-2/5 = 11k - k
-2/5 = 10k
-2/5/10 = k
-2/5 * 10 = k
-2/50 = k
k = -1/25.
-1/25 - 2/5 = 11k is true.
Answer:
Hello your question is poorly written below is the complete question
Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?
answer :
a) 231.48 days
b) n = 3.125 * 10^15
Explanation:
Battery moved 10,000 coulombs
current rate = 0.5 mA
<u>A) Determine how long the clock run on the battery. use the relation below</u>
q = i * t ----- ( 1 )
q = charge , i = current , t = time
10000 = 0.5 * 10^-3 * t
hence t = 2 * 10^7 secs
hence the time = 231.48 days
<u>B) Determine how many electrons per second flowed </u>
q = n*e ------ ( 2 )
n = number of electrons
e = 1.6 * 10^-19
q = 0.5 * 10^-3 coulomb ( charge flowing per electron )
back to equation 2
n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )
hence : n = 3.125 * 10^15
Check the current weather map for 2 different times, and see where the center of the storm is. That tells you what direction it's moving. With its speed and direction, you have its velocity.
The radius of the sphere in meters is ,r =
Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,
r/10 = tan [arctan(1/2)] = tan (1/2)
Since, tan (θ/2) = 1-cos(θ) / sin(θ)
we find that,
r/10 = 
Hence, r = 
So, the radius of the sphere in meters is ,r =
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brainly.com/question/14100787
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F = m*a
30 N = (ma + mb) * a
30 = 5*a
a = 6 m/s ^2
F de B em A
30 - F de B,A = ma * a
30 - F de B em A = 3 * 6
30 - 18 = F de B em A
12 = F de B em A
Resposta: 6 m/s^2 e 12N
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