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Nezavi [6.7K]
2 years ago
7

I am unique even though there are billions of me no two are the same what am I

Physics
2 answers:
Aleksandr [31]2 years ago
6 0
I think the answer is Snow flakes
erastova [34]2 years ago
4 0

Answer:

Flowers

Explanation:

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How much force is required to accelerate a 800kg car by 15 m/s2
Tom [10]

Answer:

force=12000

Explanation:

F=m*a aka force equals mass times acceleration so 800*15=12000

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2 years ago
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What stores energy for a quick release in a cell?
luda_lava [24]
It is the mitochondria of a cell that stores energy for a quick release. <span>Mitochondria break down glucose to release the energy for cells to use. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>
5 0
3 years ago
What to do if vehicle struck by lightning
agasfer [191]
Lighting flows around the outside of a truck, and the majority of the current flows from the cars metal cage into the ground below. It's not very safe to be in a car or truck during bad weather.
4 0
3 years ago
You pull on a spring whose spring constant is 22 N/m, and stretch it from its equilibrium length of 0.3 m to a length of 0.7 m.
Liono4ka [1.6K]

Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x  Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

F_{a} =\frac{F_{f}+F_{i}  }{2}  Formula (2)

Ff: final force

Fi: initial force

The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx :  displacement

W = F_{a} (x_{f} -x_{i} )   :Formula (3)

x_{f} :  final deformation

x_{i}  :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N

F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N

We calculate average force applying formula (2):

F_{a} =\frac{15.4N+6.2N}{2} = 11 N

We calculate the work done on the spring  applying formula (3) :         :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

E_{pf} =\frac{1}{2} *k*x_{f}^{2}

E_{pi} =\frac{1}{2} *k*x_{i}^{2}

E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J

E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J

W=ΔEp=  5.39 J-0.99 J = 4.4J

:

4 0
2 years ago
why can scientists ignore the forces of attraction between particles in a gas under ordinary conditions?
Anna35 [415]
<span>The particles in a gas are apart and moving fast, so the forces of attraction are too weak to have a noticeable effect.</span>
5 0
3 years ago
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