Answer You need to consider that the gravity on earth is 9.8 m/s/s. This means any object you let go on the earths surface will gain 9.8 m/s of speed every second. You need to apply a force on the object in the opposite direction to avoid this acceleration. If you are pushing something up at a constant speed, you are just resisting earths acceleration. The more massive and object is, the greater force is needed to accelerate it. The equation is Force = mass*acceleration. So for a 2kg object in a 9.8 m/s/s gravity you need 2kg*9.8m/s/s = 19.6 Newtons to counteract gravity. Work or energy = force * distance. So to push with 19.6 N over a distance of 2 meters = 19.6 N*2 m = 39.2 Joules of energy. There is an equation that puts together those two equations I just used and it is E = mgh
The amount of Energy to lift an object is (mass) * (acceleration due to gravity) * (height)
:Hence, the Work done to life the mass of 2 kg to a height of 10 m is 196 J. Hope it helps❤️❤️❤️
Explanation:
Option B The thickness of the central portion of a thin conveying lens can be determined very accurately by using a micrometer screw gauge.
<h3>What can be measured using a micrometer screw gauge?</h3>
One micrometer of thickness can be measured with a micron micrometre screw gauge. A Use of Micrometer Screw Gauge as like example Upon turning the screw of the micrometer screw gauge four times, a 2 mm space is covered.
<h3>What purposes does a micrometer serve?</h3>
A tool known as a micrometer is used to measure solid objects’ lengths, thicknesses, and other dimensions precisely and linearly.
<h3>What is the micrometer screw gauge’s SI unit?</h3>
The SI symbol m is also known as a micron, which is an SI-derived unit of length equaling 1106 meters, where 106 is the SI standard prefix for the prefix “micro-.” A micrometer is one-millionth of a meter.
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The answer is A. digestion of food. The chemical change to matter means that the molecule structure has been destroyed. For B, C and D. the matter are still ice, paper and ice. So they are physical change.
Answer:
200 m\ s Ans .....
Explanation:
Data:
f = 200 Hz
w = 1.0 m
v = ?
Formula:
v = f w
Solution:
v = ( 200)(1.0)
v = 200 m\s <em>A</em><em>n</em><em>s</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
Use VFR1 = VFR2 to discover the velocity at in the hose VFR =
A * V
D hose =10 * D nozzle, R hose = 5 * D nozzle
Area of a circle = πR^2
Area h=3.14*25*D^2 = 75.5D^2
(Radius=Diameter/2) area n = 3.14*(D^2/4) = .785D^2
Use VFR = VFR v2 = 0.4m/s
0.4*.785D^2 = 75.5*D^2* v1 D^2
= .314 =75.5*V1
v1 = 0.004m/s
Now we have the velocity, we can use Bernoulli's equation.
P1+ρgh1+ρV1^2 /2 = constant
There is no atmospheric pressure before so the P1= the gauge
pressure at the pump, let’s call the height of the hose 0m and the height of
the nozzle 1m so the is no ρgh1 Likewise, there is only atmospheric pressure at
the nozzle which is 100000 PA, and lastly the density ρ of water is 1000 KG/M^3
Pg + 1000*.004^2/2 = 100000+1000*9.8*1+ 1000*0.4^2/2
Pg + .008= 100000+9800+80
Pg+.008= 109880
Pg=109880.008 PA
