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Alja [10]
3 years ago
5

PLEASE HELP!

Physics
1 answer:
luda_lava [24]3 years ago
5 0

Answer:

When a body is placed on a table top, it exerts a force equal to its weight downwards on the table top but does not move or fall. (i) Name the force exerted ...

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Which of the following is a contact force?
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A student is told to use 20.0 g of sodium chloride to make an aqueous solution that has a concentration of 10.0 g/L (grams of so
Stells [14]

Answer:

she should add solute to the solvent

Explanation:

Given data :

Mass of the sodium chloride, = 20.0 g

Concentration of the solution = 10 g/L

Volume of 20.0 g of sodium chloride = 7.50 mL

Now, from the concentration, we can conclude that for 10 g of sodium chloride volume of the solution is 1 L

thus, for 20 g of sodium chloride  volume of the solution is 2 L or 2000 mL

also,

Volume of solution = Volume of solute(sodium chloride) + volume of solvent (water)

thus,

2000 mL = 7.5 mL + volume of solvent (water)

or

volume of water = (2000 - 7.5) mL

or

volume of water = 1992.5 mL

or

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About how much time elapses between high tides on the same day?
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6 0
3 years ago
At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th
Ainat [17]

Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

 ac is 6.23*g which is 9.81

v is 101m/s

substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r  

So (86kg* 101^2)/(167) =

Fc=5253 N

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