And the density of water is 1000kgm3 plz urgent

How can juggling improve your hand eye coordination?

Ne4ueva [31]

Answer:

you can predict where the juggling ball is going to land and the move you hand to catch it

Explanation:

5 0

3 years ago

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A small metal ball is given a negative charge, then brought near to end a of the rod (figure 1). What happens to end a of the ro

erma4kov [3.2K]

What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.

<h3>Electrostatics</h3>

I have attached the image of the rod.

We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.

This means that their fields will cancel.

Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.

This also applies to a strong conducting rod and therefore it is strongly attracted.

Read more about Electrostatics at; brainly.com/question/18108470

7 0

2 years ago

See attachment for question

zloy xaker [14]

The Mercury's mass for the given acceleration due to gravity is 0.3152 x 10²⁴ kg.

The ratio of the calculated and accepted value of the Mercury's mass is 0.95.

Mass is the amount of matter present in the object.

The mass of the object is always constant, anywhere it is on the Earth or Moon or any other planet.

Given is the acceleration due to gravity of Mercury planet at North pole is g = 3.698 m/s² and the radius of Mercury planet is 2440 km.

The acceleration due to gravity is related with mass as

g = GM/R²

Substitute the values, we have

3.698 = 6.67 x 10⁻¹¹ x M/(2440 x1000)³

M = 2.2016 x 10¹³ / 6.67 x 10⁻¹¹

M = 0.3152 x 10²⁴ kg

Thus, the mercury's mass is 0.3152 x 10²⁴ kg.

(b) Accepted value of Mercury's mass is 3.301 x 10²³ kg

Ratio of the value of mass calculated and accepted is

Mcalc/M accep = 0.3152 x 10²⁴ kg / 3.301 x 10²³ kg

= 0.95

Thus, the ratio is 0.95

Learn more about mass.

brainly.com/question/19694949

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8 0

2 years ago

In a home's central heating system, when the room temperature measured by the thermostat reaches a set value, the thermostat sen

shutvik [7]

These actions are an example of feedback.
Given that the room has reached the desired temperature, there is no more need for it to be heated, at least until the temperature drops a bit. This is why the thermostat sends feedback about this situation to the heater, which immediately switches off until it is needed again.

3 0

3 years ago

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Assume the Earth is a ball of perimeter 40, 000 kilometers. There is a building 20 meters tall at point a. A robot with a camera

torisob [31]

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

No intersection: There's nothing that blocks the camera's view of the top of the building.
Two intersections: The planet blocks the camera's view of the top of the building.
One intersection: The point at which the top of the building appears or disappears.

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle $\angle \mathrm{B\hat{C}D}$ which corresponds to this minor arc.

This angle comes can be split into two parts:

$\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}$ .

Also,

$\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}$ .

The radius of this circle is:

$\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}$ .

The lengths of segment DC, AC, BC can all be found:

$\rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m$ ;
$\rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m$ ;
$\rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m$ .

In the two right triangles $\triangle\mathrm{DAC}$ and $\triangle \rm BAC$ , the value of $\angle \mathrm{B\hat{C}A}$ and $\angle \mathrm{A\hat{C}D}$ can be found using the inverse cosine function:

$\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}$

$\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}$

$\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}$ .

The length of the minor arc will be:

$\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km$ .

5 0

3 years ago

And the density of water is 1000kgm3 plz urgent​

And the density of water is 1000kgm3 plz urgent