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VashaNatasha [74]
3 years ago
7

If a charge at 60c flow in a conductor for 30 second then the current that flow in a conductor is​

Physics
1 answer:
saw5 [17]3 years ago
3 0

Explanation:

<h3>Given</h3>

- Charge = 60c

time = 30 sec

<h3>To find -</h3>

current

<h3>Solution </h3>

Current = Charge/time

I = V/T

I = 60/30

I = 2 ampere

More to know -

I = Current

V = Charge

T = Time

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Can someone help me with this question
Mademuasel [1]

Answer:

Net force: 20 N to the right

mass of the bag: 20.489 kg

acceleration:  0.976  m/s^2

Explanation:

Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:

195 N - 175 N = 20 N

So net force on the bag is 20 N to the right.

The mass of the bag can be found using the value of the weight force: 201 N:

mass = Weight/g = 201 / 9.81 = 20.489 kg

and the acceleration of the bag can be found as the net force divided by the mass we just found:

acceleration = 20 N / 20.489 kg = 0.976  m/s^2

8 0
3 years ago
Why is it important for an engineer who is designing a new home-cooling/heating system to have knowledge about the transfer of h
julia-pushkina [17]

Answer:

In the explanation :)

Explanation:

Heat is a concept that is important to understand in various engineering fields. It is particularly relevant for civil, mechanical and chemical engineers because heat transfer plays a key role in material selection, machinery efficiency and reaction kinetics, respectively.

3 0
3 years ago
Read 2 more answers
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

6 0
3 years ago
a current of 1.80 a flows in a wire. how many electrons are flowing past any point in the wire per second?
liubo4ka [24]

Any point in the wire has 1.12 x 10^{19}. electrons flow per second.

<h3>What causes a current in a wire?</h3>
  • Electric current in a wire, where electrons serve as the charge carriers, is a measurement of the amount of charge that moves through any point of the wire in a given amount of time.
  • A free electron is drawn to a proton to become neutral if an electron is added to the wire.
  • Lack of electrons can result from pushing electrons out of their orbits.
  • Electric current is the name given to the constantly moving electrons in wire.

The current is the quantity of charge Q flowing through a certain point of the wire in a time interval of \Delta t.

I = \frac{Q}{\Delta t}.

by using this relationship

I=1.80 A, we can find the charge passing any point in the wire in 1 second:

Electric Charge, Q = 1.80 C.

To find how many electrons corresponds to this charge, we should divide this value by the charge of a single electron

charge of the electron = 1.6 x 10^{-19} C.

No. of Electrons = Q/q = \frac{1.80}{1.6* 10^{-19}}= 1.12 x 10^{19}.

To learn more about Electric current   refer,

brainly.com/question/9467901

#SPJ4

4 0
1 year ago
Use the following terms in the same sentence: proton,neutron,and isotope
Juliette [100K]
An isotope is one of 2+ types of the same element, which will have the same number of protons but a different number of neutrons in its nucleus. This means that the relative atomic mass may be different among isotopes, but the chemical properties will remain the same. Isotopes are usually found in radioactive forms of an element. 
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3 years ago
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