<u>Answer:</u>
1A, 2E, 3C, 4F, 5B, 6D
<u>Explanation:</u>
Tilt meter is a an equipment used to measure very small changes from vertical level. So it is a correct match for blank space in statement A.
Richter Scale is a scale of measurement of earthquake strength, So it is a correct match for blank space in statement E.
Mercalli Intensity Scale measures effect of an earthquake, like damage caused.So it is a correct match for blank space in statement C.
Seismograph is earthquake wave measuring instrument. So it is a correct match for blank space in statement F.
Correlation Spectrometer (C.O.S.P.E.C.) is used to measure sulfur dioxide content in smoke. So it is a correct match for blank space in statement B.
Moment Magnitude Scale is an earthquake measuring scale for great earthquakes. So it is a correct match for blank space in statement D.
Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight is given below.
Explanation:
Measure unstretched length of spring, L. E.g. L = 0.60m.
Set mass to a convenient value (e.g. m = 0.5kg).
Hang mass.
Measure new spring length, L'. E.g. L' = 0.70m.
Calculate extension: e = L' - L = 0.70 – 0.60 = 0.10m
Use mg = ke (in equilibrium weight = tension)
k = mg/e
Don't know what value you are using for example. Suppose it is 10N/kg (same thing as 10m/s²).
k = 0.5*10/0.10 = 50 N/m
Repeat for a few different masses. (L always stays the same.)
Take the average of your k values.
Answer:
<em>B</em><em>.</em><em> </em><em>Kinetic</em><em> </em><em>friction</em><em> </em>
Explanation:
This is definitely the correct answer because kinetic friction acts when an object is in motion and it allows the object to move without slipping, etc
<em>ALSO</em><em>,</em><em> </em><em>PLEASE DO</em><em> </em><em>MARK</em><em> </em><em>ME AS</em><em> </em><em>BRAINLIEST UWU</em><em> </em>
<em>Bonne</em><em> </em><em>journée</em><em> </em><em>;</em><em>)</em><em> </em>
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is 
Explanation:
From the question we are told that
The diameter of the wire is 
The radius of the wire is 
The resistivity of aluminum is 
The electric field change is mathematically defied as
Generally the charge is mathematically represented as
Where A is the area which is mathematically represented as
So
Therefore
substituting values
![Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5Cint%5Climits%5E%7Bt%7D_%7B0%7D%20%7B%20%5B%200.0004t%5E2%20-%200.0001t%20%2B0.0004%5D%20%7D%20%5C%2C%20dt)
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%20t%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
From the question we are told that t = 5 sec
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%205%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004%285%29%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20%285%29%5E2%7D%7B2%7D%20%2B0.0004%285%29%5D%20%7D)
Answer:
the center of mass is 7.07 cm apart from the bend
Explanation:
the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is
x₁ = L/2 = 20 cm /2 = 10 cm
when the wire is bent in a right angle the coordinates of the new centre of mass will be
x₂ = L₂/2
y₂= L₂/2
where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2
x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm
y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm
x₂=y₂=X
locating the bend in the origin (0,0) the distance to the centre of mass is
d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm
d = 7.07 cm
