<em><u>This</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>you</u></em><em><u> </u></em><em><u>can</u></em><em><u> </u></em><em><u>do</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>boy</u></em><em><u> </u></em><em><u>or</u></em><em><u> </u></em><em><u>girl</u></em><em><u> </u></em>
Answer with Explanation:
We are given that
A.Mass,m=12 kg


Speed,v=1.5m/s
Net force in x direction must be zero



Net force in y direction








Power,P=Fv

Where 
B.Substitute the values


Answer:
im very con fused on what you mean by this
Explanation:
Answer:
Resistance in the flash tube, 
Explanation:
It is given that,
Speed of the bullet, v = 500 m/s
Distance between one RC constant, d = 1 mm = 0.001 m
Capacitance, 
The time constant of RC circuit is given by :

R is the resistance in the flash tube
..........(1)
Speed of the bullet is given by total distance divided by total time taken as :




Equation (1) becomes :


So, the resistance in the flash tube is
. Hence, this is the required solution.
Answer:
The pacific floor (oceanic crust) is thinner and more denser
Explanation:
The continental crust have a thickness of about 35 to 40 km on an average, and are composed of rocks that has less denser granitic minerals such as feldspar, quartz.
On the other hand, the oceanic crust have a thickness of about 7 to 10 km on an average, and it is comprised of denser mafic rocks that contains high amount of olivine and pyroxene minerals.
Due to this, the oceanic crust subducts below the continental crust during the time of collision.
Thus, the pacific floor (oceanic crust) is thinner and more denser, in comparison to the South American continental crust.