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3 years ago

If an element has a charge of 1+, how many more protons does it have than electrons?

1 answer:

7 0

If an element has a charge of +1, there is 1 more proton than electrons.

A proton has a charge of +1
A neutron has a charge of 0
A electron has a charge of -1.

For there to be a charge of 0, there would be the same amount of charges for both proton and neutron. To get a charge of 1, you will need 1 more proton.

hope this helps

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The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the

Inga [223]

Answer:

27.5 days

0.92 month

Explanation:

$r$ = radius of the orbit of moon around the earth = $3.85\times10^{8} m$

$M$ = Mass of earth = $5.98\times10^{24} m$

$T$ = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

$T^{2} = \frac{4\pi ^{2} r^{3} }{GM}$

inserting the values, we get

$T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3} }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec$

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

$T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days$

1 month = 30 days

$T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month$

6 0

3 years ago

Which one is correct and why?

Gelneren [198K]

'A' and 'C' are exactly the same circuit, except the voltmeter's terminals are flipped.

'A' is the correct way to hook everything up.

If you start at the positive terminal of the battery, and follow the flow of current through the circuit and around to the negative terminal, you're following the path where the voltage gets lower and lower and lower all the way.

So each time you come to any device in the circuit ... whether it's a resistor or a meter ... you would be hitting the positive side of it first, and then the voltage where you come out on the other side of it would be lower.

So the left side of the resistor is more positive, and the right side is more negative. The voltmeter is connected correctly in 'A', but it's backwards in 'C'. If you connect the voltmeter like in 'C' and turn things on, the voltmeter will try to go <em>down</em> from zero. You can't read the number on it, and It's possible that the voltmeter might be damaged.

6 0

3 years ago

If you travel from Yakima to Ellensburg (Yakima to Ellensburg is 50 miles) with a speed of 60 miles/hour for half of the

koban [17]

Answer:

$\displaystyle \frac{480}{7}\approx 68.6\; \rm mph$ .

Explanation:

The average speed of an object is equal to total distance over total time.

Distance traveled: $\rm 50 \; mi$ .

How much time is taken? This trip is divided into two halves, each of distance $\displaystyle \frac{50}{2} = 25\;\rm mi$ .

Time spent on the first half of the trip:

$\displaystyle t_1 = \frac{s_1}{v_1} = \frac{25}{60} = \frac{5}{12}\; \rm hours$ .

Similarly, time spent on the second half of the trip:

$\displaystyle t_2 = \frac{s_2}{v_2} = \frac{25}{80} = \frac{5}{16}\; \rm hours$ .

In total:

$\displaystyle \frac{5}{12} + \frac{5}{16} = \frac{35}{48} \; \rm hours$ .

Average speed:

$\begin{aligned} \text{Average speed} &= \frac{\text{Total Distance}}{\text{Total Time}}\\ &= 50 \left/\frac{35}{48}\right.\\ &= 50 \cdot \frac{48}{35} \\&= \frac{480}{7}\approx 68.6\; \rm mph \end{aligned}$ .

This value turned out to be slightly different from the average of the speed during the two halves of the journey. The reason is that the object traveled at each speed for a different amount of time. It spent more time at the slower speed, which gives that speed a greater weight in the average. That explains why the average speed is closer to $\rm 60\; mph$ rather than $\rm 80\; mph$ .