The simplest way to do this is to set up equivalent fractions, like this-
=
Solve for x by using cross multiplication.
40*2.2= 88
1*x=88
x=88
Therefore, the boy weighs 88lbs.
Answer:
3.57 m/s
Explanation:
The sum of the 2 momentums Is equal the finale momentums. so if momentums Is q, v Is velocity and m Is Mass, q3=m1*v1+m2**v2=16+9=25 m*kg/s
q3=m3*v3
v3=q3/m3=25/(4+3)=3.57m/s
Answer:
Explanation:
Path difference due to a transparent slab is given as
here we know that
now total shift in the bright fringe is given as
Also we know that the fringe width of maximum intensity is given as
now we have
now the shift is given as
given that the shift is
here we have
now plug in all values in it
Ca(NO3)2(aq) + Na2CO3(aq) → 2NaNO3 + CaCO3⬇. NaNO3 is solution so CaCO3 is the precipitate formed.
Answer:
The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Explanation:
Given:-
- The diameter of the drill bit, d = 98 cm
- The power at which drill works, P = 5.85 hp
- The rotational speed of drill, N = 1900 rpm
Find:-
What Torque And Force Is Applied To The Drill Bit?
Solution:-
- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).
- The relation between these quantities is given:
T = 5252*P / N
T = 5252*5.85 / 1900
T = 16.171 Nm
- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):
T = F*r
Where, r = d / 2
F = 2T / d
F = 2*16.171 / 0.98
F = 33 N
Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
