The tiny planet Mercury has a radius of 2400 km and a mass of 3.3 times 10^23 kg.
a) What would be the gravitational acceleration of an astronaut standing on
the surface of Mercury? b) Compare the motion of a ball dropped on the
surface of Mercury to that of a ball dropped on Earth.

Answer:  his total speed would be 3 seconds

Explanation: because if you mulimply 10 x 3 you get 30 meaning that your answer would be 3

Answer:

A - Juan

Explanation:

I took the test!

Answer:

False -    

F = G M1 M2 / R^2

So F depends on M1 and M2 and as long either is not zero there will be a gravitational force between them.

Answer:

It is important to use renewal resources because we use so many non- renewable resources that we may lose the ability to make a lot of things like gasoline or power for our cities.

Answer:

Fc= 47.363 N

Explanation:

First, you need tangential velocity because they don't give that to you in the problem.

So, the formula for tangential velocity is υ=2πr/t.

For this problem its v=2π4/10= 2.513 m/s

Then we plug this into our equation for centripetal force

Fc= 30*(2.513)^2/4=47.363 N

Fc=47.363 N

The centripetal force acting on Tommy is 47.36 N if Tommy, who has a mass of 30 kg, sits 4.0 meters from the center of a merry-go-round that is rotating for a period of 10 seconds.

A tangible body's mass is the amount of matter it possesses. It's also a metric of inertia or the resistance to velocity when a net force is exerted.

It is given that:

Tommy, who has a mass of 30 kg, sits 4.0 meters from the center of a merry-go-round that is rotating for a period of 10 seconds.

As we know the formula for the centripetal force:

Fc = mv²/r

m = 30 kg

r = 4 meters

First, find the tangential velocity:

The formula for tangential velocity is:

v = 2πr/t

v = 2π4/10

v = 2.513 m/s

Plug the values in the formula:

Fc= 30(2.513)²/4

Fc = 47.36 N

Thus, the centripetal force acting on Tommy is 47.36 N if Tommy, who has a mass of 30 kg, sits 4.0 meters from the center of a merry-go-round that is rotating for a period of 10 seconds.

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Answer:

Mixing metals - alloy

Mixing clay & sand - ceramic

Mixing oil & gas - polymer

Answer:

I hope this helps

Explanation:

Answer:

earth

sorry I need points but I don’t know the answer again I’m really sorry

Please find attached photographs for your answer.

Hope it helps.

Do comment if you have any query.

Answer:

Examples of balanced forces

Resting against a wall.

Lying down.

Aircraft in a steady-flight.

Floating in water.

Standing in ground.

Tug of war equally balanced teams.

Fruit hanging from a tree.

Ball hanging from a rope.

Answer:

I=0.3 A

R=25 Ohm

V=?

V=IR

0.3X25=7.5

Answer:

7.5 V

Explanation:

current × resistance = voltage

Answer:

Lol it was the last one

Explanation

something

Answer:

D

Explanation:

Just count the dots

Answer:

[tex]V_{2.2} =2.2V[/tex]

Fast and loose: that's a classic voltage divider. the drop from the i-th resistor is [tex]R_i \over {\sum R}[/tex] of the drop across the whole series. In our situation, it's [tex]2.2\cdot 10^3 \over {2.2\cdot10^3 + 6.8\cdot 10^3[/tex] of 9 V. By plugging numbers in a calculator, it's 22/90 of 9V, or 2.2V

With Ohm's Law

The series is equivalent of a single resistor of Resistance [tex]2.2\cdot 10^3+6.8\cdot 10^3 \Omega = 9.0 k\Omega[/tex]. By Ohm's first Law ([tex]V=Ri[/tex]) the current flowing through the resistor is [tex]9V = 9*10^3\Omega i \rightarrow i=1mA[/tex]. At this point, the drop across the first resistor is, again by Ohm's law[tex]V = 2.2 \cdot 10^3 \Omega \cdot 1\cdot 10^{-3} A = 2.2V[/tex]

Answer:

120km

Explanation:

To get speed you divide distance by time.

Answer:

• Average current = 25,000 Amperes

• Time = 0.80 seconds

a)

[tex]{ \rm{quantity \: of \: charge = current \times time}} \\ { \rm{q = 25000 \times 0.80}} \\ { \boxed{ \mathfrak{answer : \: { \rm{quantity \: of \: charge = 20000 \: coulombs}} }}}[/tex]

b)

[tex]{ \rm{number \: of \: electrons = \frac{current}{charge \: quantity} }} \\ \\ { \rm{n = \frac{25000}{20000} }} \\ \\ { \boxed{ \mathfrak{ \: answer : \: { \rm{n = 1.25 \: }}}}}[/tex]

[tex]{ \underline{ \mathfrak{ \green{Good \: grace}} \: \:⚜ \: \: { \blue{ \mathfrak{Good \: God}}}}}[/tex]

Answer:

the net force is the same direction as the acceleration

Explanation:

so toward the center of the circle about which the object is constantly moving.

Answer:

true

Explanation:

this is fair to other people

Answer:

Centripetal acceleration = 46.26m/s2

Tension = 4.6N

Explanation:

Centripetal acceleration is given by the following formula: [tex]a_c = \frac{v^2}{r}[/tex]

So, to calculate acceleration we first need velocity and radius.

The radius is given to us as 75cm = 0.75m, and we can find velocity from the following formula:[tex]v=r\omega[/tex] where w = the angular velocity given by: [tex]\omega = \frac{2\pi}{T}[/tex]

We are given a period of 0.80s, meaning that our angular velocity is:

[tex]\frac{2\pi}{0.80} = 7.85 rad s^{-1}[/tex]

Now we just need to multiply is by the radius to find the velocity:

[tex]0.75*7.85=5.89ms^{-1}[/tex]

Finally we square the velocity and divide by the radius to find the final centripetal acceleration: [tex]\frac{5.89^{2} }{0.75} =46.26ms^{-2}[/tex]

[tex]F=\frac{mv^{2} }{r}[/tex] so we just need to multiply the centripetal acceleration by mass to get the tension force of the string: [tex]46.26*0.1 = 4.6N[/tex]

Hope this helped!

Answer:

The entire electromagnetic spectrum, from the lowest to the highest frequency (longest to shortest wavelength), includes all radio waves (e.g., commercial radio and television, microwaves, radar), infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.

Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.

Explanation: Hope this helped :)

longitudinal wave, wave consisting of a periodic disturbance or vibration that takes place in the same direction as the advance of the wave. ... Sound moving through air also compresses and rarefies the gas in the direction of travel of the sound wave as they vibrate back and forth.

FORMULA

y = displacement of the point on the traveling sound wave

x = distance the point has traveled from the wave's source

y_0 = amplitude of the oscillations

\omega = angular frequency of the wave

t = time elapsed

c = speed of the wave

Since the period of the supposed planet is less than the period of the earth orbiting the sun, That means the mass of the central star is greater than the mass of the sun. The answer is option C

According to Kepler's third law of planetary motion for planets and satellites, he described circular orbit about the same center body stating that:

The squared of the period is directly proportional to the cube of the radius. That is,

[tex]T^{2} \alpha r^{3}[/tex]

The orbital period formula is expressed as :

[tex]T^{2}[/tex] = 4[tex]\pi ^{2}[/tex] [tex]r^{3}[/tex]  ÷ GM

Suppose a planet orbits a star at the same distance as Earth (1 AU) but with a period of roughly 200 Earth days. The mass of the central star will be greater than the mass of our sun based on the formula given above.

In a geostationary orbit, whatever the mass of the satellite may be, they will travel at the same speed in a particular orbit. mass M in the above formula becomes the mass of the sun or the center mass.

Since the period of the supposed planet is less than the period of the earth orbiting the sun, That means the mass of the central star is greater than the mass of the sun.

Therefore, the correct answer is option C

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Answer:

The sum of vectors P and Q is given by the vector R, the resultant sum vector using the parallelogram law of vector addition. If the resultant vector R makes an angle ϕ with the vector P, then the formulas for its magnitude and direction are: |R| = √(P2 + Q2 + 2PQ cos θ) β = tan-1[(Q sin θ)/(P + Q cos θ)]

Answer:

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

GPE = 4000 × 9.8 × 20 = 784,000

We have the final answer as

Hope this helps you

Explanation:

All motions are relative to some frame of reference. Saying that a body is at rest, which means that it is not in motion, merely means that it is being described with respect to a frame of reference that is moving together with the body

Answer:

Motion is always described in relation to a reference point.

Explanation:

For example, if a car is driving at the speed of 30 miles per hour and you are standing by the road, to you, the car appears to be moving 30 miles per hour down the road. However, if you were in a car also driving at 30 miles per hour next to the truck, the same truck wouldn't appear to move.

The given kinematic equations of motion are derived from Newton's Laws

of Motion.

Reasons:

The direction the ball is kicked = Upwards in the air

Maximum height to which the ball rises = 6.25 m

A) The speed of the ball the instant it left the soccer player's foot

Solution:

The equation to use is; v² = v₀² + 2·[tex]a_y[/tex]·(y - y₀)

[tex]a_y[/tex] = The acceleration due to gravity ≈ -9.81 m/s²

At the maximum height, y - y₀ = 6.25 m

The velocity at the maximum height, v = 0

v₀ = The speed of the ball the instant it left the soccer player's foot

Which gives;

v² = v₀² + 2·[tex]a_y[/tex]·(y - y₀)

0 = v₀² - 2 × 9.81 × (6.25) = v₀² - 122.625

v₀² = 122.625

v₀ = √(122.625) ≈ 11.074

The ball's speed the instant it left the soccer player's foot, v₀ ≈ 11.074 m/s.

B) Required:

The time it takes the soccer ball to rise to the maximum height.

Solution:

The equation of motion required is v = v₀ + [tex]a_y[/tex]·t

At the maximum height, v = 0

v₀ ≈ 11.074 m/s

v = v₀ + [tex]a_y[/tex]·t

Which gives;

0 = 11.074 - 9.81 × t

[tex]t \approx \dfrac{11.074}{9.81} \approx 1.129[/tex]

The time it took the ball to rise to its maximum height, t ≈ 1.129 seconds

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By Newton's third law, we will see that the force that block A exerts on block B is equal in magnitude but in opposite direction than the force that block B exerts on block A.

Remember that Newton's third law says that whenever two objects interact, they exert equal and opposite forces on each other.

The force that block B exerts on block A is equal to block B's weight, which is:

[tex]W_b = 0.05kg*9.8m/s^2 = 0.49 N[/tex]

Then the force that block A exerts on block B is equal in magnitude but in opposite direction, thus it will be equal to:

F' = -0.49N

Then the diagram is the one that you can see below.

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Answer:

Sorry I need points [tex] \boxed{}[/tex]