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snow_tiger [21]

2 years ago

11

A 3.75 bucket of water attached to a rope is swung in a 0.555 m circle. At the bottom, the tension in the rope is 172 N. What is

the speed of the bucket at that point?

1 answer:

julsineya [31]2 years ago

8 0

Idk i meednthenpoints desparetly or ill be stucknon this question for hours with no points to ask

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A bus is traveling with a uniform acceleration of 2.75 meters/second2. If the initial velocity of the bus is 16.5 meters/second,

Kaylis [27]

<span>The velocity will be 41.25 m/s2 after 9 seconds. To find velocity after a specific time period, multiply the acceleration (2.75) times the number of seconds (9) to receive 24.75 m/s, then add that to the initial velocity of 16.5 m/s. 24.75 + 16.5 = 41.25 m/s2.</span>

8 0

3 years ago

A zebra drinks from a watering hole and is ambushed by a crocodile. What's more important and why: the zebra having a high maxim

elena55 [62]

Answer:

before this type of attack, high acceleration is the most important thing.

Explanation:

As the zebra is ambushed by the crocodile the most important thing is a quick reaction, in this attack the most likely is that the crocodile is in the water so it cannot run after the zebra.

Consequently, before this type of attack, high acceleration is the most important thing.

5 0

2 years ago

A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m

Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force $F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k$

Mass of object = 2.00 kg

Initial position $d=(2.75)i-(2.00)j+(5.00)k$

Final position $d=-(5.00)i+(4.50)j+(7.00)k$

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

$W=F\cdot d$

$W=F\cdot(d_{f}-d_{i})$

Put the value into the formula

$W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)$

$W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)$

$W=2.75\times-7.75+7.50\times6.50+6.75\times2.00$

$W=40.93\ J$

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

4 0

2 years ago

The primary colors of light are A. red, yellow, and blue. B. red, white and blue. C. blue, green, and red. D. cyan, magenta, and

zhuklara [117]

I think it is A I hope I helped you

3 0

2 years ago

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A boxcar at a rail yard is set into motion at the top of a hump. The car rolls down quietly and without friction onto a straight

enyata [817]

Answer:

I have answer with explanations in each of the six (6) special cases.

Explanation:

Please find attached the detailed solution to all your questions.

3 0

3 years ago