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Nina [5.8K]
2 years ago
12

Calculate the gravitational force between two elephants, one has a

Physics
1 answer:
Virty [35]2 years ago
8 0

Answer:

2.177088 * 10^-8 N

Explanation:

solution;

mass(m1) = 1 * 10 kg

mass(m2) = 8 * 102 kg

distance(d) = 5m

universal gravitational constant (G) =6.67 * 10^-11

we know that,

gravitational force = (G*m1*m2)/d^2

=((6.67 *10^-11) *(1 *10)*(8 *102))/5^2

=(5.44272 * 10^-7)/25

=2.177088 *10^-8

Gravitational force = 2.177088 * 10^-8

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Is O2 classified as a compound?
dybincka [34]

Answer: NO.

Explanation: Oxygen is not a compound. It has only one element in it.

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Much of the pollution in out oceans comes from:
BabaBlast [244]

Answer:

c.) sewage and industrial waste

3 0
3 years ago
1: In a longitudinal wave, the particle displacement is __________ the direction of wave movement
jenyasd209 [6]
The answer is parallel

5 0
3 years ago
You are holding one end of an elastic cord that is fastened to a wall 3.0 m away. You begin shaking the end of the cord at 2.3 H
Karo-lina-s [1.5K]

Answer:

Time take to fill the standing wave to the entire length of the string is 1.3 sec.

Explanation:

Given :

The length of the one end x= 3m, frequency of the wave f = 2.3 Hz, wavelength of the wave λ = 1 m.

Standing wave is the example of the transverse wave, standing wave doesn't transfer energy in a medium.

We know,

∴ v = fλ

Where v = speed of the standing wave.

also, ∴ v=\frac{x}{t}

where t = time take to fill entire length of the string.

Compare above both equation,

⇒   t = \frac{3}{2.3} sec

     t = 1.3sec

Therefore, the time taken to fill entire length 0f the string is 1.3 sec.

7 0
3 years ago
You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner
omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

Q_1 = Q_2

V_1A_1=V_2A_2

Our values are given as,

A_1=\frac{1}{2}^2*\pi=0.785 in^2

A_2=\frac{5}{8}^2*\pi=1.227 in^2

Re-arrange the equation to find the first ratio of rates we have:

\frac{V_1}{V_2}=\frac{A_2}{A_1}

\frac{V_1}{V_2}=\frac{1.227}{0.785}

\frac{V_1}{V_2}=1.56

The second ratio of rates is

\frac{V2}{V1}=\frac{A_1}{A2}

\frac{V2}{V1}=\frac{0.785}{1.227}

\frac{V2}{V1}=0.640

3 0
3 years ago
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