Answer:
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Answer:
18 Ω
Explanation:
As K and F are at the same voltage, we can redraw the diagram as in figure 2
Series resistances add directly, so we get figure 3
Adding parallel resistances gets us to figure 4
Now we can move two 6Ω resistances for clarification in figure 5
As the voltage between C and J will be identically split between D and H, there will be no voltage drop across the middle 6Ω resister and no current through it, identical to an infinite resistance, so that 6Ω can be eliminated as in figure 6
Add series resistances to get to figure 7
Add parallel resistances to get to figure 8
Add series resistances to get to figure 9
Answer:
The chunk went as high as
2.32m above the valley floor
Explanation:
This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.
Applying the principle of energy conservation for the two ice we have based on the scenery
Momentum before impact = momentum after impact
M1U1+M2U2=(M1+M2)V
Given data
Mass of ice 1 M1= 5.20kg
Mass of ice 2 M2= 5.20kg
velocity of ice 1 before impact U1= 13.5 m/s
velocity of ice 2 before impact U2= 0m/s
Velocity of both ice after impact V=?
Inputting our data into the energy conservation formula to solve for V
5.2*13.5+5.2*0=(10.4)V
70.2+0=10.4V
V=70.2/10.4
V=6.75m/s
Therefore the common velocity of both ice is 6.75m/s
Now after impact the chunk slide up a hill to solve for the height it climbs
Let us use the equation of motion
v²=u²-2gh
The negative sign indicates that the chunk moved against gravity
And assuming g=9.81m/s
Initial velocity of the chunk u=0m/s
Substituting we have
6.75²= 0²-2*9.81*h
45.56=19.62h
h=45.56/19.62
h=2.32m
By definition, the work done by a force is given by:

Where,
F: magnitude of force
d: distance traveled.
Substituting values we have:
Answer:
the work w1 done on the block by the force of magnitude f1 = 60.0 n is:
W1 = 2.40 J