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Blababa [14]
2 years ago
14

According to Newton's third law, two objects interacting under a force (such as gravity) both feel the same force. If the planet

s pull on the Sun as much as the Sun pulls on the planets, why are we able to approximate the Sun as a fixed position when studying the planetary orbits?
Physics
1 answer:
asambeis [7]2 years ago
8 0

Explanation:

That's because the Sun's acceleration is much smaller

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The propeller on a boat motor is initially rotating at 8 revolutions per second. As the boat captain reduces the boat speed, the
I am Lyosha [343]

Answer: 5.77\ rps

Explanation:

Given

Initial angular velocity is \omega_i=8\ rps

rate of reduction \alpha=0.9 rev/s^2

after 17 revolution i.e. \theta =17\ rev

using \Rightarrow \omega_f^2-\omega_i^2=2\alpha\theta

Insert the values

\Rightarrow \omega_f^2=8^2-2\times (0.9)\times17\\\Rightarrow \omega_f^2=33.4\\\Rightarrow \omega_f=5.77\ rps

5 0
3 years ago
A basebal (radius = .036 m, mas = .145kg) is droped from rest at the top of the Empire State Building (height = 1250ft). Calcula
lora16 [44]

Answer:

a) the initial potential energy = 541.95J

b) the final kinetic energy = 87.991 J

Explanation:

<u>Step 1</u>: Data given

The ball has a mass of 0.145 kg and is at a height of 381m

<u>Step 2</u>: Calculate potential energy

The potential energy = m * g * h

with m = the mass of the ball = 0.145 kg

with g = Gravitational acceleration = 9.81 m/s²

with h = the height of the building = 381m

The potential energy = 0.145 Kg * 9.81 m/s² * 381m = 541.95 J

FD = 1/2ρCDAv²

⇒ with FD = the drag force = the force component in the direction of the flow velocity

⇒ with ρ = density of the fluid (air in our case: ρ≈1.1839 Kg/m3 at 1 atm and 25 °C)

⇒ with v = velocity of the ball

⇒ with A = reference area, which in our case is just the cross sectional area of the ball: A=πr2

⇒ with CD is the drag coefficient - a dimensionless coefficient, that in the case of a sphere, CD=0.47

Following Newton's second law:

ΣFy = may = -mg +Dv²

Here is D=1/2ρCDA ( for convenience) = 0.001172

The terminal speed we can define as the speed of the ball where ay = 0

Therefore: -mg + Dvt² = 0

⇒vt = √(mg/D)

vt = √(0.145 * 9.8 / 0.001172) ≈ 34.837

Uinitial=mgh≈541.951 J (see first question)

Kfinal=1/2*mvt²=(m²g)/2D ≈ 87.991 J

The final kinetics energy is 87.991 J

4 0
4 years ago
Which of these is an example of potential energy?
Mama L [17]

Answer:

A person on top of a hill

Explanation:

7 0
2 years ago
according to newton's law of universal gravitation, in which of the following situations does the gravitational attraction betwe
Alex17521 [72]

Explanation:

The force acting between two masses is given by :

F=G\dfrac{m_1m_2}{d^2}........(1)

Where

G is the universal gravitational constant

m_1\ and\ m_2 are masses

d is the distance between two masses

It is clear from equation (1) that the gravitational attraction between the bodies always increases if the masses of bodies increases and when the separation between masses decreases.

So, the correct answer is "the masses increase, and the distance between the centers of mass decreases". This is because the force of gravitation is directly proportional to the masses and inversely proportional to the separation.

4 0
3 years ago
A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the e
gayaneshka [121]

Answer:

  V_inside = 36 V

Explanation:

<u>Given  </u>

We are given a sphere with a positive charge q with radius R = 0.400 m Also, the potential due to this charge at distance r = 1.20 m is V = 24.0 V.  

<u>Required</u>

We are asked to calculate the potential at the centre of the sphere  

<u>Solution</u>

The potential energy due to the sphere is given by equation

V = (1/4*π*∈o) × (q/r)                                          (1)

Where r is the distance where the potential is measured, it may be inside the sphere or outside the sphere. As shown by equation (1) the potential inversely proportional to the distance V  

V ∝ 1/r

The potential at the centre of the sphere depends on the radius R where the potential is the same for the entire sphere. As the charge q is the same and the term (1/4*π*∈o) is constant we could express a relation between the states , e inside the sphere and outside the sphere as next

V_1/V_2=r_2/r_1

V_inside/V_outside = r/R

V_inside = (r/R)*V_outside                               (2)

Now we can plug our values for r, R and V_outside into equation (2) to get  V_inside

V_inside = (1.2 m )/(0.600)*18

               = 36 V

  V_inside = 36 V

7 0
3 years ago
Read 2 more answers
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