrac{ \lambda D}{y_1}[/tex]

replacing the value from the given question, we get:

[tex]d = \dfrac{ (597 \ nm )\times (3.00 \ m)}{4.84 \ mm} \\ \\ d = \dfrac{ (597 \ nm \times (\dfrac{1 \ m}{10^9\ nm}) )\times (3.00 \ m)}{4.84 \ mm(\dfrac{1 \ m}{1000 \ mm })} \\ \\ d = 3.7 \times 10^{-4} \ m[/tex]

In the double-slit study, the formula which illustrates the position of dark fringes [tex]y_m[/tex] on-screen can be illustrated as:

[tex]y_m = (m+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]

The value of m in the dark fringe first order = 0

∴

[tex]y_0 = (0+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]

[tex]y_0 = (\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]

making λ the subject of the formula, we have:

[tex]\lambda = \dfrac{2y_o d}{D} \\ \\ \lambda = \dfrac{2(4.84 \ mm) \times \dfrac{1 \ m}{1000 \ mm} (3.7 \times 10^{-4} \ m) }{3.00 \ m}[/tex]

[tex]\lambda = 1.19 \times 10^{-6} \ m ( \dfrac{10^6 \mu m }{1\ m}) \\ \\ \lambda = 1.19 \mu m[/tex]

Answer:

Δd = 7.22 10⁻² m

Explanation:

For this exercise we must use the dispersion relationship of a diffraction grating

           d sin θ = m λ

let's use trigonometry

           tan θ = y / L

how the angles are small

           tant θ = sinθ  /cos θ = sin θ

we substitute  

           sin θ = y / L

          d y / L = m λ

          y = m λ L / d

let's use direct ruler rule to find the distance between two slits

If there are 500 lines in 1 me, what distance is there between two lines

         d = 2/500

        d = 0.004 me = 4 10⁻⁶ m

diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1

let's calculate for each wavelength

λ = 656 nm = 656 10⁻⁹ m

         d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶

         d₁ = 2.788 10⁻¹ m

λ = 486 nm = 486 10⁻⁹ m

         d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶

         d₂ = 2.066 10⁻¹ m

the distance between the two lines is

         Δd = d1 -d2

         Δd = (2,788 - 2,066) 10⁻¹

         Δd = 7.22 10⁻² m

Answer:

W = 354.9 J

Explanation:

Given that,

The mass of a bale of hay, m = 23 kg

The displacement, d = 3.9 m

The horizontal force exerted on the hay, F = 91 N

We need to find the work done. We know that,

We know that,

Work done, W = Fd

So,

W = 91 N × 3.9 m

W = 354.9 J

So, the required work done is 354.9 J.

Answer:

The wavelength of the wave is 1.419 cm.

Explanation:

wavelength emitted by the one micro wave = 1.41 cm

beat frequency =  170 M Hz

Let the frequency of the wave is f.

[tex]f=\frac{c}{\lambda }\\\\f =\frac{3\times 10^8}{0.0141}\\\\f = 2.13\times 10^{10} Hz[/tex]

Let the frequency of the other wave  is f'.

[tex]f' = 2.13\times 10^{10}-170\times10^6\\\\f' = 21130\times 10^6 Hz[/tex]

The wavelength is given by

[tex]f' = 21130\times 10^6 Hz\\\lambda = \frac{3\times 10^8}{21130\times 10^6}\\\\\lambda = 0.01419 m = 1.419 cm[/tex]

Answer:

The product of mass and velocity is the correct answer

Explanation:

Momentum is defined as mass × velocity

p = mv

Answer:

The product of mass and velocity

Explanation:

I  just did it and got it right with a 100%

Posted 1/3/23

Answer:

T = 188.5 s, correct is  C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

initial instant. Before the crash

        L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

        L_f = I₀ w + m r v

        L₀ = L_f

        m r v₀ = I₀ w + m r v

angular and linear velocity are related

        v = r w

        w = v / r

        m r v₀ = I₀ v / r + m r v

         m r v₀ = (I₀ / r + mr) v

       v = [tex]\frac{m}{\frac{I_o}{r} +mr} \ r v_o[/tex]

let's calculate

       v = [tex]\frac{0.020}{\frac{1.4}{0.6 } + 0.020 \ 0.6 } \ 0.6 \ 4[/tex]

       v = [tex]\frac{0.020}{2.345} \ 2.4[/tex]

       v = 0.02 m / s

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

        v = x / T

         T = x / v

the distance of a circle with radius r = 0.6 m

         x = 2π r

we substitute

         T = 2π r / v

let's calculate

         T = 2π 0.6/0.02

         T = 188.5 s

reduce

         t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is  C

Answer:

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Answer:

a. 59 ev. helpful answer

Answer:

Explanation:

From the given information:

a)

Let's have an imaginary view of the rod located at a given distance r from he the mass (m) of the sphere.

Then the equation for the potential energy as related to the small area of the dr of the rod can be computed as:

[tex]dU = -\dfrac{GMm}{L}*\dfrac{dr}{r}[/tex]

where,

G = gravitational constant

[tex]U = - \int^{x+L}_{x}Gm\dfrac{M}{L}*\dfrac{dr}{r}[/tex]

[tex]U = - \dfrac{GMm}{L}\int^{x+L}_{x}\dfrac{dr}{r}[/tex]

By taking the integral within the limit

[tex]U = - \dfrac{GMm}{L} \Big[In \ r\Big]^{x+L}_{x}[/tex]

[tex]\mathbf{\implies - \dfrac{GmM}{L} In \Big(\dfrac{{x+L}}{{x}}\Big)}[/tex]

b)

By using [tex]F= -\dfrac{dU}{dx}[/tex], the magnitude of the gravitational force can be determined as follows:

Here, we have:

[tex]F = -\dfrac{d}{dx} \Big [\dfrac{-GmM}{L}In(\dfrac{x+l}{x}) \Big ] \\ \\ = \dfrac{GmM}{L}\times \dfrac{x}{x+L}\times (0-\dfrac{L}{x^2}) \\ \\ By \ solving \\ \\ \mathbf{ =-\dfrac{GmM}{x(x+L)}}[/tex]

From above, the negative sign indicates an attractive force

Answer:

I don't see anything on your question?

False

It's not straight up

Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also exerts an equal and opposite force on object A. In other words, forces result from interactions.

Hope it helps

Answer:

Newton's third law state that every action there is equal but opposite reactions

hope this will help you more

Answer:

the mass of the bullet is 10.5 g

Explanation:

Given;

initial velocity, u₁ = 280 m/s

final velocity of the bullet, v₁ = 70 m/s

final velocity of the block, v₂ = 0.2 m/s

mass of the block, m₂ = 11 kg

initial velocity of the block, u₂ = 0

let the mass of the bullet = m₁

Apply the principle of conservation of linear momentum for elastic collision to calculate the mass of the bullet.

m₁u₁ + m₂u₂ = m₁v₁  + m₂v₂

280m₁  +  11(0)  = 70m₁  +  11 x 0.2

280m₁ = 70m₁  + 2.2

280m₁ - 70m₁  = 2.2

210m₁ = 2.2

m₁ = 2.2/210

m₁ = 0.0105 kg

m₁ = 10.5 g

Therefore, the mass of the bullet is 10.5 g

Answer:

Explanation:

There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.

Explanation:

Specifically his leg muscles. As the leg muscles expand, they push down on the ground. Newton's 3rd law says that for any action, there's an opposite and equal reaction. That means a downward push into the ground will have the ground push back, more or less, and that's why the kangaroo will jump. The ground (and the earth entirely) being much more massive compared to the animal means that the ground doesn't move while the kangaroo does move. Perhaps on a very microscopic tiny level the ground/earth does move but it's so small that we practically consider it 0.

This experiment can be done with a wall as well. Go up to a wall and lean against it with your hands. Then do a pushup to move further away from the wall, but you don't necessarily need to lose contact with the wall's surface. As you push against the wall, the wall pushes back, and that causes you to move backward. If the wall was something flimsy like cardboard, then you could easily push the wall over and you wouldn't move back very much. It all depends how much mass is in the object you're pushing on.  

Answer:

[tex]a=0.2*10^{-5}g[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=180=>0.18kg[/tex]

Charge [tex]Q=18mC=18*10^-^3C[/tex]

Velocity [tex]v=2.2m/s[/tex]

Length of Wire [tex]L=8.6cm=>0.086[/tex]

Current [tex]I=30A[/tex]

Generally the equation for Magnetic Field of Wire B is mathematically given by

 [tex]B=\frac{\mu_0*I}{2\pi*l}[/tex]

 [tex]B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}[/tex]

 [tex]B=6.978*10^{-5}T[/tex]

Generally the equation for Force on the plane F is mathematically given by

 [tex]F=qvB[/tex]

Therefore

 [tex]ma=qvB[/tex]

 [tex]a=\frac{qvB}{m}[/tex]

 [tex]a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}[/tex]

 [tex]a=2.37*10^{-5}[/tex]

Therefore in Terms of g's

 [tex]a=\frac{2.37*10^{-5}}{9.8}[/tex]

 [tex]a=0.2*10^{-5}g[/tex]

Answer:

The required ratio is 1.99.

Explanation:

We need to find the atio of speeds of a proton and an alpha particle accelerated through the same voltage.

We know that,

[tex]eV=\dfrac{1}{2}mv^2[/tex]

The LHS for both proton and an alpha particle is the same.

So,

[tex]\dfrac{v_p}{v_a}=\sqrt{\dfrac{m_a}{m_p}} \\\\\dfrac{v_p}{v_a}=\sqrt{\dfrac{6.64\times 10^{-27}}{1.67\times 10^{-27}}} \\\\=1.99[/tex]

So, the ratio of the speeds of a proton and an alpha particle is equal to 1.99.