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2 years ago

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Anna pushes a box with a force of 8.00 newtons. she generates a power of 3.00 watts. how much time does it take for anna to move

the box the distance of 9.72 meters? a. 12.4 seconds b. 20.1 seconds c. 25.9 seconds d. 30.8 seconds e. 35.8 seconds

1 answer:

mote1985 [20]2 years ago

7 0

The time taken for anna to move the box the distance of 9.72 meters is 25.9 seconds. The correct option is c.

<h3>What is force?</h3>

The force is the action of push or pull to move or stop an object.

Anna pushes a box with a force of 8.00 newtons. she generates a power of 3.00 watts.

Force is related to power by the relation

Force = Power / Velocity

Put the values, we get

8 = 3/V

V = 0.375 m/s

Velocity is defined as the ratio of distance and the time taken.

The box is moved by the distance of 9.72 meters. So, time taken is

V = d/t

0.375 = 9.72/t

t =25.9 seconds.

Thus, time taken for anna to move the box the distance of 9.72 meters is 25.9 seconds.

Learn more about force.

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At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q

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Answer:

$\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}$

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Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

$\displaystyle F=\frac{KQ_1Q_2}{d^2}$

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

$\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}$

$\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}$

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

$\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l$

The force F3 is

$\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}$

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

$\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle$

F2 is pointed upwards and its horizontal component is zero

$\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle$

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

$\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle$

$\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

Now we compute the total force

$\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

$\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle$

$\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}$

Now we compute the magnitude

$\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}$

The direction of the total force is given by

$\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558$

$\boxed{\displaystyle \theta =68^o}$

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