Answer:
gₓ = 23.1 m/s²
Explanation:
The weight of an object is on the surface of earth is given by the following formula:
where,
W = Weight of the object on surface of earth
m = mass of object
g = acceleration due to gravity on the surface of earth = strength of gravity on the surface of earth
Similarly, the weight of the object on Jupiter will be given as:
where,
Wₓ = Weight of the object on surface of Jupiter = 34.665 N
m = mass of object = 1.5 kg
gₓ = acceleration due to gravity on the surface of Jupiter = strength of gravity on the surface of Jupiter = ?
Therefore,
<u>gₓ = 23.1 m/s²</u>
This question involves the concepts of orbital velocity and orbital radius.
The orbital velocity of ISS must be "7660.25 m/s".
The orbital velocity of the ISS can be given by the following formula:
where,
v = orbital velocity = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m
R = 67.86 x 10⁵ m
Therefore,
<u>v = 7660.25 m/s</u>
Learn more about orbital velocity here:
brainly.com/question/541239
Answer:
15 meters
Explanation:
The inicial energy of the ball is just potencial energy, and its value is:
E = m * g * h = m * g * 20,
where m is the ball mass, and g is the value of gravity.
In the moment that the ball strickes the ground, all potencial energy transformed into kinetic energy, and 25% of this energy is lost, so the total energy at this moment will be:
E' = 0.75 * E = 0.75 * m * g * 20 = 15*m*g
This kinetic energy will make the ball goes up again, and at the maximum height, all kinetic energy is transformed back into potencial energy.
So, as the mass and the gravity are constants, we can calculate the height the ball will reach:
E' = m*g*h = 15*m*g -> h = 15 meters
<h2>
The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = 3 s
Substituting
s = ut + 0.5 at²
s = 0 x 3 + 0.5 x 9.81 x 3²
s = 44.145 m
The seagull's approximate height above the ground at the time the clam was dropped is 4 m
Answer:
He did not multiply the chlorine and oxygen atoms by the coefficient 4
Explanation:
