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ehidna [41]
3 years ago
6

Help again i need it quick

Physics
2 answers:
ale4655 [162]3 years ago
4 0

Answer:

15mph

Explanation:

Hope this helps

Romashka [77]3 years ago
3 0

Answer:

it seems that the answer should be 7.5 mph as the average

Explanation:

sorry i did (s x t) when it was (s/t)

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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\%,

4 0
3 years ago
An object that has negative acceleration is definitely doing what?
krek1111 [17]

If an object is consist of a negative acceleration or composed of it, it is likely engaging of having to accelerate in a certain direction in which the direction is heading towards the opposite of the direction that is stated and that is likely to be a positive direction.

3 0
4 years ago
Copper wire is 1. 9 mm in diameter and carries a current of 20 a. What is the electric field stregnth inside this wire?
Tju [1.3M]

The electric field strength will be 0.6252 V/m. It is the strength at which the field is created by charges.

<h3>What is electric file strength?</h3>

The electric field strength is defined as the ratio of electric force and charge.

The electric field strength is found as;

\rm E = \frac{I \rho }{A} \\\\ \rm E = \frac{20  \times 1.68 \times 10^{-8} }{ (0.6385 \times 10^{-6}} \\\\ E= 0.5262 \  V/m

Hence, the electric field strength will be 0.6252 V/m.

To learn more about the electric field strength, refer to the link;

brainly.com/question/4264413

#SPJ4

3 0
2 years ago
What does an electromagnet do?
butalik [34]
An electromagnet is a device that sends electricity through a coil of wire to produce a magnetic field. This leads to a magnet that can be controlled - turned on and off with the flip of a switch, or increased or decreased in strength. The coils are often wrapped around a regular magnet to make it stronger.
8 0
3 years ago
A 100 N force is applied to a 2.00 Kg mass<br> What is the acceleration
Law Incorporation [45]

Answer:

force=mass ×acceleration

100=2a

a=50mls²

3 0
3 years ago
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