A student uses a meter to measure 120 coulombs flowing through a circuit in 60 seconds. The electric current in this circuit will be 2 A
Current is a flow of electrical charge carriers, usually electrons or electron-deficient atoms. The common symbol for current is the uppercase letter I. The standard unit is the ampere, symbolized by A.
current = charge / time
given
time = 60 seconds
charge = 120 Coulombs
current = Q / T = 120 / 60 = 2 A
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B-thymine nucleotides is the answer i am sure
Answer:
120 deg Unknown = 100 deg Celsius since we know that the ice point is at
0 deg Celsius and the steam point is 100 deg Celsius
So you have 1.2 deg U / 1 deg C
70 deg U / (1.2 deg U / 1 deg c) = 58.3 deg C (temperature in deg Celsius)
Or 58.3deg C * 1.2 deg U / 1 deg C = 70 deg U
Answer:
the acceleration of an object on an incline is the value of the parallel component
Explanation:
Answer:
V(3) =14i^ -34j^ +8.57 k^
S(3) =(25,-45,3.97)
Explanation:
We know that
V =a dt
from t=0 to 3s
V = 4i - 6tJ + sin(.2t)k m/s² dt
V =4t i^ - 3t^2j^ - cos(2t)/2 k^ +C
So we have
V(0) =-1/2 k^ +C =2i -7j +8.4 k
C=2i -7j +8.9k^
V =4t i^ - 3t^2j^ - cos(2t)/2 k^ + 2i -7j +8.9k^
Then putting 3s
V(3) =14i^ -34j^ +8.57 k^
Also
S(t) =V(t) dt
S(t) = 4t i^ - 3t^2j^ - cos(2t)/2 k^ + 2i -7j +8.9k^] dt
S(t)= (2t^2 +2t)i^ - (t^3 +7t) j^ -[sin(2t)/4 - 8.9t] k^ +C =i+3j-5k
So at t= 0 we have
S(0) = C= i+3j-5k
So S(t) =(2t^2 +2t)i^ - (t^3 +7t) j^ -[sin(2t)/4 - 8.9t] k^ +i+3j-5k
When t= 3
S(3) =25i -45j + 3.97k
S(3) =(25,-45,3.97)