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FrozenT [24]
3 years ago
5

Please solce thiss quesstion​

Physics
1 answer:
noname [10]3 years ago
4 0

Answer:

Can you give a clearer question

Explanation:

pls give me points i have questions

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A step-down transformer has 2500 turns on its primary and 5.0 x 10' tums on its secondary. If the potential difference across th
Brilliant_brown [7]

Answer:

I dont know sorry

Explanation:

hehe

4 0
3 years ago
A single loop of current is immersed in an externally applied uniform magnetic field of 3 Tesla oriented in the positive y direc
vlabodo [156]

Answer:

mu=12Tm^2

Explanation:

the magnetic moment mu of a single loop is given by:

\mu = I A B

where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:

\mu=(0.5A)(4m*2m)(3T)=12Tm^2

hope this helps!!

6 0
3 years ago
The diagram does not represent a real electric field because the field lines, can someone help explain this for me
Papessa [141]

electric field lines are graphical presentation of electric field intensity

It is the graphical way to represent the electric field variation

If we draw the tangent to electric field line then it will give the direction of net electric field at that point

So whenever we draw the electric field lines of a charge distribution then it will always follow this basic properties

here we will always follow these basic properties of field lines

now as we can see that here two positive charges are placed nearby so the electric field must be like it can not intersect at any point because at intersection of two lines the direction of electric field not defined

As we have two directions of tangents at that point

So here the incorrect presentation is the intersection of two field lines which is not possible


4 0
3 years ago
If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int
vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system on the surrounding


In this problem, the work done by the system is

W=-225 kcal=-941.4 kJ

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

Q=-5 \cdot 10^2 kJ=-500 kJ

with a negative sign as well because it is released by the system.


Therefore, by using the initial equation, we find

\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ

8 0
3 years ago
What's the definition energy?​
Anit [1.1K]
Energy is the quantitative property that must be transferred to an object in order to perform work on, or to heat, the object.
4 0
3 years ago
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