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erastovalidia [21]
3 years ago
13

What two factors affect the rate of acceleration of an object?

Physics
1 answer:
Masja [62]3 years ago
5 0

For help with this answer, we look to Newton's second law of motion:

     Force = (mass) x (acceleration)

Since the question seems to focus on acceleration, let's get
'acceleration' all alone on one side of the equation, so we can
really see what's going on.

Here's the equation again:

                                                 Force = (mass) x (acceleration)

Divide each side by 'mass',
and we have:                            Acceleration = (force) / (mass) .

Now the answer jumps out at us:  The rate of acceleration of an object
is determined by the object's mass and by the strength of the net force
acting on the object.


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A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
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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."

F_{x}=-1N.m

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

T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

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]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

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

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

8 0
3 years ago
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