E=hf C=wavelength*F
E=hC/wavelength
E=(6.626*10^-34)*(3.00*10^8)/670*10^-9
E=(6.626*10^-34)*(3.00*10^8)/450*10^-9
For an ideal transformer power loss is assumed to be zero
i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage
this can be written in form of equation

here we know that


![i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1](https://tex.z-dn.net/?f=i_1%20%3D%2010%20A%7B%2Ftex%5D%3C%2Fp%3E%3Cp%3Enow%20we%20will%20use%20above%20equation%3C%2Fp%3E%3Cp%3E%5Btex%5D140%2A3.5%20%3D%2010%20%2A%20V_1)

So primary coil voltage is 49 Volts
Answer:
Mass of the steel cube = 7800 kg
Volume of the steel = 1.025 cubic centimetre
Explanation:
Given:
The density of the steel = 7.8
Side of the cube = 12 cm
<u>(1)The mass of steel cube :</u>
We know that,

We are given with density and sides of the cube
then volume of the cube
=
= 
= 1000 cubic centimetre
Now


mass = 7800 kg
<u>(2)volume of steel:</u>
Given the mass = 8 kg

Substituting the values


volume = 1.025 cubic centimetre
Answer: Mars
Explanation:
Mars is a planet similar to the earth which the rest can't sustain life either because it's too hot or cold or too much gas maybe even toxic acid.
Answer:
The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.
Determine Fx."

Explanation:
We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.
torque=cross product of force and position . mathematically this can be express as

Where
and the position vector

using the determinant method to expand the cross product in order to determine the torque we have
![\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C2%26-3%262%5C%5C%20F_%7Bx%7D%20%267%26-5%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C)
by expanding we arrive at

since we have determine the vector value of the toque, we now compare with the torque value given in the question

if we directly compare the j coordinate we have
