By Newton's second law, the net force on the object is
∑ <em>F</em> = <em>m</em> <em>a</em>
∑ <em>F</em> = (2.00 kg) (8 <em>i</em> + 6 <em>j</em> ) m/s^2 = (16.0 <em>i</em> + 12.0 <em>j</em> ) N
Let <em>f</em> be the unknown force. Then
∑ <em>F</em> = (30.0 <em>i</em> + 16 <em>j</em> ) N + (-12.0 <em>i</em> + 8.0 <em>j</em> ) N + <em>f</em>
=> <em>f</em> = (-2.0 <em>i</em> - 12.0 <em>j</em> ) N
Answer:
0.6944 m/sec^2
Explanation:
The computation of the average acceleration is given below:
a = v - u ÷ t
where
a denotes average acceleration
v denotes final velocity
u denotes initial velocity
t denotes time
So, the average acceleration is
= (25 - 0) ÷ 10
= 0.6944 m/sec^2
Answer:
Explanation:
Just start with the trivial. If gravity was a constant then E = mgh and 1/2 m v^2 = E
so v=sqrt(2gh)=sqrt(2*9.8*1590000) = 5582m/s
Now this will be too high as gravity reduces with distance.
However it is still true that 1/2mv^2 = loss of gravitational potential energy
so 1/2 v^2 = loss of gravitational potential ( i.e a field without considering mass )
As g = GM/ Ro^2 and P = - GM/R
the Po = - 9.8 * (6370*10^3)= - 62.4 * 10 ^ 6 J/kg
P1= Po * 6370/(6370+1590) = - 49.93 * 10 ^ 6 J/kg
find the CHANGE and then from that the velocity
ie v = sqrt(2*( P1 - Po)) = 5094 m/s
Note how it is a bit smaller than the first estimate but not by such a margin that they are unrecognizably different.
Answer:
B) 0.0456
Explanation:
It is given that :
Rationale :
UCL = 480
LCL = 480
∴ Mean ,
The standard deviation of sample (Sn) = 11.55
Z (for UCL)
= 2
Similarly,
Z (for LCL)
= -2
Now using the z table for finding the confidence level between Z value of -2 and 2.
Confidence level = 0.4772 + 0.4772
= 0.9544
Risk (alpha) = 1 - confidence level
= 1 - 0.9544
= 0.0456