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Hitman42 [59]
3 years ago
12

While playing football, Chris runs 4.5 m at 20⁰ south of west. If an opponent was trying to tackle him how far west and south wo

uld he have to run to reach him?
Physics
1 answer:
rewona [7]3 years ago
5 0

Answer:

South = 1.5m

West =4.2m

Explanation:

Kindly see attached a rough draft of the situation

Step one

Given data

From the sketch the direction of the player is along the resultant of the triangle, corresponding to the Hypotenuse

Step two:

Hence for an opponent to tackle him towards the south, he must be at

sin θ= opp/hyp

sin 20=x/4.5

x=sin 20*4.5

x=0.342*4.5

x= 1.5m

Also, for an opponent to tackle him towards the south, he must be at

cos θ= adj/hyp

cos 20=y/4.5

y=cos 20*4.5

y=0.93*4.5

y= 4.2m

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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
Geologists apply various methods to study the layers of the Earth. Which of the following is NOT a method used to study the Eart
77julia77 [94]

Answer:

A. Scientists use seismic computer models to measure the atmospheric conditions above the Earth's crust

Explanation:

why would use atmosphere to study the layers of earth? dont think thats possible

8 0
3 years ago
What happens to a visible light wave when you increase the frequency
Anastasy [175]
When frequency increases, the wavelength halves.
6 0
3 years ago
Christopher drives into the city to buy new new hockey equipment. Because of traffic conditions, he averages only 15 mph. On the
alina1380 [7]

Answer:

36 minutes

Explanation:

Distance= Time*Speed

Let time taken to go to city be x hours, therefore, time to travel back is (2-x) hours

The distance to and from the city are equal hence

15x=35(2-x)

15x=70-35x

50x=70 hence x=70/50=1.4 hours

Time to drive home will be 2-x=2-1.4=0.6 hours

0.6*60=36 minutes

Therefore, time to travel back home is 36 minutes

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3 years ago
Why does he have maximum potential energy at this point?
yan [13]
He is going the fastest speed
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