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2 years ago

A box experiences a force of 2 N to the left and 3 N to the right. Which is true of the box's motion?

2 answers:

5 0

Below are the choices:

<span> A) The box will slow down.
B) The box's velocity will be 1 m/s.
C) The box's velocity will not change.
D) The box will experience acceleration
</span>
The answer is D) The box will experience acceleration

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jok3333 [9.3K]2 years ago

4 0

Answer:

The motion of box is accelerated and the direction of acceleration is rightwards.

Explanation:

F1 = 2 n left

F2 = 3 N right

The net force acting on the box = F = F2 - F1 = 3 - 2 = 1 N right

As there is a net force acting on the box, so there is an acceleration in the box. If there is an acceleration, the box is in motion under the constant acceleration, it means the velocity of box is increasing.

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Robert is riding the Giant Drop at Great America. If

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Explanation:

i hope this helps, its not the same person but its the same equation.

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2 years ago

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

dmitriy555 [2]

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area $= \sigma = 1.99*10^{-7} C/m^2$

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates is $E = \frac{\sigma}{\epsilon}$

force acting on charge is F = q E

Work done by charge q id $\Delta X =\frac{ q\sigma \Delta x}{\epsilon}$

this work done is converted into kinectic enerrgy

$\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}$

solving for v

$v = \sqrt{\frac{2q\Delta x}{\epsilon m}$

$\epsilon = 8.85*10^{-12} Nm2/C2$

$v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}$

v = 1.15*10^{7} m/s

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2 years ago

Do these four animals have invertebrates?

ira [324]

The worm is the invertebrate. All the other animals are vertebrates.

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2 years ago

A metal ball has a net charge of 4.5x10-7 C

netineya [11]

a) the number of protons is $2.81\cdot 10^{12}$ more than the electrons

b) $4.69\cdot 10^{-15} kg$

Explanation:

The net electric charge on the ball is

$Q=+4.5\cdot 10^{-7}C$

This electric charge is given by the algebraic sum of the charge of the protons and of the charge of the electrons.

The charge of one proton is:

$q_p =+e= +1.6\cdot 10^{-19}C$

While the charge of one electron is

$q_e = -e=-1.6\cdot 10^{-19}C$

So the net charge on the metal ball will be given by

$Q=N_p q_p + N_e q_e = (N_p -N_e)e$

where

$N_p$ is the number of protons

$N_e$ is the number of electrons

So we find:

$N_p-N_e=\frac{Q}{e}=\frac{4.5\cdot 10^{-7}}{1.6\cdot 10^{-19}}=2.81\cdot 10^{12}$

This means that the number of protons is $2.81\cdot 10^{12}$ more than the electrons.

In this case, we want to make the ball neautral, so we have to remove a net charge of Q' such that the new charge is zero:

$Q-Q'=0$

This implies that the charge that we must remove is

$Q'=Q=4.5\cdot 10^{-7}C$

To do that (and to make the ball losing mass at the same time), we have to remove protons, since they have positive charge.

The number of protons that must be removed is:

$N_p = \frac{Q'}{q_p}=\frac{4.5\cdot 10^{-7}}{1.6\cdot 10^{-19}}=2.81\cdot 10^{12}$

The mass of one proton is

$m_p = 1.67\cdot 10^{-27}kg$

Therefore, the total mass that must be removed from the ball is

$M=m_p N_p = (1.67\cdot 10^{-27})(2.81\cdot 10^{12})=4.69\cdot 10^{-15} kg$

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2 years ago

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Aneli [31]

Explanation:

Below are attachments containing the solution.

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2 years ago