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3 years ago

A car traveling at +1.5 m/s accelerates at the rate of 0.65 m/s for an interval of 2.6a find vf

Physics

1 answer:

mario62 [17]3 years ago

5 0

Vf ( final velocity) = ?

Vo ( initial velocity) = 1.5 m/s

a (acceleration) = .65 m/s^2

t (time) = 2.6 seconds

The equation you want is Vf = Vo + (a)(t).

Plugging in the variables gives Vf = 1.5 + (.65)(2.6).

The final velocity is 1.5 + 1.69, which comes out to +3.19 m/s.

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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

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3 years ago

What is the force of a ball if it has a mass of 10 kg and accelerates at an acceleration of 20 m/s^2? *

vampirchik [111]

Answer:

200N

Explanation:

mass(m) = 10 kg

acceleration(a) = 20 m/s^2

Force = mass * acceleration

= 10*20

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PtichkaEL [24]

During that final period of time,
his acceleration is
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100 °C is a greater temperature than which of the following?

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