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3 years ago

A runner achieves a velocity of 12.1 m/s 8.2 sec after he begins. What is

Physics

1 answer:

Karo-lina-s [1.5K]3 years ago

7 0

Given - 

Velocity (v) = 12.1 m/s
Time (t) = 8.2 sec

To find -

Acceleration

Solution - 

★ Acceleration = velocity/ time

→ A = 12.1/8.2

→ 1.47.... m/s²

Therefore, the acceleration of the runner will be 1.47 m/s².

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According to Boyle’s law, PV= K, what was the volume at the time of the first measurement given the following information? Round

alexandr402 [8]

Answer:

Iam a learning

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2 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

Given:

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

3 years ago

What is the difference between a data table and a graph?

alexandr402 [8]

At the same time, however, you get less detail or less precision in a chart or graph than you do in the table. Imagine the difference between a table of sales figures for a ten-year period and a line graph for that same data. You get a better sense of the overall trend in the graph but not the precise dollar amount.

3 0

3 years ago

Read 2 more answers

True or false please help me now.

jolli1 [7]

Answer:

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Explanation:

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2 years ago

You measure the power delivered by a battery to be 4.26 W when it is connected in series with two equal resistors. How much powe

gtnhenbr [62]

Answer:

The power delivered by the battery is 17.04 W

Explanation:

Power through a circuit is given as

P = IV ....1

where P is the power

I is the current through the circuit

V is the voltage through the circuit

The voltage in a circuit is given as

V = IR ....2

Let us take the value of each resistor as equal to R

when connected in series, the total resistance will be

$R_{t}$ = R + R = 2R

If we assume constant voltage through the circuit, then from equation 2, the current in this case is

I = V/2R

If the resistors are connected in parallel, then the total resistance will be

$\frac{1}{R_{t} }$ = $\frac{1}{R}$ +

$R_{t}$ = R/2

The current in this case will be increased since the resistance is reduced

I = 2V/R

comparing the two situations, we can see that the current increased when connected in parallel to a ratio of

$\frac{2V}{R}$ ÷ $\frac{V}{2R}$ =

This means that the current increased 4 times

From equation 1, we can see that electrical power is proportional to the current at a constant voltage, therefore, the power will also increase by four times to

P = 4 x 4.26 = 17.04 W

3 0

3 years ago