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Cloud [144]
3 years ago
14

In an 8.00 km race, one runner runs at a steady 11.0 km/h and other runs at 14.8 km/h. How far from the finish line is the slowe

r runner when the fast runner fisnishes the race?
Physics
1 answer:
Serga [27]3 years ago
8 0
The fast runner will cover 8 km by time = t = d/v =8/14.8 = 0.54 hour
at that time ( 0.54 hr ) the slower runner would  cover d = v*t = 11*.0.54 = 5.94 km
So the slower would be at a distance = 8km - 5.94km =2.06 km from the finish line.
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xxTIMURxx [149]

Explanation:

given,

mass of one planet (m1)=2*10^23 kg

mass of another planet (m2)=5*10^22kg

distance between them(d)=3*10^16m

gravitational constant(G)=6.67*10^-11Nm^2kg^-2

gravitational force between them(F)=?

we know,

F=Gm1m2/d^2

or, F=6.67*10^-11*2*10^23*5*10^22/(3*10^16)^2

or, F=6.67*2*5*10^-11+23+22/3*3*10^32

or, F=66.7*10^34/9*10^32

or, F=7.41*10^34-32

•°• F=7.41*10^2

thus, the gravitational force between them is 7.14*10^2

3 0
3 years ago
What is the net charge (overall or sum of charges) of the nucleus and what number is it equal to?
polet [3.4K]
Because the nucleus is made up of positively charged protons and neutrally charged neutrons, and no negatively charged particles, the charge of the nucleus will always be equal to the sum of the charges of its protons. A simpler way to say it is because each proton has a +1 charge, the charge of the nucleus will be the same as the number of protons in it.
7 0
3 years ago
An engine has an energy input of 125j and 35 j of that energy is transformed into useful energy
Tatiana [17]
The engine's efficiency is (35J)/125J) = 28% .

Do you have a different question to ask ?
5 0
3 years ago
Emmit is lifting a box vertically which forces are necessary for calculating the total force
zavuch27 [327]

Answer:

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Explanation:

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I hope this helps!

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3 years ago
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choli [55]

The solution for the problem is:

1 Watt = 1 Joule per second 
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a Kilowatt is 1,000 Watts 
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54,000,000 Watt*second = ? Joules 
54,000,000 Joules / second = 54,000,000 Watts

3 0
3 years ago
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