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vaieri [72.5K]
3 years ago
13

HELP PLEASE : ) IT'S SCIENCE

Physics
1 answer:
nekit [7.7K]3 years ago
3 0

Answer:

1. A and C are balanced forces as all the forces get cancelled by the opposite and equal force.

2. B and D as in B the Net force is 5N to the right and in D the Net force is 20N to the right.

3. This would be all the unbalance forces which are B and D

4. B would be moving 5N to the right and D would move 20N to the right.

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The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the
FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

B=\frac{2\pi*a }{u*I}

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a  = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

I=\frac{2\pi*a*B }{u} Formula(2)

B=0.2*10^{-5}  T = 0.2*10^{-5} \frac{weber}{m^{2} }

a  =8cm=0.08m

u=4*\pi *10^{-7} \frac{Weber}{A*m}

We replace the known information in the formula (2)

I=\frac{2\pi*0.08*0.2*10^{-5}  }{4\pi *10x^{-7} }

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0
3 years ago
Describe the steps of aerobic cellular respiration and the amount of energy produced by each step.
Setler [38]

Aerobic cellular respiration have 3 parts, this is Glycolysis, Pyruvate Oxidation and Krebs cycle.

<h3>How is the aerobic breathing process?</h3>

Aerobic respiration consists of carrying out the process of degradation of organic molecules, reducing them to molecules with practically no releaseable energy. The products of the initial degradation of the organic molecule are combined with oxygen in the air and transformed into carbon dioxide and water.

In this case, Aerobic cellular respiration have 3 parts:

  • Glycolysis(yeilds 2ATP & 2NADH).
  • Pyruvate Oxidation(yeilds 2NADH).
  • Krebs cycle(yeilds 2GTP,2FADH2 & 6NADH).

So the total =4ATPs (2GTP equivalent to 2ATP) +10NADH (equivalent to 30ATPs) + 2FADH2(4 ATPs) =38 ATPs.

See more about Aerobic cellular respiration at brainly.com/question/22531444

#SPJ4

3 0
2 years ago
How do defy gravity?
kaheart [24]
Exert force upward.
Like when you pick something up from the floor, or walk up the stairs.
6 0
4 years ago
A moving company collects data about couches they move and displays the data in a table. The force along the ramp, the angle of
USPshnik [31]

Answer:

the correct answers are d, e, & f

Explanation:

just took the edge test

7 0
4 years ago
Read 2 more answers
A 7.0 mm -diameter copper ball is charged to 40 nC. What fraction of its electrons have been removed? The density of copper is 8
mylen [45]

Answer:

f = 2.6 \times 10^{-13}

Explanation:

Let the mass of copper ball is "m" gram

now the total number of copper atom present in the ball is given as

N = \frac{m}{29} \times 6.02 \times 10^{23}

now the total number of electrons in one copper atom is 29

so total number of electrons in given sample of copper ball is

N_e = m(6.02 \times 10^{29})

now diameter of the ball is 7.0 mm

density of the ball = 8900 kg/m^3

now we have

m = (\frac{4}{3}\pi r^3)(8900)

m = (\frac{4}{3}\pi(\frac{0.007}{2})^3)(8900)

m = 1.6 gram

now we have

N_e = 9.63 \times 10^{23}

now the charge on the copper ball is 40 nC

so the number of electrons removed

Q = ne

40 \times 10^{-9} = n(1.6 \times 10^{-19}

n = 2.5 \times 10^{11}

so the fraction of number of electrons removed is given as

f = \frac{n}{N_e}

f = \frac{2.5 \times 10^{11}}{9.63 \times 10^{23}}

f = 2.6 \times 10^{-13}

7 0
3 years ago
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