Answer: h = u^2 / 2g
Explanation:
Given the following :
Horizontal Velocity of projection= u
If :
magnitude of horizontal = magnitude of vertical Displacement
u = u
Minimum height of tower (h) =?
Horizontal Velocity = u
Gravity potential energy = mgh - - - (1)
Kinetic energy = 1/2 mu^2 - - - (2)
m = mass
Where u = magnitude of velocity
g = acceleration due to gravity
h = height
Equating (1) and (2)
mgh = 1/2 mu^2
gh = 1/2 mu^2
2gh = u^2
u = √2gh
Vertical component of Velocity 'v' will be:
u = √2gh
u = √2 × g × h
Square both sides
u^2 = 2 × g × h
h = u^2 / 2g
Answer:
59.19%
Explanation:
Step 1:
Data obtained from the question. This include the following:
Input temperature = 495 K.
Output temperature = 293 K.
Efficiency =..?
Step 2:
Determination of the efficiency of the engine.
Efficiency is simply defined as the ratio of output to input times 100. Mathematically, it is expressed as:
Efficiency = output /input x 100
With the above formula, we can calculate the efficiency of the engine as follow:
Input temperature = 495 K.
Output temperature = 293 K.
Efficiency =..?
Efficiency = output /input x 100
Efficiency = 293/495 x 100
Efficiency = 59.19%
Therefore, the efficiency of the engine is 59.19%
Answer:
a) Qh= 6750 kJ
b) e = 0.296
c) = 390.625° C
Explanation:
Given:
Work done, W = 2000 kJ
Heat flow, Q = 4750 kJ
Temperature at which heat flows out, = 275° C
a) Now, the heat flow through the engine (Qh)
Qh = W + Q
or
Qh = 2000 + 4750
or
Qh= 6750 kJ
b) The efficiency (e) is given as:
on substituting the values, we get
or
e = 0.296
c)
where,
is the temperature at which heat flow
on substituting the values, we get
or
= 390.625° C
Answer:
a) v = 54.7m/s
b) v = (58 - 1.66a) m/s
c) t = 69.9 s
d) v = -58.0 m/s
Explanation:
Given;
The height equation of the arrow;
H = 58t - 0.83t^2
(a) Find the velocity of the arrow after two seconds. m/s;
The velocity of the arrow v can be given as dH/dt, the change in height per unit time.
v = dH/dt = 58 - 2(0.83t) ......1
At t = 2 seconds
v = dH/dt = 58 - 2(0.83×2)
v = 54.7m/s
(b) Find the velocity of the arrow when t = a. m/s
Substituting t = a into equation 1
v = 58 - 2(0.83×a)
v = (58 - 1.66a) m/s
(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s
the time when H = 0
Substituting H = 0, we have;
H = 58t - 0.83t^2 = 0
0.83t^2 = 58t
0.83t = 58
t = 58/0.83
t = 69.9 s
(d) With what velocity will the arrow hit the surface? m/s
from equation 1;
v = dH/dt = 58 - 2(0.83t)
Substituting t = 69.9s
v = 58 - 2(0.83×69.9)
v = -58.0 m/s
Answer:
D. 5m
Explanation:
fλ = c, where f is frequency, λ is wavelength and c is speed.
6λ=30
λ=30/6=5