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Answer:

the altitude of these satellites above the surface of the earth;

35790 km ( SI units )

22243.63 miles ( U.S. customary unit )

Explanation:

Given the data in the question;

Time taken by the satellite to complete on revolution is 23.934 hours

= 23.934 × 60 × 60 = 86162.4 seconds

now, let h represent altitude, r represent Orbit radius, v represent Orbit speed.

we know that

v² = GM/r

= gR²/r

= (9.81m/s² × ( 6.37 × 10⁶ m)²) / r

v = 19.95 × 10⁶ / √r   ---------- let this be equation

also;

time t = 2πr/v

86162.4 s = 2πr/v -------- let this be equation 2

a) Determine the altitude of these satellites above the surface of the earth in both SI and U.S. customary units

we substitute v in equation into equation 2

so

86162.4 s = 2πr / (19.95 × 10⁶ / √r)

r = 42.16 × 10⁶ m

so, altitude h = r - R

h = 42.16 × 10⁶ m - 6.37 × 10⁶ m

h = 35790000 m

convert to kilometer

h = 35790000 / 1000

h = 35790 km

Convert to miles

h = 35790 / 1.609

h = 22243.63 miles

Therefore, the altitude of these satellites above the surface of the earth;

35790 km ( SI units )

22243.63 miles ( U.S. customary unit )

Answer:

See explanation

Explanation:

From the formula;

0.693/t1/2 = 2.303/t log (Ao/A)

t1/2 = half life of Sodium-24

Ao = initial activity of Sodium-24

A= activity of Sodium-24 at time = t

So,

0.693/15 = 2.303/15 log (800/A)

0.0462 = 0.1535  log (800/A)

0.0462/0.1535 =  log (800/A)

0.3 = log (800/A)

Antilog(0.3) =  (800/A)

1.995 =  (800/A)

A = 800/1.995

A = 401 Bq

ii) 0.693/15 = 2.303/30 log (800/A)

0.0462 = 0.0768 log (800/A)

0.0462/0.0768 =  log (800/A)

0.6 =  log (800/A)

Antilog (0.6) =  (800/A)

3.98 =  (800/A)

A = 800/3.98

A = 201 Bq

iii)

0.693/15 = 2.303/45 log (800/A)

0.0462 = 0.0512  log (800/A)

0.0462/0.0512  =  log (800/A)

0.9 = log (800/A)

Antilog (0.9) =  (800/A)

7.94 = (800/A)

A = 800/7.94

A= 100.8 Bq

iv)

0.693/15 = 2.303/60 log (800/A)

0.0462 = 0.038 log (800/A)

0.0462/0.038 = log (800/A)

1.216 = log (800/A)

Antilog(1.216) = (800/A)

16.44 = (800/A)

A = 800/16.44

A = 48.66 Bq

Answer:

69.7 cm

Explanation:

What is the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe?

For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m, m = order of fringe and λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.

Also, tanθ = x/D where x = distance of nth order fringe from central maximum and D = distance of screen from grating = 115 cm = 1.15 m

Now sinθ = d/mλ, Since θ is small, sinθ ≅ tanθ

So, d/mλ = x/D for a second order bright fringe, m = 2.

So, d/2λ = x/D

x = dD/2λ

So, x =

For a dark fringe, we have

d/(m + 1/2)λ = x'/D where x' is the distance of the fringe from the central maximum.

For a second-order dark fringe, m = 2. So,

d/(2 + 1/2)λ = x'/D

d/(5/2)λ = x'/D

2d/5λ = x'/D

x' = 2dD/5λ

So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is x" = dD/2λ - 2dD/5λ

x" = dD/10λ

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

x"= 1/3 × 10⁻⁵ m × 1.15 m/(10 × 550 × 10⁻⁹ m)

x" = 1.15/165 × 10² m

x" = 0.00697 × 10² m

x" = 0.697 m

x" = 69.7 cm

The distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe 69.7 cm

Diffraction of light occurs when a light wave passes by a corner or through an opening or slit that is physically the approximate size of, or even smaller than that light's wavelength

For a diffraction grating, dsinθ = mλ

where,

d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m,

m = order of fringe  

λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.

Also,   [tex]tan\theta=\dfrac{x}{D}[/tex]t   where

x = distance of nth order fringe from central maximum  

D = distance of screen from grating = 115 cm = 1.15 m

Now   [tex]Sin\theta =\dfrac{d}{m\lambda}[/tex] ,  Since θ is small, sinθ ≅ tanθ

So, [tex]\dfrac{d}{m\lambda}=\dfrac{x}{D}[/tex]

for a second order bright fringe, m = 2.

So,   [tex]\dfrac{d}{2\lambda}=\dfrac{x}{D}[/tex]

[tex]x=\dfrac{dD}{2\lambda}[/tex]

For a dark fringe, we have

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

where x' is the distance of the fringe from the central maximum.

For a second-order dark fringe, m = 2. So,

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

[tex]\dfrac{d}{ \dfrac{5}{2}\lambda}=\dfrac{X'}{D}[/tex]

[tex]X'=\dfrac{2dD}{5\lambda}[/tex]

So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is

[tex]X''=\dfrac{dD}{2\lambda}-\dfrac{2dD}{5\lambda}[/tex]

[tex]X''=\dfrac{dD}{10\lambda}[/tex]

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

[tex]x''=\dfrac{1}{3\times10^{-5}}\times \dfrac{1.15}{10\times 550\times 10^{-9}}[/tex]x

x" = 0.00697 × 10² m

x" = 0.697 m

x" = 69.7 cm

Hence the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe 69.7 cm

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Answer:

(a) The volume of the liquid helium at 25 K is 5.13 L

(b)  The volume of the liquid helium at 293 K is 60.14 L.

Explanation:

Given;

mass of the liquid helium, m = 10 g

initial temperature of the liquid helium, T₁ = 4.2 K

pressure of the liquid helium, P = 1.00 atm

Atomic mass of Helium, = 4 g

number of moles of Helium, n  = 10 / 4 = 2.5 moles

The initial volume of the liquid helium is calculated as;

[tex]PV_1 = nRT_1\\\\V_1 = \frac{nRT_1}{P} \\\\[/tex]

where;

R is ideal gas constant,  = 0.08205 L.atm./mol.K

[tex]V_1 = \frac{2.5 \times 0.08205 \times 4.2}{1 } \\\\V_1 = 0.862 \ L[/tex]

(a) The volume of the liquid helium at 25 K.

Apply Charles law;

[tex]\frac{V_1}{T_1} =\frac{V_2}{T_2} \\\\V_2 = \frac{V_1T_2}{T_1} \\\\V_2 = \frac{0.862 \times 25 }{4.2} \\\\V_2 = 5.13 \ L[/tex]

(b)  The volume of the liquid helium at 293 K.

[tex]\frac{V_1}{T_1} =\frac{V_2}{T_2} \\\\V_2 = \frac{V_1T_2}{T_1} \\\\V_2 = \frac{0.862 \times 293 }{4.2} \\\\V_2 = 60.14 \ L[/tex]

Explanation:

Given that,

Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 570 N/C.

We need to find their speeds after 47.6 ns.

For electron,

The electric force is given by :

[tex]F=qE\\\\F=1.6\times 10^{-19}\times 570\\\\=9.12\times 10^{-17}\ N[/tex]

Let a be the acceleration of the electron. So,

F = ma

m is mass of electron

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{9.12\times 10^{-17}}{9.1\times 10^{-31}}\\\\a=10^{14}\ m/s^2[/tex]

Let v be the final velocity of the electron. So,

v = u +at

u = 0 (at rest)

So,

[tex]v=10^{14}\times 47.6\times 10^{-9}\\\\v=4.76\times 10^6\ m/s[/tex]

For proton,

Acceleration,

[tex]a=\dfrac{9.12\times 10^{-17}}{1.67\times 10^{-27}}\\\\=5.46\times 10^{10}\ m/s^2[/tex]

Now final velocity of the proton is given by :

[tex]v=5.46\times 10^{10}\times 47.6\times 10^{-9}\\\\v=2598.96\ m/s[/tex]

Hence, this is the required solution.

Answer:

pretty sure A is correct, as they are pushing in opposite directions, hence canceling each other

Explanation:

Answer:

See explanation

Explanation:

Since the original count rate is 600 Bq,

i) after 1 half life, the count rate decreases to 1/2 of 600 Bq = 300 Bq

ii) after 2 half lives, the count rate decreases to 1/4 of 600 = 150 Bq

iii) after 3 half lives, the count rate decreases to 1/6 of 600 = 100 Bq

iv) after 4 half lives, the count rate decreases to 1/8 of 600 = 75 Bq

I'm assuming the question is supposed to be like this:

How  does the Orbital velocity of a satellite depends on the mass of the satellite?

Answer

It is independent of mass of satellite.

Answer:

Are you sure it was soccer ball? Or meine hearts

Explanation:

Answer:

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

Answer:

Uhh 2 one

Explanation

Answer:

the kinetic energy gained by the toy is 6J.

Explanation:

Given;

net applied to the toy by dog, F = 2 N

distance moved by the toy, d = 3 m

Apply the principle of work-energy theorem to determine the kinetic energy gained by the toy.

ΔK.E = W

         = F x d

         = 2 x 3

         = 6 J

Therefore, the kinetic energy gained by the toy is 6J.

Answer:

IMA = 8

Explanation:

Given the following data:

Length of plank = 160cm

Length of resistance = 20cm

To find the ideal mechanical advantage (IMA);

[tex] IMA= \frac {length \; of \; effort}{length \; of \; resistance} [/tex]

[tex] IMA= \frac {160}{20} [/tex]

IMA = 8

Answer:

I think it’s c

Explanation:

Answer:

a. spring constant = 125 N/m

b. This spring is less stiff than the one in Sample Problem A.

Explanation:

P.S - The Sample Problem A is as follows :

Given - Sample Problem A - A load of 45 N attached to a spring that is

            hanging vertically stretches the spring 0.14 m. What is the spring

             constant?

            Suppose the spring in Sample Problem A is replaced with a spring  

            that stretches 36 cm from its equilibrium position.

To find - a. What is the spring constant in this case?

            b. Is this spring stiffer or less stiff than the one in Sample Problem A.

Proof -

As given,

Load = 45 N

Amplitude = 0.14 m

Let the spring constant = k

As we know that,

Load = k (Amplitude)

⇒45 = k(0.14)

⇒k = [tex]\frac{45}{0.14}[/tex] = 321.43

∴ we get

Spring constant in Sample problem A = 321.43

Now,

a.)

Given, Amplitude = 36 cm = 0.36 m

Let the spring constant = k₁

⇒45 = k₁ (0.36)

⇒k₁ = [tex]\frac{45}{0.36}[/tex] = 125 N/m

b.)

AS we can see that k₁  < k

⇒ This spring is less stiff than the one in Sample Problem A.

a. spring constant = 125 N/m

b. This spring is less stiff than the one in Sample Problem A.

The spring constant generally shows the stiffness of the spring and is the ratio of the force applied to the deflection of the spring.

It is given in the question that:

Sample Problem A - A load of 45 N attached to a spring that is

hanging vertically stretches the spring 0.14 m.  

Suppose the spring in Sample Problem A is replaced with a spring  

that stretches 36 cm from its equilibrium position.

a. What is the spring constant in this case?

As given,

Load F = 45 N

Amplitude  x= 0.14 m

Let the spring constant = k

As we know that spring force will be

[tex]F=k\times x[/tex]

[tex]45=k\times 0.14[/tex]

⇒k = 321.43N/m

∴ we get

Spring constant in Sample problem A is   [tex]k=321.43\ \frac{N}{m}[/tex]

Now,

Given, Amplitude = 36 cm = 0.36 m

Let the spring constant = k₁

⇒45 = k₁ (0.36)

⇒k₁= 125 N/m

b.) Is this spring stiffer or less stiff than the one in Sample Problem A.

AS we can see that k₁  < k  new spring is less stiff than the one in Sample Problem A.

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Answer:

The force must be applied during 8 seconds to reach trhe same final speed.

Explanation:

By Impulse Theorem, a change in the magnitude of linear momentum of a system with constant mass can be done by applying a force during a given time. That is:

[tex]m\cdot (v_{f}-v_{o}) = F \cdot \Delta t[/tex]

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speed, measured in meters per second.

[tex]F[/tex] - Net external foce, measured in newtons.

[tex]\Delta t[/tex] - Time, measured in seconds.

We can eliminate mass and speeds by constructing the following relationship:

[tex]F_{1}\cdot \Delta t_{1} = F_{2}\cdot \Delta t_{2}[/tex] (2)

If we know that [tex]F_{1} = F[/tex], [tex]\Delta t_{1} = 4\,s[/tex] and [tex]F_{2} = \frac{F}{2}[/tex], then the time is:

[tex]4\cdot F = \frac{F\cdot \Delta t_{2}}{2}[/tex]

[tex]\Delta t_{2} = 8\,s[/tex]

The force must be applied during 8 seconds to reach trhe same final speed.

Answer:

The electric potential is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field with negligible acceleration of the test charge to avoid producing kinetic energy or radiation by test charge.

SI unit: volt

Other units: statvolt

In SI base units: V = kg⋅m^2⋅A^−1⋅s^−3

Dimension: M L^2 T^−3 I^−1

Explanation:

Hope it is helpful....

Explanation:

common symbol - V

SI unit - volt

other unit - statvolt

Answer:

Lol

Explanation:

Answer:

T = 0.375 s, f = 2.66 Hz and ω = 16.71 rad/s

Explanation:

Given that,

The mass of a harmonic oscillator, m = 0.5 kg

The force constant of the spring, k = 140 N/m

The frequency of a harmonic oscillator is given by :

[tex]f=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}[/tex]

Substitute all the values,

[tex]f=\dfrac{1}{2\pi }\sqrt{\dfrac{140}{0.5}} \\\\f=2.66\ Hz[/tex]

Time period is given by :

[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{2.66}\\\\T=0.375\ s[/tex]

The angular frequency is given by :

[tex]\omega=2\pi f\\\\\omega=2\pi \times 2.66\\\\\omega=16.71\ rad/s[/tex]

Hence, this is the required solution.

Answer:

    d₃ = 37,729 km,     θ=  5.1º North of West

Explanation:

This is a velocity addition problem, the easiest way to solve it is to decompose the velocities in a Cartesian system, the x-axis coincides with the West-East direction and the y-axis with the South-North direction

* first displacement is

           d₁ₓ = 11 km

* second offset is

          cos 6 = d₂ₓ / d₂

          sin 6 = d_{2y} / d₂

          d₂ₓ = d₂ cos 6

          d_{2y} = d₂ sin 6

          d₂ₓ = 6 cos 6 = 5.967 km

          d_{2y} = 6 sin 6 = 0.6272 km

* third displacement is unknown

* fourth and last displacement

          cos (-11) = d₄ₓ / d₄

          sin (-11) = d_{4y} / d₄

          d₄ₓ = d₄ cos (-11)

          d_{4y} = d₄ sin (-11)

          d₄ₓ = 21 cos (-11) = 20.61 km

          d_{4y} = 21 sin (-11) = -4.007 km

They tell us that at the end of the tour you are back on the island, so the displacement must be zero

X axis

           x = d₁ₓ + d₂ₓ + d₃ₓ + d₄ₓ

           0 = 11 +5.967 + d₃ₓ + 20.61

           d₃ₓ = -11 - 5.967 - 20.61

           d₃ₓ = -37.577 km

Y axis  

          y = d_{1y} + d_{2y} + d_{3y} + d_{4y}

          0 = 0 + 0.6272 + d_{3y} -4.007

          d_{3y} = 4.007 - 0.6272

          d_{3y} = 3.3798 km

This distance can be given in the form of module and angle

Let's use the Pythagorean theorem for the module

           d₃ = [tex]\sqrt{d_{3x}^2 + d_{3y}^2}[/tex]

           d₃ = [tex]\sqrt{37.577^2 + 3.3798^2}[/tex]

           d₃ = 37,729 km

Let's use trigonometry for the angle

            tan θ = d_{3y} / d₃ₓ

            θ = tan⁻¹ [tex]\frac{d_{3y}}{d_{3x}}[/tex]

            θ = tan-1 (-3.3798 / 37.577)

            θ = 5.1º

Since the y coordinate is positive and the x coordinate is negative, this angle is in the second quadrant, so the direction given in the form of cardinal coordinates is

            θ=  5.1º North of West

Is earth’s freshwater

Only about 3 percent of Earth’s water is fresh water.

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

Answer:

Mass = 1133.33 kg (Approx.)

Explanation:

Given:

Momentum = 2.04 x 10⁴ kg[m/s]

Velocity = 18 m/s

Find:

Mass

Computation:

Mass = Momentum / Velocity

Mass = [2.04 x 10⁴] / 18

Mass = 1133.33 kg (Approx.)

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: