A satellite of mass 5460 kg orbits the Earth and has a period of 6520 s
A)Determine the radius of its circular orbit.
B)Determine the magnitude of the Earth's gravitational force on the satellite.
C)Determine the altitude of the satellite.

Answer:

the answer is well known as gravity

Answer:

[tex]Q=50.3[/tex]

Explanation:

From the question we are told that:

Frequency [tex]F=330Hz[/tex]

Sound intensity drop [tex]I_d=2[/tex]

Time [tex]T=4s[/tex]

Therefore

Sound intensity Ratio

 [tex]\frac{I}{I_x}=\frac{1}{2}[/tex]

Generally the equation for Sound intensity is mathematically given by

 [tex]\frac{I}{I_x}=e^{-4\ \=t}[/tex]

 [tex]\frac{1}{2}=e^{-4\ \=t}[/tex]

 [tex]\=t =5.8s[/tex]

Generally the equation for Quality Factor is mathematically given by

 [tex]Q=2 \pi\frac{E}{\triangle E}[/tex]

 [tex]Q=2 \pi\frac{E}{\frac{E}{2*4}}[/tex]

 [tex]Q=50.3[/tex]

Answer:

C voltage

Explanation:

Voltage is the change in electric potential so basically current flows from high potential to low potential due to voltage.

Answer:

[tex]f = m \frac{v1 - v2}{t} \\ = 120 \times \frac{25 - 5}{2} \\ = 120 \times \frac{20}{2} \\ = 120 \times 10 \\ = 1200N \\ thank \: you[/tex]

26. D. crushing the sugar cube and dissolving it in water.

27. A. atom

28. B. molecule

29. B. plum pudding model of Joseph John Thomson

30. B. He used cathode ray tubes which showed that all atoms contain tiny negatively charged subatomic particles or electrons.

31. D. protons and neutrons are relatively heavier than electrons.

Answer:

Explanation:

As far as the displacement goes, we have 2 displacement vectors. If we didn't have the angles to deal with, this would be a much simpler process, but then that wouldn't be any fun at all, would it? I'll deal with the average speed first, then the displacement, which is a vector addition problem.

The average speed is found by adding together the distances the student traveled and then dividing this sum by the total time he spent traveling. If we are told that the student runs at 4.5 m/s for 3.0 minutes, we can use this to find out the distance he ran during that time interval. However, the units are not the same. We will find the distance the student traveled by convering the time to seconds.

3.0 minutes = 180 seconds, and

4.1 minutes = 246 seconds.

That means that the distance he ran in 180 seconds is found by multiplying this time be the speed at which he ran:

4.5 m/s(180 s) = 810 m and

3.5 m/s(246 s) = 860 m (rounded to follow the rules of sig dig).

This makes the speed equation look like this:

[tex]s=\frac{810+861}{180+246}=\frac{1671}{426}=3.9\frac{m}{s}[/tex] That's the average speed, which is NOT at all the same as the displacement. Displacement is where he ended up in reference to where he started. The angles play a huge part in this math (that is very involved, to say the least). We begin by restating the displacement of each "leg" of this journey.

The first leg took him 810 m at 207 degrees and

the second leg took him 860 m at 325 degrees

To find the x and y components of these 2 legs, or parts, we have to use the cos and sin formulas. We will call the first leg A and the second leg B. First the x components of both A and B:

[tex]A_x=810cos207[/tex] and

[tex]A_x=-720[/tex]

[tex]B_x=860cos325[/tex] and

[tex]B_x=704[/tex] and we add these to get the x-component of the resultant vector, C:

  -720

+  704

   -10 (rounded, as needed, to the tens place).

Now for the y-components of the resultant vector:

[tex]A_y=810sin207[/tex] and

[tex]A_y=-370[/tex]

[tex]B_y=860sin325[/tex] and

[tex]B_y=-490[/tex] and we add these to get the y-component of the resultant vector, C:

  -370

+ -490

 -860

Since the x component is negative and so is the y, we are in QIII, so when we finally find our angle, we will have to add 180 to it.

For the magnitude of the displacement vector, in m:

[tex]C_{mag}=\sqrt{(-10)^2+(-860)^2}[/tex] which gives us

[tex]C_{mag}=860m[/tex]

Now, because displacement is vector, we also need the angle. We find that is the formula

[tex]\theta=tan^{-1}(\frac{C_y}{C_x})[/tex] and filling in:

[tex]\theta=tan^{-1}(\frac{-860}{-10})=90[/tex] (rounded correctly), and then we add 180 to give us a final direction of 270 degrees.

So the final displacement of the student is 860 m at 270 degrees

Answer:

(a) t = 0.75 s

(b) d = 26 m

Explanation:

Height, h = 12 m

Angle, A = 30 degree

initial velocity, u = 40 m/s

(a) Let the time is t.

Use second equation of motion

 [tex]h = u t + 0.5 at^2\\\\12 = 40 sin 30 t - 0.5 \times10 \times t^2\\\\12 = 20 t - 5 t^2\\\\5 t^2 - 20 t + 12 =0 \\\\t = \frac{20\pm \sqrt{400-240}}{10}\\\\t = \frac{20\pm 12.65}{10}\\\\t = 3.3 s, 0.75 s[/tex]

(b) d = u cos A t

d = 40 x cos 30 x 0.75 = 26 m

Answer:

The sun uses hydrogen for nuclear fusion.

Answer:

Hydrogen

Explanation:

Answer:

products would appear after the raw materials

Explanation:

raw material + raw material = product (anything deriving from combining two materials)

Answer:

A. The size of the ice patches

Explanation:

In an experiment, the control variable also known as the CONSTANT is the variable that must be kept uniform or the same for all groups throughout the experiment in order not to influence the outcome of the experiment.

According to the experiment described in this question, the effect of different salts on melting points is investigated by a student. Sodium chloride (NaCl), Calcium Chloride (CaCl2), Potassium Carbonate (KCO3) and inert sand are the four types of salt used. The volume of the ice used and melted was finally measured. This means that the SIZE OF THE ICE PATCHES USED is the control variable of the experiment as the same size was used for all groups throughout.

Answer:

the size of the ice patches

Explanation:

Since fundamental units are elementary in nature and cannot be reduced further, so it can not be reduced further, so it cannot be expressed in other units

Answer:

0.25s

Explanation:

5/20 = 0.25s

This might be correct

Answer:

Firstly, science helps our understanding of the world around us. Everything we know about the universe, from how trees reproduce to what an atom is made up of, is the result of scientific research and experiment.

Scientists use the scientific method to collect measurable, experimental evidence in an experiment related to a term (often in the form of an if/then statement), the results aiming to support or differ a theory. In other words, scientific method helps scientists get accurate, repeatable results.

Answer:

900C

Explanation:

Change 2.5 mins to secs by multiplying by 60 which is 150secs

hope this helps please like and mark as brainliest

26. D. crushing the sugar cube and dissolving it in water.

27. A. atom

28. B. molecule

29. B. plum pudding model of Joseph John Thomson

30. B. He used cathode ray tubes which showed that all atoms contain tiny negatively charged subatomic particles or electrons.

31. D. protons and neutrons are relatively heavier than electrons.

After half of a revolution ...

==> Distance = π•r

==> Displacement = 2•r

Answer:

 v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

Starting point. On the roof of the building

         Em₀ = K + U = ½ m v₀² + m g y₀

Final point. On the floor

         Em_f = K = ½ m v_f²

         Emo = Em_g

         ½ m v₀² + m g y₀ = ½ m v_f²

        v_f² = v₀² + 2 g y₀

let's calculate

        v_f = √(10² + 2 9.8 25)

        v_f = 24.3 m / s

Answer:

answer is E

Explanation:

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

Answer:

the answer to this would be D.

Answer: Its D but it might swap to another letter so keep your eye out bud pass this test good luck

Explanation:

A group of two or more atoms.

Answer:

b

Explanation:

bc

Answer:

Data given.

focal length (f)=15m÷2=7.5m

Distance of the object(U)=10m

Image distance (v)=?

Magnification (M)=?

Solution:

From:

1/f=1/u+1/v

1/7.5=1/10+1/v=75

then v=75m

Magnification, M=u/v

=75/10=7.5

Then magnification=7.5

Answer:

v = 30 m and m = 3

Explanation:

Given that,

The radius of curvature of the mirror, R = 15 m

Focal length, f = 7.5 m

Object distance, u = -10 m

We need to find the image distance and the magnification of the object.

Using mirror's formula,

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(7.5)}+\dfrac{1}{(-10)}\\\\v=30\ m[/tex]

The magnification of the object in mirror is given by :

[tex]m=\dfrac{-v}{u}\\\\m=\dfrac{-30}{-10}\\\\m=3[/tex]

So, the distance of the image from the mirror and the magnification of the object are 30 m and 3 respectively.

Answer:

The package should be dropped 244.7 meters before the target

Explanation:

We need to find how long the package will take to hit the ground then solve for position.

y=0

-4.9t^2=-3000

t= sqrt (3000/4.9)

t= 24.74 seconds.

x= 247.4 meters.

Answer:

The magnetic field through the wire must be changing

Explanation:

According to Faraday's law, the induced emf, ε in a metallic conductor is directly proportional to the rate of change of magnetic flux,Φ  through it. This is stated mathematically as ε = dΦ/dt.

Now for the wire, the magnetic flux through it is given by Φ = ABcosθ where A = cross-sectional area of wire, B = magnetic field and θ = angle between A and B.

So, dΦ/dt = dABcosθ/dt

Since A and B are constant,

dΦ/dt = ABdcosθ/dt = -(dθ/dt)ABsinθ

Since dθ/dt implies a change in the angle between A and B, since A is constant, it implies that B must be rotating.

So, for an electric current (or voltage) to be produced in the wire, the magnetic field must be rotating or changing.

Answer:

12m/s

Explanation:

Using the equation of  motion

v = u + at

Given that

u = 0m/s

a = 2m/s²

time t = 6s

Substitute the given values into the expression

v = 0 + 2(6)

v = 0+12

v = 12m/s

Hence the required velocity is 12m/s