Answer:
Gold
Explanation:
Water is H20
Salt is NaCl
Sugar is C12H22O11
Gold is Au with no other elements
Answer:
because only two electrons can fit in the first orbit around the nucleus, and each period on the table is organized by number of orbits
Answer: 16.3 seconds
Explanation: Given that the
Initial velocity U = 80 ft/s
Let's first calculate the maximum height reached by using third equation of motion.
V^2 = U^2 - 2gH
Where V = final velocity and H = maximum height.
Since the toy is moving against the gravity, g will be negative.
At maximum height, V = 0
0 = 80^2 - 2 × 9.81 × H
6400 = 19.62H
H = 6400/19.62
H = 326.2
Let's us second equation of motion to find time.
H = Ut - 1/2gt^2
Let assume that the ball is dropped from the maximum height. Then,
U = 0. The equation will be reduced to
H = 1/2gt^2
326.2 = 1/2 × 9.81 × t^2
326.2 = 4.905t^2
t^2 = 326.2/4.905
t = sqrt( 66.5 )
t = 8.15 seconds
The time it will take for the rocket to return to ground level will be 2t.
That is, 2 × 8.15 = 16.3 seconds
1.8 is the mechanical advantage of the lever.
<h3>Definition of mechanical advantage</h3>
The theoretical mechanical advantage of a system is the ratio of the force that performs the useful work to the force applied, assuming there is no friction in the system.
The advantage gained by the use of a mechanism in transmitting force specifically the ratio of the force that performs the useful work of a machine to the force that is applied to the machine.
Mechanical advantage is given by the ratio of the load lifted to the force applied to lift the load.
In this case, Mechanical advantage=L/E where L is the load and E is the effort applied.
Mechanical advantage= 90/50 =1.8
Question-you use a lever to lift a heavy tree branch. you apply a force of 50 n and the lever lifts the branch with a force of 90 n. what is the mechanical advantage of the lever?
To learn more about the Mechanical advantage visit the link
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The equation for electrical power is<span>P=VI</span>where V is the voltage and I is the current. This can be rearranged to solve for I in 6(a).
6(b) can be solved with Ohm's Law<span>V=IR</span>or if you'd like, from power, after substituting Ohm's law in for I<span>P=<span><span>V2</span>R</span></span>
For 7, realize that because they are in parallel, their voltages are the same.
We can find the resistance of each lamp from<span>P=<span><span>V2</span>R</span></span>Then the equivalent resistance as<span><span>1<span>R∗</span></span>=<span>1<span>R1</span></span>+<span>1<span>R2</span></span></span>Then the total power as<span><span>Pt</span>=<span><span>V2</span><span>R∗</span></span></span>However, this will reveal that (with a bit of algebra)<span><span>Pt</span>=<span>P1</span>+<span>P2</span></span>
For 8, again the resistance can be found as<span>P=<span><span>V2</span>R</span></span>The energy usage is simply<span><span>E=P⋅t</span></span>
