Answer:
Maximum amount of heat = 10002151.38J
Explanation:
Workdone by motor in 86.1 minutes I given by:
W = power × time
W= 294 × 86.1×60
W= 1439424 Joules
W= 1.4 ×10^6Joules
The amount of heat extracted is given by:
/QL /= K/W/ = TL/W/ /(TH - TL)
Where TL= freezing compartment temperature
TH = Outside air temperature
/QL /= 271 × 1439424 / (310 - 271)
/QL/ = 390083904/39
/QL/ = 10002151.38 Joules
According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:

where F is the magnitude of the force, m is the mass of the object and a its acceleration.
In this problem, the object is the insect, with mass

. The acceleration of the insect is

, therefore we can calculate the force exerted by the car on the insect:

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.
Answer:
1.3 m/s
Explanation:
average speed = total distance/ total time
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We should continue to send robotic spacecrafts into space
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In other words, they help us learn things that we never knew before.
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
