The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a needle, the tip of which is a circle 0.200 mm in radius, what pressure is exerted on the record in N/m2?
pressure is force/Area, in this case, your force is the weight of the needle (remember to take gravity into account) F = .001 * 9.8. Your needle tip has a radius of .002m and the area of a circle is A= pir^2 = 3.14(.002)^2. Your final equation should look like this Pressure = (.001 * 9.8)/(3.14(.002)^2)
7937
To calculate the pressure exerted by the phonograph needle, we can use the formula:
Pressure = Force / Area
First, we need to find the force exerted by the needle on the record. We can calculate this using the weight of the 1.00 g object:
Force = Mass * Acceleration due to gravity
The mass is given as 1.00 g, which is equal to 0.001 kg. The acceleration due to gravity is approximately 9.8 m/s^2.
So, the force exerted by the needle is:
Force = 0.001 kg * 9.8 m/s^2
Next, we need to calculate the area of the circular tip of the needle. The radius of the circle is given as 0.200 mm, which is equal to 0.0002 m.
Area = π * Radius^2
Area = π * (0.0002 m)^2
Once we have the force and the area, we can substitute them into the pressure formula to calculate the pressure exerted by the needle on the record.
Pressure = Force / Area
Pressure = (0.001 kg * 9.8 m/s^2) / (π * (0.0002 m)^2)
Simplifying this equation will give us the answer in N/m^2.
Elana almost got it right
=0.001/(3.14*(0.0002)^2)=7961.78 or 7962 Pa
p=F/A= mg/πr²=
=0.001/3.14•0.0002²=7958 Pa