Answer: T= 715 N
Explanation:
The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:
T = mv² / r
At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.
So the kinetic energy will be the following:
K = 1/2 m v² = 15. 0 J
Solving for v², and replacing in the expression for T:
T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N
Answer:
i hope it will be useful for you
Explanation:
F=5.6×10^-10N
R=93cm=0.93m
let take m1 and m2 =m²
according to newton's law of universal gravitation
F=m1m2/r²
F=m²/r²
now we have to find masses
F×r²=m²
5.6×10^10N×0.93m=m²
5.208×10^-9=m²
taking square root on b.s
√5.208×10^-9=√m²
so the two masses are m1=7.2×10^-5
and m2=7.2×10^-5
Answer:
The maximum volume is 1417.87 
Explanation:
<u>Optimization Using Derivatives</u>
We have a 24x30 inch piece of metal and we need to make a rectangular box by cutting a square from each corner of the piece and bending up the sides. The width of the piece is 24 inches and its length is 30 inches
When we cut a square of each corner of side x, the base of the box (after bending up the sides) will be (24-2x) and (30-2x), width and length respectively. The volume of the box is
Operating
To find the maximum value of V, we compute the first derivative and equate it to zero
Simplifying by 12
Completing squares
We have two values for x
The first value is not feasible because it will produce a negative width (24-2(13.58))=-6.16
We'll keep only the solution
The width is
The length is
And the height
The maximum volume is
Figure B. The second option.
Answer:
Ecu/Eag = 0.46
Explanation:
E = PI/A
Ecu = Pcu × I/A
Pcu = 1.72×10^-8 ohm-meter
r = 0.8 mm = 0.8/1000 = 8×10^-4 m
A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π
Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1
Eag = Pag × I/A
Pag = 1.47×10^-8 ohm-meter
r = 0.5 mm = 0.5/1000 = 5×10^-4 m
A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π
Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π
Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46
