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katen-ka-za [31]

3 years ago

11

If the magnitude of a positive charge is tripled, what is the ratio of the original value of the electric field at a point to th

e new value of the electric field at that same point

1 answer:

dybincka [34]3 years ago

7 0

Answer:

Here is the complete question.

If the magnitude of a positive charge is tripled, what is the ratio of the original value of the electric field at a point to the new value of the electric field at that same point.

a) 1:2

b) 1:3

c)1:6

d)1:9

b) 1:3 is the correct option.

Explanation:

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The length of a planet's day is determined by it's

MatroZZZ [7]

Answer: "People will say the length of a day is determined by just the rotation rate of a planet, but that isn’t entirely true. The orbit around its sun also factors in. What we define as a day is the time between when the Sun is highest in the sky until the next time the Sun is at the highest point in the sky. This isn’t just the time it takes the Earth to rotate once on its axis, though. This is because, as the Earth is rotating, it’s also moving relative to the Sun"

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3 years ago

8. How many calories of heat energy (Q) would be released as a 2,500 g iron (Fe) frying pan at 190˚C cools to 25˚C?

Sedaia [141]

Answer:

$Q=197505J$

Explanation:

From the question we are told that:

Mass $m=2500g=2.5kg$

Initial heat of pan $i_h=190 \tetxtdegree$

Final heat of pan $f_h=25˚C \tetxtdegree$

Generally the equation for Heat energy released is mathematically given by

$Q=mC \triangle T\\\\Q=2.5*462 *(190-26)\\\\Q=1155 *(190-26)\\\\Q=1155 *(171)\\$

$Q=197505J$

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3 years ago

Effciency of a lever is never 100% or more. why?Give reason

Troyanec [42]

Answer:

Ideally, the work output of a lever should match the work input. However, because of resistance, the output power is nearly always be less than the input power. As a result, the efficiency would go below $100\%$ .

Explanation:

In an ideal lever, the size of the input and output are inversely proportional to the distances between these two forces and the fulcrum. Let $D_\text{in}$ and $D_\text{out}$ denote these two distances, and let $F_\text{in}$ and $F_\text{out}$ denote the input and the output forces. If the lever is indeed idea, then:

$F_\text{in} \cdot D_\text{in} = F_\text{out} \cdot D_\text{out}$ .

Rearrange to obtain:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}$

Class two levers are levers where the perpendicular distance between the fulcrum and the input is greater than that between the fulcrum and the output. For this ideal lever, that means $D_\text{in} > D_\text{out}$ , such that $F_\text{in} < F_\text{out}$ .

Despite $F_\text{in} < F_\text{out}$ , the amount of work required will stay the same. Let $s_\text{out}$ denote the required linear displacement for the output force. At a distance of $D_\text{out}$ from the fulcrum, the angular displacement of the output force would be $\displaystyle \frac{s_\text{out}}{D_\text{out}}$ . Let $s_\text{in}$ denote the corresponding linear displacement required for the input force. Similarly, the angular displacement of the input force would be $\displaystyle \frac{s_\text{in}}{D_\text{in}}$ . Because both the input and the output are on the same lever, their angular displacement should be the same:

$\displaystyle \frac{s_\text{in}}{D_\text{in}} =\frac{s_\text{out}}{D_\text{out}}$ .

Rearrange to obtain:

$\displaystyle s_\text{in}=s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}$ .

While increasing $D_\text{in}$ reduce the size of the input force $F_\text{in}$ , doing so would also increase the linear distance of the input force $s_\text{in}$ . In other words, $F_\text{in}$ will have to move across a longer linear distance in order to move $F_\text{out}$ by the same $s_\text{out}$ .

The amount of work required depends on both the size of the force and the distance traveled. Let $W_\text{in}$ and $W_\text{out}$ denote the input and output work. For this ideal lever:

$\begin{aligned}W_\text{in} &= F_\text{in} \cdot s_\text{in} \\ &= \left(F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}\right) \cdot \left(s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}\right) \\ &= F_\text{out} \cdot s_\text{out} = W_\text{out}\end{aligned}$ .

In other words, the work input of the ideal lever is equal to the work output.

The efficiency of a machine can be measured as the percentage of work input that is converted to useful output. For this ideal lever, that ratio would be $100\%$ - not anything higher than that.

On the other hand, non-ideal levers take in more work than they give out. The reason is that because of resistance, $F_\text{in}$ would be larger than ideal:

$\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}} + F(\text{resistance})$ .

As a result, in real (i.e., non-ideal) levers, the work input will exceed the useful work output. The efficiency will go below $100\%$ ,

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fomenos

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4 years ago