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3 years ago

Plz I really need this question!!!!

1 answer:

7 0

I have a hunch that you've got the question totally covered, and what you actually need is the answer.

Kinetic energy = (1/2) (mass) (speed²)

Multiply each side by 2 : 2 x KE = (mass) ( speed²)

Divide each side by (mass): Speed² = 2 x KE / mass

Square root each side: Speed = √(2 KE/mass)

Look at that ! The question GIVES you the KE and the mass. All you have to do is plug those 2 numbers into the right side of that equation, turn the crank, do the arithmetic, and the speed falls out.

I get 200 m/s . You need to check my work.

(IF that's correct or anywhere close, it's equivalent to something around 447 miles per hour, which is very reasonable for a cruising airliner.)

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A 1.2 KG rubber ball is being thrown in the air if the ball is traveling at 2.0 M/S when it is 3.0 M off the ground what is the

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3 years ago

Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour

Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = $\frac{t}{\sqrt{1-\frac{v^2}{c^2} } }$

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

$\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }$

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

$\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }$

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

$\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }$

=1.67 x 10

t₁= 16.7 years

D ) Given t = 10 years , t₁ = ? v = .4c

$\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }$

=1.09 x 10

t₁= 10.9 years

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2 years ago

Question 7 (2 points)

Arlecino [84]

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Explanation:

3 0

3 years ago

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0

3 years ago