Given:
F_gravity = 10 N
F_tension = 25 N
Let's find the net centripetal force exterted on the ball.
Apply the formula:
From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.
Hence, the tensional force is positive while the gravitational force is negative.
Thus, we have:
Therefore, the net centripetal force exterted on the ball is 15 N.
ANSWER:
15 N
Answer:
zero
Explanation:
For a solid conducting sphere, charges are present on the surface of the sphere due to a phenomenon known as electrostatic sheilding. This affects the charge present in the body and makes it zero. However, the electrostatic potential appears to be equal to the whole present point that shows on the surface. The surface of a spherical conducting solid sphere is known as an equipotential surface. Thus, the potential difference between the two opposite points on the surface of the sphere will also be zero.
Work = (force) x (distance)
40,000 J = (20 N) x (distance)
Distance = (40,000 J) / (20 N)
= 2,000 meters
= 2 kilometers.
(20 N is not a huge force when it's being used to move a car.
It's only about 4.5 pounds.)
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
