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3 years ago

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A student pushed a box 32.0 m across a smooth, horizontal floor using a constant force of 124 N. If the force was applied for 8.

00 s, how much power was developed, to the nearest watt?

2 answers:

Vika [28.1K]3 years ago

6 0

You need to calculate the work needed. This is force x distance in this case.
After you do that, divide the work by the time.

Brilliant_brown [7]3 years ago

3 0

The power developed is 500 W ( to the nearest Watt)

Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.

$P = \frac{W}{t}$

$W= F*s$

Therefore,

$P=\frac{F*s}{t}$

Substitute the given values of force , displacement and time

$F = 124 N,s = 32.0 m,t = 8.0 s$

$P =\frac{W*s}{t} =\frac{124N*22.0s}{8.0s} =496 W$

Thus the Power can be rounded off to the nearest value of 500 W

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A point charge is at the origin. With this point charge as the source point, what is the unit vector r^ in the direction of (a)

KiRa [710]

Answer:

a. $\hat{r} =- \hat{j}$
b. $\hat{r} = \ \frac{1}{\sqrt{2}} \ \hat{i} + \ \frac{1}{\sqrt{2}} \ \hat{j}$
c. $\hat{r} = \ -0.3871 \ \hat{i} + \ 0.91501\ \hat{j}$

Explanation:

Using Coulomb's Law we know that the electric field E at point $\vec{r}$ is:

$\vec{E(\vec{r})} = k_e \frac{q}{d^2} \frac{\vec{r}-\vec{r'}}{d}$

where $k_e$ is the Coulomb's Constant, q is the source charge, d is the distance between point and position of the source point charge, and $\vec{r}'$ is the position of the source point charge.

Taking all this in consideration, the unit vector clearly is:

$\hat{r} =\frac{\vec{r}-\vec{r'}}{d}$

For our problem, $\vec{r'} = (0,0)$ , as the charge is located at the origin.

So

$\hat{r} =\frac{\vec{r}}{d}$

and d will be the magnitude of $\vec{r}$

Now, we can take the values for each point.

<h3>a.</h3>

$\vec{r}= (0,-1.35 \ m)$

and, the magnitude of the vector is

$|\vec{r}| = \sqrt{r_x^2 + r_y^2}$

$|\vec{r}| = \sqrt{(0 \ m)^2 + (-1.35 \ m )^2}$

$|\vec{r}| =1.35 \ m$

So, the unit vector is:

$\hat{r} =\frac{(0,-1.35 \ m)}{1.35 \ m}$

$\hat{r} =(0,-1,0)$

$\hat{r} =- \hat{j}$

<h3>b.</h3>

$\vec{r}= (12 \ cm,12 \ cm)$

and, the magnitude of the vector is

$|\vec{r}| = \sqrt{r_x^2 + r_y^2}$

$|\vec{r}| = \sqrt{(12 \ cm)^2 + (12 \ cm )^2}$

$|\vec{r}| = \sqrt{2} \ 12 \ cm$

So, the unit vector is:

$\hat{r} =\frac{(12 \ cm,12 \ cm)}{\sqrt{2} \ 12 \ cm}$

$\hat{r} =(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$

$\hat{r} = \ \frac{1}{\sqrt{2}} \ \hat{i} + \ \frac{1}{\sqrt{2}} \ \hat{j}$

<h3>c.</h3>

$\vec{r}= (-1.10 \ m, 2.60 \ m)$

and, the magnitude of the vector is

$|\vec{r}| = \sqrt{r_x^2 + r_y^2}$

$|\vec{r}| = \sqrt{(-1.10 \ m)^2 + (2.60 \ m)^2}$

$|\vec{r}| = 2.8415 \ m$

So, the unit vector is:

$\hat{r} =\frac{ (-1.10 \ m, 2.60 \ m)}{2.8415 \ m}$

$\hat{r} =(-0.3871 ,0.91501)$

$\hat{r} = \ -0.3871 \ \hat{i} + \ 0.91501\ \hat{j}$

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Force on electron due to electric field is given by $F=qE=1.6\times 10^{-19}\times 450=720\times 10^{-19}N$

Mass of electron $m=1.67\times 10^{-27}kg$

Now according to second law of motion $F=ma$

So $720\times10^{-19}=1.67\times 10^{-27}a$

$a=431.137\times 10^8m/sec^2$

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