Response:R=1607556m
θ=180degrees
Clarification:
d1=74.8m
d2=160.7km=160.7km*1000
d2=160700m
d3=80m
d4=198.1m
Utilizing an analytical approach:
Rx=-(160700+75*cos(41.8))= -160755.9m
Ry= -(74.8+75sin(41.8))-198.1=73m
Magnitude, R:
R=√Rx+Ry
R=√160755.9^2+20^2=160755.916
R=160756m
Direction,θ:
θ=arctan(Rx/Ry)
θ=arctan(-73/160755.9)
θ=-7.9256*10^-6
It is worth noting that since θ is in the second quadrant, 180 is added
θ=180-7.9256*10^6=180degrees
Answer:
All observers are accurate.
Explanation:
This situation reflects a matter of reference frames regarding the book's motion as perceived by different observers.
From their distinct frames of reference, each observer's perspective is valid.
Observer A is in an inertial reference frame.
Observers capable of explaining the book's behavior and its relationship to the car through the interplay of forces and changes in velocity are classified as being in inertial reference frames.
Observer A's observations illustrate this, for she pointed out the relative motion between the book and the car, indicating her position in an inertial reference frame.
Likewise, observers in these inertial reference frames can elucidate object velocity changes based on the forces affecting them from other objects.
This is exemplified by observer B, who notes the car's force impacting the book's velocity.
Observer C occupies a non-inertial reference frame, as Newton's laws of motion do not apply. This scenario arises within non-inertial frames.
Answer:
155.38424 K
2.2721 kg/m³
Explanation:
= Reservoir pressure = 10 atm
= Reservoir temperature = 300 K
= Exit pressure = 1 atm
= Exit temperature
= Specific gas constant = 287 J/kgK
= Specific heat ratio = 1.4 for air
Assuming isentropic flow
Flow temperature at exit is 155.38424 K
Density at exit can be derived using the ideal gas equation
Flow density at exit measures 2.2721 kg/m³
I will analyze each option. My assumption is that the answer is C.
Option A states that gravity acts downward on the box but does not affect its horizontal acceleration, provided there is no friction.
Option B indicates that the normal force goes upward on the box, which also does not influence horizontal acceleration.
In option C, the reaction force discussed relates to Newton’s 3rd law. This reaction force acts on Lien rather than the box itself, meaning she must overcome this force to set the box in motion. I believe this is the correct choice.
Option D refers to the push force applied by her; she wouldn’t have to counteract her own force regarding the box, but must address the reaction force as I mentioned in option C.
