Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>
Answer:
The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.
Explanation:
Let us call the mass of the balloon 
and the mass of the basket 
, then according to newton's second law:
,
where 
is the upward acceleration, and 
is the net propelling force (counts the gravitational force).
Also, the tension 
in the rope is 79.8 N more than the basket's weight; therefore,
and this tension must equal
Combining equations (2) and (3) we get:
since , we have
Putting this into equation (1) and substituting the numerical values of 
and 
, we get:
Thus, the mass of the balloon and the basket is 2295 kg and 301 kg respectively.
Answer: W = 294 J
Explanation: Solution:
Work is expressed as the product of force and the distance of the object.
W = Fd where F = mg
W= Fd
= mg d
= 15 kg ( 9.8 m/s²) ( 2m )
= 294 J
Answer:
The answer to your question is:
a) t = 3.81 s
b) vf = 37.4 m/s
Explanation:
Data
height = 71.3 m = 234 feet
t = 0 m/s
vf = ?
vo = 0 m/s
Formula
h = vot + 1/2gt²
vf = vo + gt
Process
a)
h = vot + 1/2gt²
71.3 = 0t + 1/2(9.81)t²
2(71.3) = 9,81t²
t² = 2(71.3)/9.81
t² = 14.53
t = 3.81 s
b)
vf = 0 + (9.81)(3.81)
vf = 37.4 m/s
Answer:
The total amount of heat needed will be 
.
Explanation:
We will divide the calculation in two: First, the heat needed to melt the ice, and then the heat needed to warm the resulting liquid from 0°C to 37°C.
<em>i) </em>The fusion heat will be:
<em>ii)</em> The heat needed to warm the water from 
to 
will be:
So, the total amount needed will be the sum of these two results:
.
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
