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3 years ago

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Ari is swimming a 25-meter race. After swimming 6 meters, she catches up to Amanda in a ratio of 7:3 from the 6-meter mark. At w

hat meter mark does Ari catch up to Amanda? Round to the nearest tenth, if necessary. Ari catches up to Amanda at meters.

2 answers:

taurus [48]3 years ago

7 0

Answer:

Ari catches up to Amanda at 19.3 meters

Explanation:

We have given Ari is swimming a 25-meter race.

It is given that after swimming 6 meters she catches up to Amanda in a ratio of 7:3 from the 6-meter mark.

So ,The distance between 6 meters and 25 meters = 25-6 =19 meters

It is given that she catched Amanda in ratio 7:3

So, she catched her at $\frac{7}{10}$ of the distance

So, the distance at which she catched Amanda $\frac{7}{10}\times 19=13.33meter$

Ari catches 13.3 metres from 6 m marks

So the total distance = 13.33+6 =19.33 meters

So Ari catches up to Amanda at a distance of 19.33 meters

PolarNik [594]3 years ago

7 0

Answer:19.3

Explanation:

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A 20 cm-radius ball is uniformly charged to 71 nC.

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Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

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3 years ago

B)A man walks 95 km, East, then 55 km, north. Calculate his RESULTANT

Varvara68 [4.7K]

The resultant displacement of the man is 109.77 km in the direction N60°E.

<h3>Displacement</h3>

Displacement is the distance travelled in a specified direction.

To calculate displacement, the straight line from starting point to end point of travel is taken and calculated.

<h3>Resultant displacement of the man </h3>

In the example above, a man walks 95 km, East, then 55 km, north.

The two distances form a right-angled triangle with two sides 95 and 55 units. The hypotenuse gives the resultant displacement, D.

Using Pythagoras rule:

D^2 = 95^2 + 55^2

D^2 = 12050

D = 109.77

Thus, the resultant displacement is 109.77 km

To calculate the direction:

Let the direction be y

y + x = 90°

tan x = 55/95

tanx x = 0.578

x = 30°

Then, y = 90 - 30

y = 60°

Therefore, the resultant displacement of the man is 109.77 km in the direction N60°E.

Learn more about displacement at: brainly.com/question/321442

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