The correct answer you're looking for would be Theories.
Let the mass of the person be m. Total momentum is conserved (because the exterior forces on the system are balanced), especially the component in the vertical direction.
Given that,
Mass of gallon is M
Let man mass be m
Velocity of man is v
Let velocity if ballot be Vb
When the person begin to move we have
Conservation of momentum
mv + MVb=0
MVb=-mv
Vb= -(m/M) v
Given that the mass of man is less than mass of balloon. i.e. m<M
So, if m<M, then, m/M <1
Therefore, .
Vb= -(m/M) v
Vb< -v
This implies that the velocity of balloon is less than the velocity of man and if is also moving in opposite direction
So the man is moving upward, then the balloon is moving downward and it's velocity is less than the velocity of man,
The answer is C
Down with a speed less than v
Science is an art, u cant make art without creativity
The coefficient of friction between the road and the car's tire is determined as 0.78.
<h3>Acceleration of the car</h3>
The acceleration of the car is calculated as follows;
v² = u² - 2as
0 = u² - 2as
a = u²/2s
where;
- u is the initial velocity = 97 km/h = 26.94 m/s
a = (26.94)²/(2 x 47)
a = 7.72 m/s²
<h3>Coefficient of friction</h3>
μ = a/g
μ = (7.72)/9.8
μ = 0.78
Learn more about coefficient of friction here: brainly.com/question/14121363
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Answer:
v₃ = 3.33 [m/s]
Explanation:
This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.
In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.
where:
m₁ = mass of the car = 1000 [kg]
v₁ = velocity of the car = 10 [m/s]
m₂ = mass of the truck = 2000 [kg]
v₂ = velocity of the truck = 0 (stationary)
v₃ = velocity of the two vehicles after the collision [m/s].
Now replacing: