c) You wish to put a 1000-kg satellite into a circular orbit 300 km above the earth's surface. (a)
What speed, period, and radial acceleration will it have? (b) How much work must be done to the
satellite to put it in orbit? (c) How much additional work would have to be done to make the

zero

Explanation:

Work W is defined as

W = F•d = Fdcos(theta)

and it is a dot product of the force and displacement and theta is angle between F and d Since the force is perpendicular to d, angle is 90° thus cos90 = 0. Hence work is zero.

Answer:

Explanation:

one electron has [tex]1.60217662*10^{-19}~coulombs~then\\\\1.02*10^{20}~electrons------->1.02*10^{20}*1.60217662*10^{-19}~coulombs= 16.3422~coulombs[/tex]

Answer:

THE ANSWER IS SOMETHING LIKE 55

Answer:

[tex]E=576.5V/m[/tex]

Explanation:

From the question we are told that:

Length [tex]l=56.0cm=0.56m[/tex]

Distance apart [tex]d=7.0mm=0.007m[/tex]

Electron Transferred [tex]n=10^{-10}[/tex]

Therefore

Total Charge

Since Charge on each electron is

[tex]e=1.602*10^{-19}[/tex]

Therefore

[tex]T=1.602*10^{-19} *10^{10}[/tex]

[tex]T=1.602*10^{-9}[/tex]

Generally the equation for Charge density is mathematically given by

[tex]\rho=T/A[/tex]

Where

Area

[tex]A=0.56*0.56[/tex]

[tex]A=0.3136[/tex]

Therefore

[tex]\rho=1.602*10^{-9}/0.3136[/tex]

[tex]\rho=5.10*10^{-9}[/tex]

Generally the equation for Electric Field in the capacitor is mathematically given by

[tex]E=\frac{\rho}{e_0}[/tex]

[tex]E=\frac{5.10*10^{-9}}{8.85x10{-12}}[/tex]

[tex]E=576.5V/m[/tex]

Answer:

66.83 meters

Explanation:

After a quick online search, it seems that scientists calculate the average reaction time of individuals as 2.3 seconds between seeing an obstacle and putting their foot on the brakes. Now that we have this reaction time we need to turn the miles/hour into meters/second.

1 mile = 1609.34 meters  (multiply these meters by 65)

65 miles = 104,607 meters

1 hour = 3600 seconds

Therefore the car was going 104,607 meters every 3600 seconds. Let's divide these to find the meters per second.

[tex]\frac{104,607}{3600} = \frac{29.0575 meters}{1 second}[/tex]

Now we simply multiply these meters by 2.3 seconds to find out the distance covered before the driver puts his/her foot on the brakes...

29.0575m * 2.3s = 66.83 meters

Answer:

400

Explanation:

Answer:

Explanation:

The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation

[tex]v_f=v_0+at[/tex] where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...

0 = 21.5 + (-9.8)t and

-21.5 = -9.8t so

t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)

Now we will use that time to find out the max height of the object in the equation

Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:

Δx = [tex]21.5(2.19)+\frac{1}{2}(-9.8)(2.19)^2[/tex] which simplifies down a bit to

Δx = 47.1 - 23.5 so

Δx = 23.6 meters.

Now we can plug that in to the PE equation to find the PE of the object:

PE = (.19)(9.8)(23.6) so

PE = 43.9 J

Answer:

The answer is (D): When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.

Explanation:

Answer:

The Answer is D

Explanation:

Hope this helps!!!!

Explanation:

the efficiency of a machine can be increased by reducing the friction

please mark the brainliest

Answer:

the required speed with which the missile move relative to the Earth is -0.727c

Explanation:

Given the data in the question;

relative velocity relation;

u' = u-v / 1 - [tex]\frac{uv}{c^2}[/tex]

so let V[tex]_B[/tex] represent the velocity as seen by an external reference frame; u=V[tex]_B[/tex]

and let V[tex]_A[/tex] represent the speed of the secondary reference frame; v=V[tex]_A[/tex]

hence, u' is the speed of B as seen by A

so

u' = V[tex]_B[/tex]-V[tex]_A[/tex] / 1 - [tex]\frac{V_BV_A}{c^2}[/tex]

now, given that; V[tex]_A[/tex] = 0.9c  and V[tex]_B[/tex]  = 0.5c

we substitute

u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{(0.5c)(0.9c)}{c^2}[/tex]

u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{c^2(0.5)(0.9)}{c^2}[/tex]

u' = ( 0.5c - 0.9c ) / 1 - (0.5 × 0.9)

u' = ( -0.4c ) / 1 - 0.45

u' = -0.4c / 0.55

u' = -0.727c

Therefore, the required speed with which the missile move relative to the Earth is -0.727c

Answer:

For liquids: A measuring cylinder is used.

For solid: Over flow can is used

Answer:

i think a measuring cylinder

Answer:

a) F₁ = F₂,  F₃ = F₄,  b)  the correct answer is 3

Explanation:

a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies

Forces 1 and 2 are action and reaction forces F₁ = F₂

Forces 3 and 4 are action and reaction forces F₃ = F₄

as it indicates that the

b) how the car increases if speed implies that force 1> force3

      F₁ > F₃

therefore the correct answer is 3

Answer:

A) 20 N, B) 20 N, & C) 8 N

Explanation:

For the object to be in equilibrium, the upward forces must be equal to the downward forces and the forward forces must be equal to the backward forces.

1. Determination of A and B.

Forward forces = Backward forces

A + 10 + B = 25 + 25

A + 10 + B = 50

Collect like terms

A + B = 50 – 10

A + B = 40

Assume A and B to be equal. Thus, A is 20 N and B is 20 N.

2. Determination of C

Upward forces = Downward forces

C + 112 = 20 + 100

C + 112 = 120

Collect like terms

C = 120 – 112

C = 8 N

Thus, for the object to be in equilibrium, A must be 20 N, B must be 20 N and C must be 8N.

Answer:

b) 1 m/s

I am sure...........

Answer:

1.2 x 10^-12

Explanation:

3/2.5 x 10^-6/10^6

1.2 x 10^-6 x 10^-6

1.2 x 10^-12

Answer:

cause they give u milk

Explanation:

Answer:

Cows are important as they provide humans many things for survival. They provide milk, meat, and leather, all of these are important resources.

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

[tex]T = 2 \pi \sqrt\frac{m}{k}[/tex]

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

[tex]T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s[/tex]

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Answer:

a. At a distance greater than r

b. T_m is greater than T_e.

Explanation:

a. Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed vat a distance r from the center of the earth. The second satellite travels at a speed that is less than v. At what distance from the center of the earth does the second satellite orbit?

Since the centripetal force on any satellite, F equals the gravitational force F' at r,

and F = mv²/r and F' = GMm/r² where m = mass of satellite, v = speed of satellite, G = universal gravitational constant, M = mass of earth and r = distance of satellite from center of earth.

Now, F = F'

mv²/r = GMm/r²

v² = GM/r

v = √GM/r

Since G and M are constant,

v ∝ 1/√r

So, if the speed decreases, the radius of the orbit increases.

Since the second satellite travels at a speed less than v, its radius, r increases since v ∝ 1/√r.

So, the distance the second satellite orbits is at a distance greater than r

b. An identical satellite is orbiting the moon at the same distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make one revolution compare to the time T_e it takes the satellite orbiting the earth to make one revolution?

Since the speed of the satellite, v = √GM/r where M = mass of planet

Since the satellite is orbiting at the same distance, r is constant

So, v ∝ √M

Since mass of earth M' is greater than mass of moon, M", the speed of satellite circling moon, v_m is less than v the speed of satellite circling earth at the same distance, r

Now, period T = 2πr/v where r = radius of orbit and v = speed of satellite

Since r is constant for both orbits, T ∝ 1/v

Now, since the speed of the speed of the satellite on earth orbit v  is greater than the speed of the satellite orbiting the moon, v_m, and T ∝ 1/v, it implies that the period of the satellite orbiting the earth, T_e is less than the period of the satellite orbiting the moon, T_m since there is an inverse relationship between T and v. T_e is less T_m implies T_m is greater than T_e

So, T_m is greater than T_e.

Answer:

[tex]v=3.49\times 10^4\ m/s[/tex]

Explanation:

Given that,

Mass of Venus, [tex]M_V=4.87\times 10^{24}\ kg[/tex]

We know that,

Mass of Sun, [tex]M_s=1.98\times 10^{30}\ kg[/tex]

The distance between the center of Sun and the center of Venus is [tex]1.08\times 10^{11}\ m[/tex]

We need to find the peed of the planet Venus in its essentially circular orbit around the sun. using the formula,

[tex]v=\sqrt{\dfrac{GM_s}{r}}[/tex]

Put all the values,

[tex]v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.98\times 10^{30}}{1.08\times 10^{11}}}\\\\v=3.49\times 10^4\ m/s[/tex]

So, the speed of the planet venus is [tex]3.49\times 10^4\ m/s[/tex].

Answer:

The same as its translational KE.

The easy way to do this is to make up numbers and use them.

So, I'll say m=2 and r=3. I will also say v=3 .

Rot. Inertia of a hoop is mr^2. So the rot KE is: 1/2 (mr^2)(w^2)

note: (1/2*I*w^2)

Translational kinetic energy is basically normal KE, so 1/2(m)(v^2)

Now, lets plug our made up values in:

Rot Ke : 1/2 (9*2)(3/3) *note w = v/r

Tran Ke: 1/2(2)(9)

Rot Ke: 9

Tran Ke: 9

9=9, same.

Answer:

The amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice

Explanation:

let the latent heat of fusion of ice  = L

then, the latent heat of vaporization of water = 10L

The heat of fusion of 1 kg of ice = 1 x L = L

The heat of vaporization 1 kg of water = 1 x 10L = 10L

Therefore, the amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice

Answer:

concave curves inward like an hourglass and convex is an outward curve like a football

Explanation:

hope this helps

Answer:

there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.

Explanation:

The pressure measurement is carried out by calibrating the force exerted by the air on a surface of known area, suppose a small area 1 mm² = 0.01 cm²

To find out if the random movement of air molecules affects the pressure reading, let's calculate the number of molecules that reaches the pressure gauge.

In a system at atmospheric pressure and in a volume of 1 m³ (walls of 1 m each) there is one mole of air molecules, this mole is evenly distributed, so how many molecules fall on our surface

           # _molecule = 6.02 10²³ 0.01 10⁻⁴ / 1

           #_molecular = 6.02 10¹⁷ molecules per second

therefore the variation of the number of molecules is not very important

Consequently there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.

Answer:

the particles vibrate inside the tyre

Explanation:

as the car moves kinetic energy is transfered in the tyres which causes the particles to vibrate inside the tyre so the kinetic store is. transferred into thermal

Answer:

9.42 m/s

Explanation:

Applying,

V' = Aω.............. Equation 1

Where V' = maximum speed of the string, A = Amplitude of the wave, ω = angular velocity.

But,

ω = 2πf................. Equation 2

Where f = frequency, π = pie

And,

f = v/λ................ Equation 3

Where, λ = wave length, v = velocity

Also,

v = √(T/μ)................. Equation 4

Where T = Tension, μ = linear density.

From the question,

Given: T = 50.0 N, μ = 5.0 g/m = 0.005 kg/m

Substitute into equation 4

v = √(50/0.005)

v = √(10000)

v = 100 m/s

Also Given: λ = 2.0 m

Substitute into equation 3

f = 100/2

f = 50 Hz.

Substitute the value of f into equation 2

Where π = constant = 3.14

ω = 2(3.14)(50)

ω = 314 rad/s

Finally,

Given: A = 3.0 cm = 0.03 m

Substitute into equation 1

V' = 0.03(314)

V' = 9.42 m/s

Answer:

francium

Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.