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Nataly [62]
3 years ago
13

A car weighing 11.1 kN and traveling at 13.4 m/s without negative lift attempts to round an unbanked curve with a radius of 61.0

m. (a) What magnitude of the frictional force on the tires required to keep the car f static friction between the tire or not ("yes" or "no")? on its circular If the coefficient ro s 0.35, is the attempt at taking the curve successf (a) Number Units (b)
Physics
1 answer:
klasskru [66]3 years ago
8 0

Answer:

Well..

Explanation:

That's impossible. I know because I once weighed 11.1 kN, and I was temporarily immobile. It's probably the same for a car, and therefore it can not be "traveling" anywhere at all.. unless you put the car on an airplane or a boat or something.

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   =  (1.9 x 365.24 x 86,400 x 10⁹) nanosec

   =  6.00 x 10¹⁶ nanoseconds

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Answer:

Answer is given in the attachment.

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Vsevolod [243]

Explanation:

- Newton's first law of motion:

"An object at rest (or in uniform motion) remains at rest (or in uniform motion) unless acted upon an unbalanced force

In this situation, we can apply Newton's first law to the keys of the keyboard that are not hit by the fingers of the man. In fact, as no force act on the keys, they remain at rest.

- Newton's second law of motion:

"The acceleration experienced by an object is proportional to the net force exerted on the object; mathematically:

F=ma

where F is the net force, m is the mass of the object, and a its acceleration"

In this case, we can apply Newton's second law to the keys of the keyboard that are hit by the man: in fact, as they are hit, they experience a downward force, and therefore they experience a downward acceleration.

"Newton's third law of motion:

"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

Here We can apply Newton's third law to the pair of objects finger-key: in fact, as the finger apply a force on the key (action force), then the key exerts a force back on the finger (reaction force), equal and opposite.

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3 years ago
By what factors does the speed of the elctron exceed that of the proton?<br>​
____ [38]

Answer:

» An electron is lighter than a proton.

<u>explanation</u><u>:</u>

{  =  \: \sf{an \: electron \: has \: formula \:  \: }}{ \bf{ {}^{0}_{  - 1}e }}

hence it's mass number is zero

{  =  \: \sf{an \: electron \: is \: helium \: particle \:  \: }}{ \bf{ {}^{4} _{2}He  }}

hence it's mass number is 4

<u>Therefore</u><u>,</u><u> </u><u>proton</u><u> </u><u>is</u><u> </u><u>heavier</u><u> </u><u>than</u><u> </u><u>electron</u>

» An electron has a small charge magnitude than a proton.

<u>Explanation</u><u>:</u>

An electron has charge of -1 while proton has charge of +2, therefore electron is less deflected by any energetic fields than a proton

8 0
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