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

Answer:
Explanation:
Electric field due to a point charge Q at a point at distance d is given by the relation
E = 
Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.
The charges are situated on the corners of a square in such a way that
equal charges of Q1 and Q3 are situated on the diametrically opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two charges will be zero.
On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero
Overall, net field due to all the four charges will be zero
Scientists could investigate this idea by making cathode ray tubes out of different materials to see if the ray was the same.
Explanation:
the morning of the birthday party balloon filled with the 2.5 Litre of helium
temperature is 294kelvin
the party starts at the 4 p.m.
temperature rises 305 Kelvin.
the new volume = 4 litre.
At same temperature,
P
1
V
2
=P
2
V
2
(Boyle's law)
P
1
=10atm;P
2
=1atm
V
1
=4l=V
2
=8l
But while filling balloons from cylinder when pressure in cylinder becomes 1 atm then further filling is not possible (P
1
′
=9atm)
Let n be the number of balloons that can be filled.
∴P
1
′
V
1
=n(P
1
V
1
)
9×8=n(4×1)
n=
4
9×8
=18balloons