From the bladder and goes through the male genitalia.
Answer:
t = 1.4[s]
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

where:
P = impulse or lineal momentum [kg*m/s]
m = mass = 50 [kg]
v = velocity [m/s]
F = force = 200[N]
t = time = [s]
Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

where:
m₁ = mass of the object = 50 [kg]
v₁ = velocity of the object before the impulse = 18.2 [m/s]
v₂ = velocity of the object after the impulse = 12.6 [m/s]
![(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]](https://tex.z-dn.net/?f=%2850%2A18.2%29-200%2At%3D50%2A12.6%5C%5C910-200%2At%3D630%5C%5C200%2At%3D910-630%5C%5C200%2At%3D280%5C%5Ct%3D1.4%5Bs%5D)
The star with apparent magnitude 2 is more brighter than 7.
To find the answer, we have to know about apparent magnitude.
<h3>What is apparent magnitude?</h3>
- 100 times as luminous as a star with an apparent brightness of 7 is a star with a magnitude of 2.
- The apparent magnitude of bigger stars is always smaller.
- The brightest star in the night sky is Sirius.
- The brightness of a star or other celestial object perceived from Earth is measured in apparent magnitude (m).
- The apparent magnitude of an object is determined by its inherent luminosity, its distance from Earth, and any light extinction brought on by interstellar dust in the path of the observer's line of sight.
Thus, we can conclude that, the star with apparent magnitude 2 is more brighter than 7.
Learn more about the apparent magnitude here:
brainly.com/question/350008
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Answer:
Explanation:
If the work done on the cart is NET work
Then the work will result in an increase in kinetic energy
KE₀ + W = KE₁
½mv₀² + W = ½mv₁²
½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²
v₁ = 1.626991...
v₁ = 1.6 m/s