For a questions which involves constant acceleration, you can use the SUVAT equations.
S - displacement - -26.5m ( If you to take the right as a positive value - since displacement is a vector quantity involving direction and magnitude)
U
V - Final Velocity - unknown
A - Acceleration - -1.8m/s^2
T - Time - 3.81s
s = vt - 1/2at^2
-26.5m = 3.81v - 1/2 • -1.8 • 3.81^2
Rearrange
-26.5m/-1/2•-1.8•3.81^2=3.81v
3.81v = -2.028~
v = -0.532~ (3sf)
D. You’d need to experiment!
The power source voltage remains the same in a parallel circuit,
And we'll have equal current in both lines
<h3>Kirchhoff's junction rule</h3>
Generally, Kirchhoff's junction rule states that when there is current flow at any junction of a circuit, the total sum of this current rushing into the junction amount to the same amount of current out of the Node.
Therefore, when the parallel circuit has two branches
i=i1+12
Since we have an equal resistor therefore we'll have equal current in both lines i.e i1=i2
And Voltage remains the same in a parallel circuit
More on Voltage
brainly.com/question/14883923
The solution would be like
this for this specific problem:<span>
KCL at Junction a. </span><span>
<span>+ I1 + I2 + I3 = 0 (1) </span>
<span>KVL
<span>+ 13 V - 0.2 R I1 - 0.025 R I1 - 5 V + 0.02 R I2 = 0 (2) </span>
<span>8 + 0.02 I2 = 0.225 I1 </span>
<span>I1 = 35.6 + 0.0889 I2 (2A) </span></span></span>
<span>KVL (bottom loop - CCW
direction) </span><span>
<span>- 0.02 R I2 + 5 V + 0.5 R I3 = 0 (3) </span>
<span>0.5 I3 = -5 + 0.02 I2 </span>
<span>I3 = -10 + 0.04 I2 (3A) </span></span>
<span>Replace
2A and 3A into 1. </span><span>
<span>+ I1 + I2 + I3 = 0 </span>
<span>( 35.6 + 0.0889 I2 ) + I2 + ( -10 + 0.04 I2 ) = 0 </span>
<span>1.129 I2 = -25.6 </span>
<span>I2 = -22.6A </span>
<span>Solve 2A and 3A for other currents. </span>
<span>I1 = 35.6 + 0.0889 I2 = 35.6 + 0.0889 * -22.6 = 33.5A </span>
<span>I3 = -10 + 0.04 I2 = -10 + 0.04 * -22.6 = -10.9A
So the answer is letter D.</span></span>
The melting of an ice cube is considered a physical change.