Answer:
12 N/cm²
Explanation:
From the question given above, the following data were obtained:
Weight (W) of block = 240 N
Area (A) = 20 cm²
Pressure (P) =?
Next, we shall determine the force exerted by the block. This can be obtained as follow:
Weight (W) of block = 240 N
Force (F) =.?
Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,
Force (F) = Weight (W) of block = 240 N
Force (F) = 240 N
Finally, we shall determine the pressure on the floor as follow:
Force (F) = 240 N
Area (A) = 20 cm²
Pressure (P) =?
P = F/A
P = 240 / 20
P = 12 N/cm²
Therefore, the pressure on the floor is 12 N/cm².
Conservation of linear momentum:
m*v inital = m*v final
0.06*0.7 + 0.03*0 = 0.06*(-0.2) + 0.03*v
(my algebra, or use ur calculator: 0.06*.07=0.042, etc ... or ur teacher may think you got some help)
0.06*(0.7+0.2)=0.03*v, v = 0.06*0.9/0.03=1.8 m/s
Answer 1.8 m/s (positive, to the right).
Answer and Explanation:
curents i = 2.9 A
i ' = 4.4 A
the magnitude (in T.m) of the path integral of B.dl around the window frame = μo * current enclosed
= μo* ( i '- i )
Since from Ampere's law
where μ o = permeability of free space = 4π * 10 ^-7 H / m
plug the values we get the magnitude (in T.m) of the path integral of B.dl = ( 4π*10^-7 ) (2.9+4.4)
= 1.884 * 10^-6 Tm
Answer:
The car strikes the tree with a final speed of 4.165 m/s
The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m
Explanation:
First we need to calculate the initial speed 
Once we have the initial speed, we can isolate the final speed from following equation:
Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.
To do that, we replace 62 m in the first formula, as follows:
