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3 years ago

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Physics

1 answer:

igomit [66]3 years ago

6 0

Reactions between atoms occur whenan atom seeks to fill its outer shell of electrons

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15 points for 2 questions

Elden [556K]

I believe the answer would be c because i think that you multiply the 2

8 0

3 years ago

Read 2 more answers

How does information travel during transcription?

Verdich [7]

Answer: c) DNA to RNA

3 0

2 years ago

Samples of different materials, A and B, have the same mass, but the sample

madam [21]

Answer:

<u><em>A.)</em></u> The particles that make up material A have more mass than the

particles that make up material B.

HOPE that helps!!! :)

4 0

2 years ago

A box with a mass of 2 kg accelerates in a straight line

mihalych1998 [28]

Answer: 16N

Explanation:

Given that:

mass of box M= 2 kg

Initial speed V1 = 4 m/s

Final speed V2 = 8 m/s

Time taken T= 0.5 s

Average strength of this force F = ?

Now, recall that Force is the rate of change of momentum per unit time

i.e Force = momentum / time

Hence, F = M x (V2 - V1)/T

F = 2kg x (8 m/s - 4 m/s) / 0.5s

F = 2kg x (4 m/s / 0.5s)

F = 2kg x 8 m/s/s)

F = 16N

Thus, the average strength of this

force is 16 newton.

7 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