Answer:
9.8m/s²
Explanation:
The acceleration of the ball thrown after leaving my hand is 9.8m/s². This will be the acceleration due to gravity on the body.
- Acceleration due to gravity is caused by the pull of the earth on a massive object.
- The value of this acceleration is 9.8m/s².
- As the ball nears the surface, it comes near zero.
If you have (1 x 10⁹) cubic meters of volume, and 1/2 gram of mass
in each cubic meter, then you have
(1/2 x 10⁹) grams = (5 x 10⁸) grams = (5 x 10⁵) kilograms of mass .
On Earth, that mass weighs 4,900,000 newtons .
(about 1,102,300 pounds)
(about 551 tons)
Answer:
Well the definition of an application is the act of putting to a special use or purpose so lam assuming that you want specific uses that scientists make of gravity in their work.
Well our first application has helped us to send satellites around the solar system with what Nasa calls gravity assist. Using a particular planets gravity to slingshot a satellite to another destination. Look it up.
The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.
Hope that helps.
Explanation:
Answer:
10
Explanation:
displacement would be 10 because knowledge
Answer:
The required angle is (90-25)° = 65°
Explanation:
The given motion is an example of projectile motion.
Let 'v' be the initial velocity and '∅' be the angle of projection.
Let 't' be the time taken for complete motion.
Let 'g' be the acceleration due to gravity
Taking components of velocity in horizontal(x) and vertical(y) direction.
= v cos(∅)
= v sin(∅)
We know that for a projectile motion,
t =
Since there is no force acting on the golf ball in horizonal direction.
Total distance(d) covered in horizontal direction is -
d =
×t = vcos(∅)×
=
.
If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -
α +β = 90° as-
d =
=
=
=
.
∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.
∴ If α = 25° , then
β = 90-25 = 65°
∴ The required angle is 65°.