C Weight is the gravitational pull on an object
Answer:
its the sound that a heart produces when beating, this can help doctors detect abnormalities
The student who did the most work is student 2 with 2500 Joules.
<u>Given the following data:</u>
To determine which of the students did the most work:
Mathematically, the work done by an object is given by the formula;

<u>For </u><u>student 1</u><u>:</u>

Work done = 600 Joules
<u>For </u><u>student 2</u><u>:</u>

Work done = 2500 Joules.
Therefore, the student who did the most work is student 2 with 2500 Joules.
Read more: Read more: brainly.com/question/13818347
Answer:
5308.34 N/C
Explanation:
Given:
Surface density of each plate (σ) = 47.0 nC/m² = 
Separation between the plates (d) = 2.20 cm
We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

Now, plug in
for 'σ' and
for
and solve for the electric field. This gives,

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C
River shore is located at distance

speed of the woman is given as

now the time taken by the woman to cover the distance is


for the same time interval the dog will run to and fro with speed 4.5 m/s
so the total distance moved by the dog is given by



<em>so the total distance that dog will move is 7200 m or 7.2 km</em>