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VashaNatasha [74]
2 years ago
10

A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is

Physics
1 answer:
sp2606 [1]2 years ago
7 0

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

           v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

          x = v₀² / 2a

let's calculate

          x = 2²/(2 0.8)

         x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

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Describe the parts of a lever. Include the following terms (fulcrum, resistance arm and effort arm).
yarga [219]

Answer:

hi here is your answer and this is a very important question.

Explanation:

A lever is a rigid bar with three parts: the fixed point around which the bar pivots is the fulcrum: the effort arm (in-lever arm) is the part of the lever to which force is applied; the resistance arm (out-lever arm) is the part that bears the load to be moved.

5 0
3 years ago
12. Unmanned Space Probe A 2500 kg unmanned space probe is moving in a straight line at a constant speed of 300 m/s. Control roc
Lena [83]

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4 0
3 years ago
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A weightlifter lifts a 1250-N barbell 2 m in 3 s.How much power was used to lift the barbell?
Vlad [161]

The power is 833.3 W

Explanation:

First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:

W=(mg)h

where

mg = 1250 N is the weight of the barbell

h = 2 m is the change in height

Substituting,

W=(1250)(2)=2500 J

Now we can calculate the power, which is equal to the work done per unit time:

P=\frac{W}{t}

where

W = 2500 J is the work done

t = 3 s is the time taken

Substituting,

P=\frac{2500}{3}=833.3 W

Learn more about power:

brainly.com/question/7956557

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5 0
3 years ago
Earth is slightly closer to the sun in january than in july. How does the area swept out by earth’s orbit around the sun during
mr Goodwill [35]

This is because both areas are the same.

6 0
3 years ago
Two of the three tuning forks have known frequencies. When a 610 Hz fork and the unknown fork are struck together, four beats pe
RUDIKE [14]

-- In combination with 610 Hz, the beat frequency is 4 Hz.

So the unknown frequency is either (610+4) = 614 Hz
or else (610-4) = 606 Hz.

In combination with 605 Hz, the beat frequency will be
either (614-605) = 9 Hz or else (606-605) = 1 Hz.

-- In actuality, when combined with the 605 Hz, the beat
frequency is too high to count accurately.  That must be
the 9 Hz rather than the 1 Hz.

So the unknown is (605+9) = 614 Hz.
 
6 0
3 years ago
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