Answer:
The pressure inside the dropper is same as the atmospheric pressure when the rubber bulb is not pressed. ... But when we press the rubber bulb the pressure inside the dropper increases and hence the water flows out. Atmospheric pressure acting from outside the dropper balances the pressure exerted by water and water does not come out of a dropper. On pressing the dropper inside pressure of water becomes more than outside atmospheric pressure and water run out. When we press the bulb of a dropper with its nozzle kept in water, air in the dropper is seen to escape in the form of bubbles. Once we release the pressure on the bulb, water gets filled in the dropper.
<u><em>Thank You</em></u>
<u> </u>Please mark me brainliest
Thermal- transfer of heat thru space
Radiation- the average amount of energy of motion in the molecules of a substance
Thermometer- a thin glass tube with a bulb on one end that contains a liquid, usually mercury or colored alcohol
Brainly?
<span> = r x F </span>
<span>| τ | = | r x F | = | r | | F | sinΘ </span>
<span>100 = | (0.3)<0, 1, 0> x (F/5)<0, 3, -4> | </span>
<span>5000/3 = | F | | <0, 1, 0> x <0, 3, -4> | </span>
<span>5000/3 = | F | | <-4, 0, 0> | </span>
<span>5000/3 = | F | * 4 </span>
<span>| F | = 1250/3 </span>
<span>| F | = 417 N </span>
<span>Check: </span>
<span>τ = r x F </span>
<span>100<-1, 0, 0> = 0.3<0, 1, 0> x 250/3<0, 3, -4> </span>
<span>100<-1, 0, 0> = 25 ( <0, 1, 0> x <0, 3, -4> ) </span>
<span>100<-1, 0, 0> = 25 ( <-4, 0, 0> ) </span>
<span>100<-1, 0, 0> = 100<-1, 0, 0> </span>
<span>What the other people forgot was that you need the angle from the y-axis, which would be Θ = arctan(O/A) = arctan(z/y) = arctan(-4/3) = -53.1°. Then you take the sine of that to get 0.8. There is more work being done on the z-vector than there is on the y-vector, so the force is distributed (0.6) in the y-direction, and (0.8) in the z-direction (0.6² + 0.8² = 1). </span>