Answer:
forever
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
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)
Question 6 (10 points)
The Washington Monument in our nation's capital is 555 ft high. Neglecting the effects of air resistance, what would be the
speed of the penny as it "hit" the ground?
оа
Ob
Oc
Od
48.3 m/s
-68.4 m/s
84.7 m/s
-57.6 m/s
Answer:
a= g = - 9.81 m/s2.
The following equations will be helpful:
a = (vf - vo)/t d = vot + 1/2 at2 vf2 = vo2 + 2ad
When you substitute the specific acceleration due to gravity (g), the equations are as follows:
g = (vf - vo)/t d = vot + 1/2 gt2 vf2 = vo2 + 2gd
If the object is dropped from rest, the initial velocity ("vi") is zero. This further simplifies the equations to these:
g = vf /t d = 1/2 gt2 vf2 = 2gd
The sign convention that we will use for direction is this: "down" is the negative direction. If you are given a velocity such as -5.0 m/s, we will assume that the direction of the velocity vector is down. Also if you are told that an object falls with a velocity of 5.0 m/s, you would substitute -5.0 m/s in your equations. The sign convention would also apply to the acceleration due to gravity as shown above. The direction of the acceleration vector is down (-9.81 m/s2) because the gravitational force causing the acceleration is directed downward.
hope this info helps you out!
The force of air particles over an area is ________
temperature.
pressure.
volume.
kelvins.
Help me pretty please with a cherry on top!!
Answer:
A. im not for sure hope it helps
Explanation:
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
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
A baseball on the Moon and an identical baseball on the Earth are thrown vertically upward with the same initial velocity of 12 m/s. The acceleration due to gravity on the Moon is 1/6 that of Earth. What is the maximum height the ball reaches on the moon? What is the maximum height the ball reaches on the earth? What is the difference in the maximum height the ball reaches on the Moon compared to on Earth?
Answer:
a. 3.68 m b. 0.61 m c. 3.07 m
Explanation:
a. What is the maximum height the ball reaches on the moon?
Using v² = u² - 2g'h where u = initial velocity of baseball = 12 m/s, v = final velocity of baseball = 0 m/s (since it stops at maximum height), g' = acceleration due to gravity on moon = g/6 where g = acceleration due to gravity on earth = 9.8 m/s², so g' = 9.8 m/s²/6 = 1.63 m/s² and h = maximum height of ball on moon.
So, making h subject of the formula, we have
h = -(v² - u²)/2g'
substituting the values of the variables, we have
h = -((0 m/s)² - (12 m/s)²)/2(1.63 m/s²)
= -(- 12 m²/s²)/2(1.63 m/s²)
= 6 m²/s²)/(1.63 m/s²)
= 3.68 m
b. What is the maximum height the ball reaches on the earth?
Using the same equation for the maximum height the baseball travels on the moon for that on the earth, the maximum height h' the baseball travels on the earth is given by
h' = -(v² - u²)/2g where u = initial velocity of baseball = 12 m/s, v = final velocity of baseball = 0 m/s (since it stops at maximum height), g = acceleration due to gravity on earth = 9.8 m/s²and h' = maximum height of ball on moon.
substituting the values of the variables, we have
h = -((0 m/s)² - (12 m/s)²)/2(9.8 m/s²)
= -(- 12 m²/s²)/2(9.8 m/s²)
= 6 m²/s²)/(9.8 m/s²)
= 0.61 m
c. What is the difference in the maximum height the ball reaches on the Moon compared to on Earth?
The difference in the maximum height the ball reaches on the Moon compared to on Earth is d = h - h' = 3.68 m - 0.61 m = 3.07 m
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]
