The average density of the material from which the coin is made is 9.67 g/cm³.
<h3>Volume of the coin</h3>
The volume of the coin at the given diameter is calculated as follows;
V = Ah
where;
- A is area of the coin
- h is the thickness of the coin
V = πd²/4 x h
V = π(2.8)²/4 x (0.21 cm)
V = 1.293 cm³
<h3>average density of the coin</h3>
The average density of the material from which the coin is made is calculated as follows;
density = mass/volume
density = 12.5 g / (1.293 cm³)
density = 9.67 g/cm³
Thus, the average density of the material from which the coin is made is 9.67 g/cm³.
Learn more about average density here: brainly.com/question/1354972
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Four of them pass a point every second.
Answer:
150000000000 m
0.0000005 seconds
33.33 ns
Explanation:
Speed of electromagnetic waves through vacuum = 
Echo time = 1000 seconds
Echo time is the time taken to reach the object and come back to the observer
Distance = Speed×Time

Venus is 150000000000 m away from Earth
Time = Distance / Speed

Echo time will be twice the time

The echo time will be 0.0000005 seconds
Difference in time = Difference in distance / Speed

The accuracy by which I will be able to measure the echo time is 33.33 ns
Assuming that the can is motionless, we can then assume that the vertical component of T = mg and that Fe = the horizontal component of T.
<span> Since T itself is larger than it's vertical or horizontal components separately, then T is greater than all the forces.</span>
First we need to find the speed of the dolphin sound wave in the water. We can use the following relationship between frequency and wavelength of a wave:

where
v is the wave speed

its wavelength
f its frequency
Using

and

, we get

We know that the dolphin sound wave takes t=0.42 s to travel to the tuna and back to the dolphin. If we call L the distance between the tuna and the dolphin, the sound wave covers a distance of S=2 L in a time t=0.42 s, so we can write the basic relationship between space, time and velocity for a uniform motion as:

and since we know both v and t, we can find the distance L between the dolphin and the tuna: