Answer:
It traveled 4 centimeters.
Explanation:
In a speed versus time graph, the distance travelled is given by the area under the graph.
In this graph we have the following:
- The speed of the object is v = 1 cm/s between time t = 0 s and t = 4 s
- The speed of the object is v = 0 cm/s between time t = 4 s and t = 8 s
Since the speed in the second part is zero, the distance travelled in the second part is zero. So, the only distance travelled by the object is the distance travelled during the first part, which is equal to the area of the first rectangle:
Answer:
W = - 5.01 10¹⁰ J
Explanation:
Work is defined by the expression
W = ∫ F.dr
Where the blacks indicate vectors, in the case the force is radial and the distance is also radial, whereby the scalar producer is reduced to an ordinary product
W = ∫ F dr
W = G m₁m₂ ∫ 1 /r² dr
W = G m₁ m₂2(-1 / r)
We evaluate between the lower limits r = Re and upper r = ∞
W = G m₁m₂ (-1 / Re + 1 / ∞)
W = - G m₁ m₂ / Re
Let's calculate
W = - 6.67 10⁻¹¹ 800 5.98 10²⁴ / 6.37 10⁶
W = - 5.01 10¹⁰ J
The amount of gold in decigrams if 450 micrograms is needed is 4.5 × 10-³ decigrams.
<h3>How to convert micrograms to decigrams?</h3>
According to this question, 450 micrograms of a sample of gold is needed but we only have a mass balance that measures in decigrams.
This means that we are to convert the amount of gold you need to decigrams by comparing the exponents.
The conversion factor of micrograms to decigrams is as follows:
1 micrograms = 1 × 10-⁵ decigrams
This means 450 micrograms is equivalent to 450 × 1 × 10-⁵ = 4.5 × 10-³ decigrams
Therefore, the amount of gold in decigrams if 450 micrograms is needed is 4.5 × 10-³ decigrams.
Learn more about decigrams at: brainly.com/question/6869599
#SPJ1
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
