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Dennis_Churaev [7]
3 years ago
13

1. A limiting factor for using nuclear energy is the

Physics
2 answers:
Goryan [66]3 years ago
8 0

Answer:

waste

Explanation:

took da test. nhgvgg

Arturiano [62]3 years ago
3 0
Pollutiondfffgggcffff fffffg
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What energy transfer will a stretched rubber band have when let go
GarryVolchara [31]

Answer:

when the rubber band is realeased the potential energy is quickly converted to kinetic energy this is equal to one mass of the the rubber band multiplied by its velocity( in meters per second)

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3 years ago
What is the difference between chemical change and a physical change?
NeX [460]

Answer:

Explanation:

There are several differences between a physical and chemical change in matter or substances. A physical change in a substance doesn't change what the substance is. In a chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed.

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3 years ago
Which type of ocean sediment contains the remains of dead organisms?
konstantin123 [22]
The <span>biogenous sediment contains the remains of dead organisms such as shells and skeletons.</span>
4 0
3 years ago
Read 2 more answers
A projectile is launched straight upwards at 75 m/s. Three seconds later, its velocity is...?
yawa3891 [41]

Answer:

V = V0 + a t

V = 75 - 9.8 * 3 = 45.6 m/s

4 0
3 years ago
You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an
a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, \mu=6\times 10^{-2}

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

KE=PE+W

\dfrac{1}{2}mv^2=mgh+W

W is the work done by the friction.

W=f\times d

f=\mu mg\ cos\theta

W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}

W=\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}

\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0
3 years ago
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