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2 years ago

What will happen to the kinetic energy of an object if its speed doubles?

1 answer:

3 0

If the speed of the object becomes double, its kinetic energy changes to four times the initial kinetic energy. Hope it help!

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Four red flags

The load being carried poses a hazard to vehicles in the proximity of the load-carrying vehicle, and so it is mandatory by law for the vehicle to make the load more visible. At night time, the load is marked using red lights, as flags would not be visible then.

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If a proton and an electron are released when they are 2.50×10^-10m apart (typical atomic distances), find the initial accelerat

katrin [286]

To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as

$F = k \frac{q_1q_2}{r^2}$

Here,

k = Coulomb's constant

q = Charge of proton and electron

r = Distance

Replacing we have that,

$F = (9*10^9)(\frac{(1.602*10^{-19})^2}{2.5*10^{-10}})$

$F = 3.6956*10^{-9}N$

The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.

The acceleration of the electron is given as

$a_e = \frac{F}{m_e}$

$a_e = \frac{3.6956*10^{-9}}{9.11*10^{-31}}$

$a_e = 4.0566*10^{21}m/s^2$

The acceleration of the proton is given as,

$a_p = \frac{F}{m_p}$

$a_p = \frac{3.6956*10^{-9}}{1.672*10^{-27}}$

$a_p = 2.21*10^{18}m/s^2$

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2 years ago

A friction force of 280 N exists between a cart and the path. If the force of reaction is 766 N, what is the minimum action forc

MatroZZZ [7]

force of friction on the cart is given as

$F_f = 280 N$

here we also know the reaction force due to surface

$R = 766 N$

so we can say reaction force is given as

$R = \sqrt{F_n^2 + F_f^2}$

$766 = \sqrt{F_n^2 + 280^2}$

$F_n^2 = 766^2 - 280^2$

$F_n = 713 N$

now by force balance we will say

$F_y + F_n = mg$

$F_y = mg - F_n$

$F_x = F_f$

also we know that

$F_f = \mu * F_n$

$280 = \mu * 713$

$\mu = 0.39$

now minimum force required to set this into motion

$F_{min} = \frac{\mu mg}{\sqrt{1 + \mu^2}}$

here we know that

$mg = F_n = 713 N$

$F_{min} = \frac{0.39* 713}{\sqrt{1 + 0.39^2}}$

$F_{min} = 259 N$

So it will require 259 N minimum force to move it

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2 years ago

Help plz I’ll mark brainliest

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I think it’s concave

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