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steposvetlana [31]
3 years ago
12

If Scobie could drive a Jetson's flying car at a constant speed of 160.0 km/hr across oceans and space, approximately how long w

ould Scoobie take to drive around Earth's equator in days
Physics
1 answer:
DochEvi [55]3 years ago
6 0

Scobie will take 10 days to drive around Earth's equator.

To calculate the time that takes Scobie to drive around Earth's equator we need to find the distance, which is given by the equation of a circumference:

d = 2\pi r

<em>Where:</em>

r: is the Earth's radius = 6371 km

Then, the distance is:

d = 2\pi r = 2\pi*6371 km = 40030.2 km

Now, if we divide the above distance by the speed of the car we can find the time:

t = \frac{d}{v} = \frac{40030.2 km}{160.0 km/h} = 250.2 h*\frac{1 d}{24 h} = 10 d

Therefore, Scobie will take 10 days to drive around Earth's equator.

     

To learn more about distance and time here: brainly.com/question/14236800?referrer=searchResults

I hope it helps you!

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It is known that the gravitational force of attraction between two alpha particles is much weaker than the electrical repulsion.
natali 33 [55]

Answer:

<em>The ratio of gravitational force to electrical force is 3.19 x 10^-36 </em>

<em></em>

Explanation:

mass of an alpha particle = 6.64 x 10^{-27} kg

charge on an alpha particle = +2e = +2(1.6 x 10^{-19} C) = 3.2 x 10^{-19} C

distance between particles = d

For gravitational attraction:

The force of gravitational attraction F = \frac{Gm^{2} }{r^{2} }

where G = gravitational constant = 6.67 x 10^{-11} m^3 kg^-1 s^-2

r = the distance between the particles = d

m = the mass of each particle

therefore, gravitational force = \frac{6.67*10^{-11}*(6.64*10^{-27} )^{2}  }{d^{2} } = \frac{2.94*10^{-63} }{d^{2} }  Newton

For electrical repulsion:

Electrical force between the particles = \frac{-kQ^{2} }{r^{2} }

where k is the Coulomb's constant = 9.0 x 10^{9} N•m^2/C^2

r = distance between the particles = d

Q = charge on each particle

therefore, electrical force = \frac{-9*10^{9}*(3.2*10^{-19} )^{2}  }{d^{2} } = \frac{-9.216*10^{-28} }{d^{2} } Newton

the negative sign implies that there is a repulsion on the particles due to their like charges.

Ratio of the magnitude of gravitation to electrical force = \frac{2.94*10^{-63} }{9.216*10^{-28} }

==> <em>3.19 x 10^-36 </em>

8 0
2 years ago
How are oceans connected to all bodies of water
inn [45]
Oceans are collected to all bodies of water because

- All Rivers, streams, or swamps end up flowing out into an ocean somehow
- The oceans are the largest bodies of water, so most of the water on earth comes from them, or has been in the oceans.
7 0
3 years ago
What Could be the control in an experiment to measure the effects of gas additives on fuel
alina1380 [7]
Well,

A control in an experiment would basically be the "normal" version of your test subjects.

In a drug testing experiment with people, the control group would be the people who don't take the drug.

In an experiment on the effects of salt on potatoes, the control group would be a potato without salt on it.

So in an experiment to measure the effects of gas additives on fuel, the control would be fuel without additives.
8 0
3 years ago
N2 + O2 → 2NO N-N triple bond: 941 kJ/mol O-O double bond: 495 kJ/mol N-O bond: 201 kJ/mol
kiruha [24]

Answer:

\large \boxed{\text{761 kJ}}

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

                        N₂  +   O₂   ⟶         2NO

                     N≡N  + O=O ⟶       2O-N=O

Bonds:         2N≡N   1O=O        2N-O + 2N=O

D/kJ·mol⁻¹:     941      495           201      607

\begin{array}{rcl}\Delta H & = & \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\& = & 2 \times 941 +1 \times 495 - (2 \times 201 + 2\times 607)\\&=& 2377 - 1616\\&=&\textbf{761 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{761 kJ}}$}.

3 0
2 years ago
To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater. it is found th
il63 [147K]

The difference in weight is due to the displacement of water (the buoyancy of water is acting on the athlete thus giving her smaller weight).<span>

The amount of weight displaced or the amount of buoyant force is: </span>

Fb = 690 N - 48 N

Fb = 642 N

From newtons law, F = m*g. Using this formula, we can calculate for the mass of water displaced:

m of water displaced = 642N / 9.8m/s^2

m of water displaced = 65.5 kg

Assuming a water density of 1 kg/L, and using the formula volume = mass/density:

V of water displaced = 65.5kg / 1kg/L = 65.5 L

The volume of water displaced is equal to the volume of athlete. Therefore:

V of athlete = 65.5 L

The mass of athlete can also be calculated using, F = m*g

m of athlete = 690 N/ 9.8m/s^2

m of athlete = 70.41 kg

 

Knowing the volume and mass of athlete, her average density is therefore:

average density = 70.41 kg / 65.5 L

<span>average density = 1.07 kg/L = 1.07 g/mL</span>

5 0
3 years ago
Read 2 more answers
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