How are the direction of the net force on an object and the direction of the object’s acceleration related?

Physics

1 answer:

Oksanka [162]2 years ago

8 0

They're the same direction.

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Your teacher gives you a piece of cardboard with two pinholes on it, saying that they are separated by 205 μm ± 3%. In order to

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m,lkj,mkn,njn

Explanation:because she is telling you to do a project

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2 years ago

A 1.50 3 103 - kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that air resistance remains const

nexus9112 [7]

Answer:

Explanation:

let force exerted by engine be F.Net force =( F-400)N, applying newton law

F-400 = 1.5 x 10³x18 =27000 ,

F = 27400 N.

velocity after 12 s = 0 + 18 x 12 = 216 m/s

Average velocity = (0 + 216 )/2 = 108 m/s

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2 years ago

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A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$