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3 years ago

What will the final velocity of a 5.0 g bullet starting from rest, if a net force of 45 N is applied over a distance of 0.80 m?

Physics

1 answer:

OLEGan [10]3 years ago

6 0

Answer:

The final velocity of the bullet is approximately 3.795 m/s

Explanation:

The given parameters are;

The mass of the bullet, m = 5.0 g

The force applied to the bullet, F = 45 N

The distance over which the force is applied, d = 0.80 m

Therefore, we have;

The final kinetic energy of the bullet = = 1/2 × m × v² = The work done on the bullet by the 45 N force = F × d

Where;

v = The final velocity of the bullet

By substitution, we get;

The final kinetic energy of the bullet = 1/2 × 5.0 × v² = 45 × 0.8

1/2 × 5.0 × v² = 45 × 0.8

v² = (45 × 0.8)/(1/2 × 5.0) = 14.4

v = √14.4 ≈ 3.795

The final velocity of the bullet = v ≈ 3.795 m/s

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Compare the magnitude of the electromagnetic and gravitational force between two electrons separated by a distance of 2.00 m. As

Lesechka [4]

First you need to know about two laws, which are:
1) Coulomb's law
2) Newton's law of gravitation

1.
According to Coulomb's law, Electric force between TWO charges is:
$F_{e} = \frac{k*q_{1}*q_{2}}{r^{2}}$ -- (A)

Where, k = 1/(4*π*epsilon_not) = $9 * 10^{9}$ $\frac{Nm^{2}}{C^{2}}$

Both $q_{1}$ and $q_{2}$ = -1.61 x $10^{-19}$ C
r = Distance between the two charges = 2.00m

Plug-in the above values in (A), you would get:

$F_{e}$ = (9 * $10^{9}$ ) (1.61 * $10^{-19}$ * 1.61 * $10^{-19}$ ) / (2*2)

$F_{e}$ = $5.83* 10^{-29}$ N

2.
According to Newton's law of gravitation:
$F_{g} = \frac{Gm_{1}m_{2}}{r^{2}}$ -- (B)

Where G = Gravitational constant = 6.674 * $10^{-11} m^{2} kg^{-1}s^{-2}$

m1 = m2 = Mass of the electron = $9.11 * 10^{-31}$ kg
r = 2.0 m

Plug-in the above values in (B), you would get:

$F_{g}$ = (6.674 * $10^{-11}$ ) ( $9.11 * 10^{-31}$ * $9.11 * 10^{-31}$ }[/tex]) / (2*2)

$F_{e}$ = $5.83* 10^{-29}$ N

$F_{e}$ = $1.38 * 10^{-71} N$

Now do Fe over Fg, you would get:
$\frac{F_{e}}{F{g}} = 4.23 * 10^{42}$

Ans: So the blanks are:
1) 5.82
2) 1.38
3) 4.23

-i

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3 years ago

Read 2 more answers

An ethernet cable is 3.80 m long and has a mass of 0.210 kg. A transverse pulse is produced by plucking one end of the taut cabl

Tasya [4]

Answer:

$T=94.54N$

Explanation:

The tension in a cable is given by:

$T=\mu v^2(1)$

Where $\mu$ is the mass density of the cable and v is the speed of the cable's pulse. These values are defined as:

$\mu=\frac{m}{L}(2)\\v=\frac{d}{t}$

The pulse makes four trips down and back along the cable, so $d=4(2L)$

$v=\frac{8L}{t}(3)$

Replacing (2) and (3) in (1), we calculate the tension in the cable:

$T=\frac{m}{L}(\frac{8L}{t})^2\\T=\frac{64mL}{t^2}\\T=\frac{64(0.21kg(3.80m))}{(0.735s)^2}\\T=94.54N$

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Answer:

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dimaraw [331]

Answer:

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