Answer:
The value is [tex]T_t = 2.5659 \ s[/tex]
Explanation:
From the we are told that
The initial speed of the object is [tex]u = 8 \ m/s[/tex]
The greatest height it reached is [tex]h = 15 \ m[/tex]
Generally from kinematic equation we have that
[tex]v^2 = u^2 + 2gH[/tex]
At maximum height v = 0 m/s
So
[tex]0^2 = 8^2 + 2 * - 9.8 * H[/tex]
=> [tex]H = 3.27 \ m[/tex]
Here H is the height from the initial height to the maximum height
So the initial height is mathematically represented as
[tex]s = h - H[/tex]
=> [tex]s = 15 - 3.27[/tex]
=> [tex]s = 11.73 \ m[/tex]
Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows
[tex]v = u + (-g) t[/tex]
At maximum height v = 0 m/s
[tex]0 = 8 - 9.8t[/tex]
=> [tex]t = 0.8163 \ s[/tex]
Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows
[tex]h = ut_1 + \frac{1}{2} gt_1^2[/tex]
Here the initial velocity is 0 m/s given that its the velocity at maximum height
Also g is positive because we are moving in the direction of gravity
So
[tex]15 = 0* t + 4.9 t^2[/tex]
=> [tex]t_1 = 1.7496[/tex]
Generally the total time taken is mathematically represented as
[tex]T_t = t + t_1[/tex]
=> [tex]T_t = 0.8163 + 1.7496[/tex]
=> [tex]T_t = 2.5659 \ s[/tex]
Electric power is to be generated by installing a hydraulic turbine-generator at a site 70 m below the free surface of a large water reservoir that can supply water at a rate of 1500 kg/s steadily. If the mechanical power output of the turbine is 800 kW and the electric power generation is 750 kW, determine the turbine efficiency and the combined turbine–generator efficiency of this plant. Neglect losses in the pipes.
Answer:
[tex]\eta_{turbine} = 0.777 = 77.7\%[/tex]
[tex]\eta_{combined} = 0.728 = 72.8\%[/tex]
Explanation:
First we calculate the power input to the turbine. The input power will be equal to the potential energy of water per unit time:
Input Power = [tex]P_{in} = \frac{Work}{Time} = \frac{Potential\ Energy\ of\ Water}{t} \\P_{in} = \frac{(mass)(g)(height)}{Time} = (mass flow rate)(g)(height)\\\\P_{in} = (1500\ kg/s)(9.81\ m/s^2)(70\ m)\\P_{in} = 1.03\ x\ 10^6\ W = 1030\ KW[/tex]
Now, for turbine efficiency:
[tex]\eta_{turbine} = \frac{Mechanical\ Power\ Out}{P_{in}}\\\\\eta_{turbine} = \frac{800\ KW}{1030\ KW}\\\\\eta_{turbine} = 0.777 = 77.7\%[/tex]
for generator efficiency:
[tex]\eta_{generator} = \frac{Power\ Generation}{Mechanical\ Power\ Out}\\\\ \eta_{generator} = \frac{750\ KW}{800\ KW}\\\\\eta_{turbine} = 0.9375 = 93.75\%[/tex]
Now, for combined efficiency:
[tex]\eta_{combined} = \eta_{turbine}\ \eta_{generator}\\\\\eta_{combined} = (0.777)(0.937)\\\eta_{combined} = 0.728 = 72.8\%[/tex]
what power is transmitted by 2A flowing across 5V
Answer:
[tex]\huge\boxed{10\:watts}[/tex]
Explanation:
Power = Current × Voltage
P = I × V
P = 2 × 5
P = 10 watts
Help please!!!
if superman at 90kg jumps a 40m building in a single bound how much work does superman perform
Answer:
5 years worth of work (aka all of the homework i currently have)
Help me pretty please with a cherry on top!!
Answer:
A. im not for sure hope it helps
Explanation:
How do I calculate how many meters are in 7.2 light years?
The exact question is:
Calculate in meters the distance between a galaxy and the Earth if the distance is equal to 7.2 light years.
Answer:
68.2 Quadrillion meters
Explanation:
A lightyear is the distance that light travels in one year.
Speed of light is [tex]3*10^8\ m/s[/tex]
So light covers 300,000,000 meters in one second.
One year has 31536000 seconds so , light covers
[tex]9.461*10^{15}\ meters\ in\ one\ year[/tex]
so 7.2 light years is
[tex]7.2*(9.461*10^{15})\\6.82*10^{16}[/tex]
so 7.2 light years is
6.82 x 10^(16) meters or
68.2 Quadrillion meters
An object weighing 49 N is pushed across a floor by a force of 12 N. What is the acceleration of the object?
Answer:
Explanation:
Given parameters:
Weight of object = 49N
Force applied = 12N
Unknown:
Acceleration of object = ?
Solution:
The acceleration of the object is found by dividing the force by the weight;
Acceleration = [tex]\frac{12}{49}[/tex] = 0.25m/s²
