this can be solve using the formala of free fall
t = sqrt( 2y/ g)
where t is the time of fall
y is the height
g is the acceleration due to gravity
48.4 s = sqrt (2 (1.10e+02 m)/ g)
G = 0.0930 m/s2
The velocity at impact
V = sqrt(2gy)
= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)
V = 4.523 m/s
<span> </span>
Given:
u = 0, initial speed (sprinter starts from rest)
v = 11.5 m/s, final speed
s = 15 m, distance traveled to attain final speed.
Let
a = average acceleration,
t = time taken to attain final speed.
Then
v² = u² + 2as
or
(11.5 m/s)² = 2*(a m/s²)*(15 m)
a = 11.5²/(2*15) = 4.408 m/s²
Also
v = u +a t
or
(11.5 m/s) = (4.408 m/s²)*(t s)
t = 11.5/4.408 = 2.609 s
Answer:
The average acceleration is 4.41 m/s² (nearest hundredth).
The time required is 2.61 s (nearest hundredth).
Again I think you did not give the right constants. So I would use the correct constants for mass of moon and distance from earth to moon.
<span>The formula for force of attraction between any two bodies in the universe
F = GMm / r^2. (Newton's Universal law of Gravitation).
G = Universal gravitational constant, G = 6.67 * 10 ^ -11 Nm^2 / kg^2.
M = Mass of Earth. = 5.97 x 10^24 kg.
m = mass of moon = 7.34 x 10^22 kg.
r = distance apart, between centers = in this case it is the distance from Earth to the Moon
= 3.8 x 10^8 m.
(Sorry I could not assume with the values you gave, they are wrong, and if we use them we would be insulting Physics).
So F = ((6.67 * 10 ^ -11)*(5.97 x 10^24)*(7.34 * 10^22)) / (3.8 x 10^8)^2.
Punch it all up in your calculator.
I used a Casio 991 calculator, it should be one of the best in the world.Really lovely calculator, that has helped me a lot in computations like this. I am thankful for the Calculator.
F = 2.0240 * 10^ 20 N.
So that's our answer.
Hurray!!</span>
If the +12V is on one side of the 2.5 ohm R then.............
V = (2.5/8) x 12 otherwise......
V = (5.5/8) x 12
Answer:
All statement are correct.
Explanation:
1. Electric field lines are the same thing as electric field vectors, electric field are mathematically vectors quantity. These vectors point in the direction in which a positive test charge would move.
2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space. Yes it is correct tangent drawn at any point on these lines gives the direction of electric filed at that point.
3. The number of electric field lines that start or end at a charged particle is proportional to the magnitude of charge on the particle, is a correct statement.
4.The electric field is strongest where the electric field lines are close together, again a correct statement as relative closeness of field lines indicate a stronger strength of electric field.
Hence we can say that all the statement are correct.
