First there are many ways to approach this problem. The simplest and most direct in my opinion is this approach:
vf^2=v1^2+2ad
so when the height reaches a maximum of 35m the velocity is zero for a split second before starting to fall. So we set vf=0 and solve for v1.
v1=sqrt(2gd)=sqrt(2*9.8*35)= 26.2m/s
for initial velocity needed to reach 35m height.
Any questions please ask!
Answer:
Input power given to the pump is 4.09 kw
Explanation:
the pictures below shows the full explanation
Answer:
Microphones are used in many applications such as telephones, hearing aids, public address systems for concert halls and public events, motion picture production, live and recorded audio engineering, sound recording, two-way radios, megaphones, radio and television broadcasting.
Power in a wire where current is flowing can be calculated from the product of the square of the current and the resistance. Resistance is equal to the product of resistivity and length divided by the area of the wire. We do as follows:
Resistance = 2.44 × 10-8 ( 0.11) / (π)(0.0009)^2 = 1.055x10^-3 <span>Ω
P = I^2R = .170^2 (</span>1.055x10^-3 ) = 3.048x10^-5 W
Answer:
a) 627.84 Joules
b) 117.72 Joules
c) 1255.68 Joules
Explanation:
<em>(See figure 1)</em>
The gravitational potential energy relative to the child’s lowest position is:
(1)
with h the vertical distance of the swing from the lowest position, m the mass of the child and g the acceleration of gravity.
a) When the ropes are horizontal, the swing is at 1.60 m from the lowest position, so by (1):
b) When the ropes make a 36.0◦ angle with the vertical, the swing is at a distance d from the lowest position, we should use trigonometric relations to find that distance. By figure 2 we have a right triangle with adjacent side and hypotenuse 1.60, so using the trigonometric relation we can solve for d:
Using d on (1):
c) Note that at the bottom of the circular arc the distance of the swing relative to the lowest position is two times the length of the rope, so h=3.20 m, using this on (1):