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Sati [7]
4 years ago
13

Electricity may be categorized as static or current. how are these two types of energy different

Physics
1 answer:
arlik [135]4 years ago
4 0
Hello!

Static electricity occurs due to an imbalance in positively and negatively charged atoms. An example of this is when you take your clothes out of the dryer, and feel a slight sting when touching them. Another example of static electricity is lightning.

Current electricity occurs when there is a constant flow of electrons, such as in plug-operated machinery or anything operated using a battery. :)
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An argument against the validity of evolutionary theory says that there are too few transitional forms representing the gap betw
ICE Princess25 [194]

Answer:They both relate in that almost every specie in the pre-mordern human fossil record can be deregarded as a transitional specie between the African Ape and the modern man.

Explanation:

6 0
3 years ago
Please answer help me
Andrews [41]
I believe the answer is x
7 0
3 years ago
Given that the concentration of bovine carbonic anhydrase is 3.3 pmol ⋅ L − 1 and R max ( V max ) = 222 μmol ⋅ L − 1 ⋅ s − 1 , d
LuckyWell [14K]

Answer:

The turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1.

Explanation:

Given:

The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^–1 = 3.3 × 10^–12 mol.L^–1

The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^–1⋅s^–1 = 222 × 10^–6 mol.L^–1⋅s^–1

The formula for the turnover number of an enzyme (kcat, or catalytic rate constant) = Rmax ÷ Et = 222 × 10^–6 mol.L^–1⋅s^–1 ÷ 3.3 × 10^–12 mol.L^–1 = 67,272,727.27 s^–1

Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1

3 0
4 years ago
What are two types of electric currents?
s2008m [1.1K]

Answer:

There are two different types of current in widespread use today. They are direct current, abbreviated DC, and alternating current, abbreviated AC. In a direct current, the electrons flow in one direction.

6 0
3 years ago
A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion s
loris [4]

To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

m*V_i = m_1V_1+m_2V_2

Here,

m = Total mass (8Kg at this case)

m_1=m_2 = Mass each part

V_i = Initial velocity

V_2 = Final velocity particle 2

V_1 = Final velocity particle 1

The initial kinetic energy would be given by,

KE_i=\frac{1}{2}mv^2

KE_i = \frac{1}{2}8*5^2

KE_i = 100J

In the end the energy increased 100J, that is,

KE_f = KE_i KE_{increased}

KE_f = 100+100 = 200J

By conservation of the moment then,

m*V_i = m_1V_1+m_2V_2

Replacing we have,

(8)*5 = 4*V_1+4*V_2

40 = 4(V_1+V_2)

V_1+V_2 = 10

V_2 = 10-V_1(1)

In the final point the cinematic energy of EACH particle would be given by

KE_f = \frac{1}{2}mv^2

KE_f = \frac{1}{2}4*(V_1^2+V_2^2)

200J=\frac{1}{2}4*(V_1^2+V_2^2)(2)

So we have a system of 2x2 equations

V_2 = 10-V_1

200J=\frac{1}{2}4*(V_1^2+V_2^2)

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: V_1 = 10m/s

Then replacing in (1) we have that

PART B: V_2 = 0m/s

8 0
3 years ago
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