Answer:
el monitor,las impresoras y las memorias portátiles
<span>To do this question, we need to know that momentum is conserved, meaning the overall velocity of the two balls has to be the same before and after the collision. </span>
<span>After collision... </span>
<span>Ball 1: 4.33m/s *cos 30 = 3.75 m/s (x-component) </span>
<span>4.33m/s * sin 30 = 2.165 m/s ( y-component) </span>
<span>Ball 2 (struck ball): 5 m/s - 3.75m/s = 1.25 m/s (x-component) </span>
<span>-2.165 m/s (y-component) note: it has to be in the opposite direction to conserve momentum </span>
<span>tan-1(2.165/1.25) = 60 degrees </span>
<span>Struck ball's velocity = sqrt(1.25^2 + 2.165^2) = 2.5 m/s at 60 degree with respect to the original line of motion. </span>
<span>Hope you understand!</span>
Given the the current flowing in the circuit and the elapsed time, the charge that passes through the LED is 1260 Coulombs.
<h3>What is Current?</h3>
Current is simply the rate of flow of charged particles i.e electrons caused by EMF or voltage.
If a charge passes through the cross-section of a conductor in a given time, the current I is expressed as;
I = Q/t
Where Q is the charge and t is time elapsed.
Given the data in the question;
- Time elapsed t = 1hr = 3600s
- Current I = 350mA = 0.35A
We substitute our given values into the expression above to determine the charge.
I = Q/t
Q = I × t
Q = 0.35A × 3600s
Q = 1260C
Therefore, given the the current flowing in the circuit and the elapsed time, the charge that passes through the LED is 1260 Coulombs.
Learn more about current here: brainly.com/question/3192435
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Answer:
(i)The current flow in black wire = 9.67 A (ii) The current low in the red wire is 9.68 A (iii) The current flow in neutral wire is 15.36 A (iv) when 240 volt were disconnected current in black wire is 7.68 A (v) when 240 volt were disconnected current in red wire is 7.68 A (vi) 15.36 A (vii) 6.34 (viii) 9.68 A (ix) 12.02 A
Explanation:
Solution
The current drawn by one amp is
I =P/V
I =200/120
I= 1.67 A
(i) The current flow in the black wire is
IBK = 4 * 1.67 A + 1A + 2A
IBK = 9.67 A
(ii) Current flow in the red wire is
IRD = 3 * 1.67 A + 1.67 A + 1A + 2A
= 8.68A + 1 A = 9.68 A
Note: Kindly find an attached copy of part of the solution to the given question above.
Answer:
runway use is 3307.8 feet
Explanation:
given data
velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s
time = 28 seconds
weight = 28000 lbs
to find out
How many feet of runway was used
solution
we will use here first equation of motion for find acceleration
v = u + at ..............1
here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time
put here value in equation 1
72.016 = 0 + a(28)
a = 2.572 m/s²
and
now apply third equation of motion
s = ut + 0.5×a×t² .......................2
here s is distance and u is initial speed and t is time and a is acceleration
put here all value in equation 2
s = 0 + 0.5×2.572×28²
s = 1008.24 m = 3307.8 ft
so runway use is 3307.8 feet
