Answer:
10 kg
Explanation:
p=ma
25=2.5m
m=25/2.5
m=10 kg
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Answer:
Before the plank will tip cat will walk 1.652 m
Explanation:
Mass of the cat along with plank 
Center of mass of the plank 
Mass of cat 
We have to find how far right of sawhorse B.
Plank will tip when weight of the cat about B is greater than the torque by the weight of the plank.
Balancing the the torque
Answer:
F > W * sin(α)
Explanation:
The force needed for the box to start sliding up depends on the incline (α).
The external forces acting on the box would be the weight, the normal reaction and the lifting force that is applied to make it slide up.
These forces can be decomposed on their normal and tangential (to the slide plane) components.
The weight will be split into
Wn = W * cos(α) (in normal direction)
Wt = W * sin(α) (in tangential direction)
The normal reaction will be alligned with the normal axis, and will be equal to -Wn
N = -W* cos(α) (in normal direction)
To mke the box slide up, a force must be applied, that is opposite to the tangential component of the weight and at least a little larger
F > |-W * sin(α)| (in tangential direction)
Answer:
(a) 333.77 J
(b) 237.85 J
(c) 4763.77 J
(d) 4667.85 J
Explanation:
Temperature of source, TH = 314 K
Temperature of A, Tc = 292 K
Temperature of B, Tc' = 298 K
heat taken out, Qc = 4430 J
Let the heat deposited outside is QH and QH' by A and B respectively.
Now
(a) Work done for A
W = QH - QC = 4763.77 - 4430 = 333.77 J
(b) Work done for B
W' = QH' - Qc = 4667.85 - 4430 = 237.85 J
(c) QH = 4763.77 J
(d) QH' = 4667.85 J
B,
The 25 N and 5 N force are acting in the same direction so we can add them together, but the 10 N force acts in the opposite direction so you subtract it.
25 N + 5 N - 10 N = 20 N
