(e) Find the height and orbital velocity of the geostationary satellite above the earth assuming earth as a sphere of radius 6370 km. [Radius of earth, R = 6370km, Acceleration due to gravity
= 10ms) ​

Answer:

It was proposed by Isaac Newton

Explanation:

The law of universal attraction of expression

          F = [tex]G \ \frac{m_1m_2}{ r^2}[/tex]G m1m2 / r ^ 2

where G is a constant, m₁ and m₂ are the masses of the bodies and r the distance between them.

It was proposed by Isaac Newton

With this law Newton explained that the force that pulls the moon towards the earth is the same as that which attracts an apple towards the earth

Answer:

R = 0.86 m

Explanation:

The formula for the range of the projectile motion can be used here:

[tex]R = \frac{v^2 Sin2\theta}{g}[/tex]

where,

R = Range of projectile = distance between Jake and Fred = ?

v = launch speed = 3 m/s

θ = Launch Angle = 35°

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]R = \frac{(3\ m/s)^2Sin[(2)(35^o)]}{9.81\ m/s^2}\\\\[/tex]

R = 0.86 m

Answer: Vehicle Balance

Explanation:

Vehicle balance simply means how the weight of a vehicle is distributed across its tires which connects to the road

It should be noted that there'll be a shift in the balance of a vehicle when the braking, acceleration, or turning bring about a scenario in which the weight of th vehicle moves from one area to another.

Answer:

31,556,736 seconds

Explanation:

The period of Earth's orbit around the Sun, T = (365 + 1/4) days (three hundred and sixty five and a quarter days

The period of Earth's orbit around the Sun in seconds is given by converting, T = [tex]365\frac{1}{4} \, days[/tex] into seconds as follows;

There are 60 seconds per minute, 60 minutes in each hour, 24 hours within each day, therefore, we get;

T = [tex]365\frac{1}{4}[/tex]×60 sec/min × 60 min/hr. × 24hr/day = 31,556,736 seconds

400 m2

Explanation:

Pressure = Force ÷ Area

5 N/m2 = 2000 N ÷ A

A = 2000 N ÷ 5

= 400 m2

If a force of 2000N is applied, the area that the force is applied to is 400 m²

The word "force" has a specific meaning in science. At this level, calling a force a push or a pull is entirely appropriate. A force is not something an object "has in it" or that it "contains."

One thing experiences a force from another. There are both living things and non-living objects in the concept of a force.

The amount of force applied to a certain region is referred to as pressure. The force per unit area is called pressure. F in this condensed version of the equation stands in for the force, which is expressed in newtons.

Given that the pressure of 5 N/m²

Force is 2000N

Pressure = Force ÷ Area

5 N/m² = 2000 N ÷ A

A = 2000 N ÷ 5 = 400 m²

Therefore, the area that the force is applied to is 400 m².

To learn more about force, refer to the link:

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Answer:

Matter is made up of atoms

Answer:

Mater is made up of atoms.

Explanation:

Atoms come together to form molecules,which are the building blocks for all types of matter.

Answer:

The mean is 79.[tex]\overline {54}[/tex]

The median is 80 - 100

The mode is 80 - 100

Explanation:

The given table is presented as follows;

[tex]\begin{array}{lcrc}Classes&Mid \ point &Frequency &Frequency \times Mid \ point\\0 - 20&10& 2&20\\20-40&30&2&60\\40-60&50&3&150\\60-80&70&12&840\\80-100&90&18&1620\\100-120&110&5&550\\120-140&130&2&260\end{array}[/tex]

The mean of a class of values, [tex]\overline x[/tex] = ∑(Frequency × Midpoint)/∑(Frequency)

Therefore, we get;

[tex]\overline x[/tex] = (20+60+150+840+1620+550+260)/(2+2+3+12+18+5+2) = 79.[tex]\overline {54}[/tex]

The mean, [tex]\overline x[/tex] =79.[tex]\overline {54}[/tex]

The median class = The middle value lass = The class at the 22 nd value = 80 - 100

The median = 80 - 100

The modal class = The class with the highest frequency = 80 - 100

The mode = 80 - 100

Answer:

a)   t = 20 s,  b)  x = 1000 m, As the runway is only 800 m long, the plane cannot land at this distance

Explanation:

This is a kinematics exercise

a) in minimum time to stop,

           v = vo + at

           v = 0

           t = -v0 / a

we calculate

          t = -100 / (5.00)

          t = 20 s

b) Let's find the length you need to stop

          v² = vo² + 2 a x

          x = -v0 ^ 2 / 2a

          x = - 100² / 2 (-5.00)

          x = 1000 m

As the runway is only 800 m long, the plane cannot land at this distance.

Answer:

asteroids

Explanation:

There is an asteroid belt between Mars and Jupiter

Answer:

4p

Explanation:

If you halve the volume the pressure will double as they are inversely proportional. If you double the temperature the particles have double the kinetic energy so the pressure will double again.

So:

p×2×2 = 4p

Answer:

P V = n R T

P2 V2 / (P1 V1) = T2 / T1

P2 = (T2 / T1) (V1 / V2)  P1   =   2 * 2 = 4

Answer:   A

Explanation:

:)

Explanation:

Obviously vector quantity

Answer:

True

Explanation:

Formula for speed is;

Speed = distance/time

Now, if an object covers an equal distance in equal time intervals, it means the speed will remain the same.

For example if an object covers 3 m every 1 second it means speed will always be; 3/1 = 3 m/s.

Thus the statement is correct.

Scalar quantity are physical quantities that have just magnitude, not direction.

Answer:

I = 0.2 A

Explanation:

Lamp is rated at 300 mA

I_lamp = 0.3 A

Voltage is; V = 3V

Thus; Resistance is given by;

R = V/I

R = 3/0.3

R = 10 ohms

Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;

R_eq = 10 + 5

R_eq = 15 ohms

Ammeter current will be;

I = V/R_eq

I = 3/15

I = 0.2 A

Answer:

Explanation:

Mass doesn't matter here because when something is falling, gravity plays fairly; an elephant falls at the same rate of acceleration as does a feather. What DOES matter is everything pertinent to the y-dimension of free-fall:

a = -9.8 m/s/s

v₀ = 0 (since the ball was held before it was dropped)

v = ??

Δx = -8 m (negative because the ball drops this far below the point from which it was released).

Putting all this together in one equation:

v² = v₀² + 2aΔx and filling in this equation:

v² = (0)² + 2(-9.8)(-8) and

v² = 156.8 so

v = 12.5 which rounds to 13 if you're using 2 sig figs, and rounds to 10 if you're only using 1 (which you should be, according to the way the numbers have been given in this problem)

Answer:

For smooth surface:The light rays bounce off the table and all move in the same direction.

Answer:

  [tex]M = \left[\begin{array}{ccc}cos \ 60&0\\0&-sin \ 60\end{array}\right][/tex]

Explanation:

To find the matrix, let's decompose the vectors, the rotated angle is (-60C) for the prime system

          x ’= x cos (-60)

          y ’= y sin (-60)

we use

          cos 60 = cos (-60)

          sin 60 = - sin (-60)

we substitute

          x ’= x cos 60

          y ’= - y sin 60

the transformation system is

         [tex]\left[\begin{array}{ccc}x'\\y'\end{array}\right] = \left[\begin{array}{ccc}cos 60&0\\0&-sin60\end{array}\right] \ \left[\begin{array}{ccc}x\\y\end{array}\right][/tex]x '

the transformation matrix is

       [tex]M = \left[\begin{array}{ccc}cos \ 60&0\\0&-sin \ 60\end{array}\right][/tex]

When fundamental units are combined, they result in derived units. RC (which means Resistance Capacitance) is a derived unit and its unit in the simplest form is Coulomb per Ampere (C/A)

Given that:

Capacitance (C) [tex]\to[/tex] Farads (f)

and

[tex]1f = 1\frac CV[/tex] ----- 1 farad = 1 capacitance per volt

Resistance (R) [tex]\to[/tex] Ohms [tex]\Omega[/tex]

[tex]1 \Omega = 1\frac{V}{A}[/tex]

The unit of RC is the product of the unit of R by the unit of C.

i.e.

[tex]RC = 1f \times 1\Omega[/tex]

Substitute [tex]1f = 1\frac CV[/tex]

[tex]RC = 1\frac CV \times 1\Omega[/tex]

Substitute [tex]1 \Omega = 1\frac{V}{A}[/tex]

[tex]RC = 1\frac CV \times 1\frac VA[/tex]

Cancel out volts (V)

[tex]RC = 1\frac CA[/tex]

[tex]\frac CA[/tex] means Coulomb per Ampere

Hence, the unit of RC is Coulomb per Ampere.

Read more about units at:

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Complete question is;

Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m each. If Margy is walking at 1.4 m/s and accelerates at 0.20 m/s² during one of the running portions, what is her final velocity at the end of the 100.0 m? Round your answer to the nearest tenth.

Answer:

6.5 m/s

Explanation:

We are told that she is walking at 1.4 m/s and accelerates at 0.20 m/s².

Thus;

Initial velocity; u = 1.4 m/s

Acceleration; a = 0.2 m/s²

Distance; s = 100 m

From Newton's equation of motion, we know that;

v² = u² + 2as

Where v is final velocity.

Thus;

v² = 1.4² + 2(0.2 × 100)

v² = 41.96

v = √41.96

v ≈ 6.5 m/s

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

[tex]F_{g0}=G \frac{mM_{E0}}{r^{2}}[/tex]

So if the mass of the earth is increased by a factor of 2, this means that:

[tex]M_{Ef}=2M_{E0}[/tex]

so:

[tex]F_{gf}=G \frac{2mM_{E0}}{r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=2[/tex]

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

[tex]M_{Ef}=3M_{E0}[/tex]

so:

[tex]F_{gf}=G \frac{3mM_{E0}}{r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=3[/tex]

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

[tex]M_{Ef}=\frac{M_{E0}}{4}[/tex]

so:

[tex]F_{gf}=G \frac{mM_{E0}}{4r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{1}{4}[/tex]

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

Answer:

[tex]K.E = 0.1905 J[/tex]

Explanation:

From the question we are told that:

Length [tex]L=1.4m[/tex]

Mass [tex]m=180g[/tex]

Angular Velocity [tex]\omega=1.80rads/s[/tex]

Generally the equation for Kinetic energy K.E is mathematically given by

 [tex]K.E =0.5 (1/3 ML^2 )w^2[/tex]

 [tex]K.E =0.5 ( 1/3 * 0.18 * 1.4^2 ) 1.8^2[/tex]

 [tex]K.E = 0.1905 J[/tex]

Answer:

[tex]W=7.56\times 10^{-19}\ J[/tex]

Explanation:

Given that,

The work function for silver is 4.73 eV.

We need to find the value of the work function from electron volts to joules.

We know that,

[tex]1\ eV=1.6\times 10^{-19}\ J[/tex]

For 4.73 eV,

[tex]4.73\ eV=1.6\times 10^{-19}\times 4.73\\\\=7.56\times 10^{-19}\ J[/tex]

So, the work function for silver is [tex]7.56\times 10^{-19}\ J[/tex].

[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]

[tex]\sf What \: is \: acceleration? \\ \\ \sf The \: rate \: of \: change \: of \: velocity \: of \: an \\ \sf object \: with \: respect \: to \: time \\ \sf is \: known \: as \: acceleration. [/tex]

Answer:

b. 12.5 mAs, 70 kVp

Explanation:

The given parameter are;

The initial exposure factors := 10 mAs and 70 kVp

The initial Grid Ratio, G.R.₁ = 8:1

The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂  = 12:1

Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;

The GCF for G.R. 8:1 = 4

The GCF for G.R. 12:1 = 5

Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;

mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs

Therefore the new exposure factors are;

12.5 mAs, 70 kVp

Answer:

Explanation:

When the spring is compressed, it is compressed to its amplitude (whereas equilibrium is the spring's natural length with no mass attached to it and displacement is the spring's reaction to a mass hung on the end of it without any "extra" pushing or pulling on the mass). It is at the amplitude where the spring experineces max potential energy, which is choice 2, E = mph

Answer:

its c

Explanation:

...

Answer:

1 km

Explanation:

Bob travels at one km/h and it takes him one hour to get there.

Answer:

C) Potential energy due to intermolecular forces.

Answer:

The angiosperms dominate Earth's surface and vegetation in more environments, particularly terrestrial habitats, than any other group of plants. As a result, angiosperms are the most important ultimate source of food for birds and mammals, including humans.

Explanation:

plz mark brainlest

Answer:

radius=4.77cm

Explanation:

given

r=? ,h=14 , v=1000

then using the formula

[tex]r = \sqrt{ \frac{v}{\pi \: h} } [/tex]

= √1000/π•14

=4.77cm