consider the motion in Y-direction
v₀ = initial velocity = 29 Sin62 = 25.6 m/s
a = acceleration = - 9.8 m/s²
t = time of travel
Y = vertical displacement = - 0.89 m
using the equation
Y = v₀ t + (0.5) a t²
- 0.89 = (25.6) t + (0.5) (- 9.8) t²
t = 5.3 sec
consider the motion along the horizontal direction :
v₀ = initial velocity = 29 Cos62 = 13.6 m/s
a = acceleration = 0 m/s²
t = time of travel = 5.3 sec
X = horizontal displacement =?
using the equation
X = v₀ t + (0.5) a t²
X = (13.6) (5.3) + (0.5) (0) t²
X = 72.1 m
d = distance traveled by the center fielder to catch the ball = 107 - x = 107 - 72.1 = 34.9 m
t = time taken = 5.3 sec
v = speed of center fielder
using the equation
v = d/t
v = 34.9/5.3
v = 6.6 m/s
25,000 Feet = 7620m
PE = mgh where m is mass, g is gravity accel: 9.8 n h is height
= 90 x 9.8 x 7620
= 6720840J
= 6.72MJ
F = ma where m is mass, a is accel = gravity = 9.8
= 90 x 9.8
= 882N
Accel = gravity = 9.8m/s^2
KE = 1/2mv^2 where m is mass n v is vel
if no wind resistance, PE leaving airplane = KE at net
6720840 = 1/2 x 90 x v^2
v^2 = 149352
v = 386.5m/s
Answer with Explanation:
We are given that mass of block=0.0600 kg
Initial speed of block=0.63 m/s
Distance of block from the hole when the block is revolved=0.47 m
Final speed=3.29 m/s
Distance of block from the hole when the block is revolved=
a.We have to find the tension in the cord in the original situation when the block has speed =

Because tension is equal to centripetal force
Substitute the values

b.

c.Work don=Final K.E-Initial K.E



To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,
Q = V*A
Where,
A= Cross-sectional Area
V = Velocity
The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,


Our values are given as,


Re-arrange the equation to find the first ratio of rates we have:



The second ratio of rates is


