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deff fn [24]
3 years ago
10

Is the eccentricity of planet orbits closer to 1 or 0?

Physics
1 answer:
jonny [76]3 years ago
8 0
The answer is closer to 0.

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The mass of the bicycle and rider is 60 kg and the total area of the tyres in contact with
AleksAgata [21]

Answer:

a little

Explanation:

First of all, it's not how you spell "tyres", it is tires.

Second of all, you already know the Mass so what you need to find out now is  contact the road. You Know that your number and letter are squared so that would turn into 6m x 2.4. Then you do the math do continue on to finish it. Have a great day!! Good luck with the answer!!

3 0
3 years ago
Two charges, - Q0 and - 4Q0 are a distance d apart. These two charges are free to move but do not because there is a third charg
TEA [102]

To solve this problem we will use the equilibrium conditions in the Electrostatic Forces. In turn, we will use the concept formulated from Coulomb's laws to determine the intensity of the Forces and make the respective considerations.

Our values for the two charges are:

q_1 = -Q_0

q_2 = -4Q_0

As a general consideration we will start by determining that they are at a unit distance (1) separated from each other. And considering that both are negative charges, they will be subjected to repulsive force. Said equilibrium compensation will be achieved only by placing a third force between the two.

Let the third charge be q_3 = +Q is placed at a distance x from q_1

F_{1,3} = \frac{k(-Q_0)(+Q)}{x^2}

The force on q_3 due to q_2 is

F_{2,3} = \frac{k(-4Q_0)(+Q)}{1-x^2}

The condition of equilibrium is

F_{1,3} = F_{2,3}

\frac{k(-Q_0)(+Q)}{x^2}= \frac{k(-4Q_0)(+Q)}{1-x^2}

\frac{1}{x^2} = \frac{4}{(1-x)^2}

x = 0.331 from q_1

To find the magnitude of q_3 we use F_{1,2} = F_{1,3}

\frac{k(-Q_0)(4Q_0)}{1^2}= \frac{k(-Q_0)(Q)}{0.331^2}

Q = 0.43Q_0

The magnitude of the third charge must be 0.43 the first charge Q_0

3 0
3 years ago
How resistance, current, and voltage behave in a series circuit.?
mylen [45]

b.the covalent circuit is held throw magnetic waves


4 0
4 years ago
I need help PLEASE HEEEEEEEEELPP
lana66690 [7]

Answer:

a = (v2 - v1) / t

From A to B    (8 - 4) m/s / 1 s = 4 m / s^2

From A to D    ( 7 - 4) m/s / 5 s = .6 m / s^2

Note these equations hold for "uniform" values

They say nothing about the acceleration at intermediate points - the equation just says that his average speed increased from 4 m/s to 7 m/s during a 5 sec period

6 0
3 years ago
O
Sergio [31]

see

below

Explanation:

refractive index = speed of light in vacuum / speed of light in medium

light travels at a speed of 3.0 x 10^8 m/s in vacuum

refractive index = 3.0 x 10^8 / 2.0 x 10^8

refractive index = 1.5

hope this helps, please mark it

4 0
3 years ago
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