Answer:
Explanation:
Initial velocity u = V₀ in upward direction so it will be negative
u = - V₀
Displacement s = H . It is downwards so it will be positive
Acceleration = g ( positive as it is also downwards )
Using the formula
v² = u² + 2 g s
v² = (- V₀ )² + 2 g H
= V₀² + 2 g H .
v = √ ( V₀² + 2 g H )
Answer:
Distance 20 km and Displacement 0 km
His displaceent is 0 km because he ends his walk where he started. The total distance of his walk is 20 km because he walks 10 km to the store + 10km back home.
Answer:
<em>The object could fall from six times the original height and still be safe</em>
Explanation:
<u>Free Falling</u>
When an object is released from rest in free air (no friction), the motion is completely dependant on the acceleration of gravity g.
If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy of
When the object strikes the ground, all the mechanical energy (only potential energy) becomes into kinetic energy
Where v is the speed just before hitting the ground
If we know the speed v is safe for the integrity of the object, then we can know the height it was dropped from
Solving for h
If the drop had occurred in the Moon, then
Where hM, vM and gM are the corresponding parameters on the Moon. We know v is the safe hitting speed and the gravitational acceleration on the Moon is g_M=1/6 g
This means the object could fall from six times the original height and still be safe
Answer:
Explanation:
Let electric potential at A ,B and C be Va , Vb and Vc respectively.
Work done = charge x potential difference
Wab = q ( Va - Vb )
Wac = q ( Va - Vc )
Given
Wac = - Wab / 3
3Wac = - Wab
Now
Wbc = q ( Vb - Vc )
= q [ ( Va-Vc ) - ( Va - Vb )]
= Wac - Wab
= Wac + 3Wac
= 4Wac
Hot combustion gases are accelerated in a 92% efficient
adiabatic nozzle from low velocity to a specified velocity. The exit velocity
and the exit temp are to be determined.
Given:
T1 = 1020 K à
h1 = 1068.89 kJ/kg, Pr1 = 123.4
P1 = 260 kPa
T1 = 747 degrees Celsius
V1 = 80 m/s ->nN = 92% -> P2
= 85 kPa
Solution:
From the isentropic relation,
Pr2<span> = (P2 / P1)PR1 = (85
kPa / 260 kPa) (123.4) = 40.34 = h2s = 783.92 kJ/kg</span>
There is only one inlet and one exit, and thus, m1 =
m2 = m3. We take the nozzle as the system, which is a
control volume since mass crosses the boundary.
h2a = 1068.89 kJ/kg – (((728.2 m/s)2 –
(80 m/s)2) / 2) (1 kJ/kg / 1000 m2/s2) =
806.95 kJ/kg\
From the air table, we read T2a = 786.3 K
