Answer:
B
Explanation:
You always want to test as many samples as possible
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:
angular momentum of the putty about the point of rotation
= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .
= .045 x 4.23 x 2/3 x .95 cos46
= .0837 units
moment of inertia of rod = ml² / 3 , m is mass of rod and l is length
= 2.95 x .95² / 3
I₁ = .8874 units
moment of inertia of rod + putty
I₁ + mr²
m is mass of putty and r is distance where it sticks
I₂ = .8874 + .045 x (2 x .95 / 3)²
I₂ = .905
Applying conservation of angular momentum
angular momentum of putty = final angular momentum of rod+ putty
.0837 = .905 ω
ω is final angular velocity of rod + putty
ω = .092 rad /s .
An object need to move in a straight line in the same direction in equal intervals of time in order for total distance traveled and displacement to be equal.
Answer:

Option A is correct
Explanation:
Let's check the options:
A: The electron has a negative charge and is found outside of the nucleus.
Yeah! It's TRUE . An electron is a <u>negatively</u><u> </u><u>charged</u><u> </u><u>particle</u><u> </u><u>and</u><u> </u><u>is</u><u> </u><u>located</u><u> </u><u>outside</u><u> </u><u>the</u><u> </u><u>nucleus</u><u> </u><u>of</u><u> </u><u>an</u><u> </u><u>atom.</u>
B : The neutron has a negative charge and is found in the nucleus.
No! It's FALSE . A neutron carries <u>no </u><u>charge</u>. i.e it is a neutral particle and found inside the nucleus.
C : The proton has no charge and is found in the nucleus.
No! It's FALSE.A proton is a <u>positively</u><u> </u><u>charged</u><u> </u><u>particle</u> present inside nucleus of an atom.
D : The neutron has no charge and is found outside of the nucleus.
I agree that the neutron has no charge. But it is found <u>inside</u> the nucleus not outside . So, this statement is FALSE .
Hence, we found our answer! :D
A. The electron has a negative charge and is found outside of the nucleus is the correct statement about an atom.
Hope I helped!
Best regards! :D
~