Answer:
A) 350 N
B) 58.33 N
C) 35 kg
D) 35 kg
Explanation:
If we use that g = 10 m/s^2, then the acceleration of gravity on the Moon will be 10/6 m/s^2 = 5/3 m/s*2
The weight of the object on Earth is given by:
Weight = mass * g = 35 * 10 = 350 N
The weight of the object on the Moon:
Weight = mass * gmoon = 35 * 5/3 = 58.33 N
The mass of the object on Earth is 35 kg
The mass of the object on the Moon is exactly the same as on the Earth (35 kg) since the mass is a quantity inherent to the object and not to its location.
The redshift of distant galaxy are larger than those of closer galaxies, which indicates that the galaxy is receding at a faster rate.
- The Universe was 5 percent its current size when light left objects now at redshift of <u>19</u>.
Reasons:
The size of the universe represented as a scale factor with relation to the redshift can be presented as follows;

Where;
a₀ = The current size of the Universe
a = The size of the early Universe = 5% of a
Therefore;


0.05 + 0.05·z = 1

- The redshift is of the observed light is, z = <u>19</u>
Learn more here:
brainly.com/question/14459434
brainly.com/question/3654558
Answer:
91.84 m/s²
Explanation:
velocity, v = 600 m/s
acceleration, a = 4 g = 4 x 9.8 = 39.2 m/s^2
Let the radius of the loop is r.
he experiences a centripetal force.
centripetal acceleration,
a = v² / r
39.2 x r = 600 x 600
r = 3600 / 39.2
r = 91.84 m/s²
Thus, the radius of the loop is 91.84 m/s².
<span>Birds which famous for performing a ritualized mating dance are: The Grebes
Grebes most commonly live near a fresh water source.
They conducted this ritual because When two Grebes with opposite sex dances near each other, they will release some sort of pheromones that make them attracted to one another. </span>
Answer:
t = 0.029s
Explanation:
In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:
(1)
m: mass of the water balloon = 1.20kg
Δv: change in the speed of the balloon = v2 - v1
v2: final speed = 0m/s (the balloon stops in my hands)
v1: initial speed = 13.0m/s
Δt: interaction time = ?
The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

The interaction time to avoid that the water balloon breaks is 0.029s