In a closed system, the loss of momentum of one object is same as________ the gain in momentum of another object
according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is
total momentum before collision = total momentum after collision
P₁ + P₂ = P'₁ + P'₂
where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.
P'₁ - P₁ = - (P'₂ - P₂)
so clearly gain in momentum of one object is same as the loss of momentum of other object
Answer:
See attached pictures.
Explanation:
See attachments for explanation.
Answer:
R = 0.0503 m
Explanation:
This is a projectile launching exercise, to find the range we can use the equation
R = v₀² sin 2θ / g
How we know the maximum height
² =
² - 2 g y
= 0
= √ 2 g y
= √ 2 9.8 / 15
= 1.14 m / s
Let's use trigonometry to find the speed
sin θ = / vo
vo = / sin θ
vo = 1.14 / sin 60
vo = 1.32 m / s
We calculate the range with the first equation
R = 1.32² sin(2 60) / 30
R = 0.0503 m
When the relationship between two variables are said to be proportional, it means that one variable is a constant multiple of the other variable. They are related by a constant of proportionality, usually denoted as k.
In this problem, the dependent variable is the distance in kilometers. Your mileage is limited with the amount of fuel you have. Thus, the independent variable is the liters of fuel. When these two are proportional, it could be expressed as
distance = k * liters of fuel, such that
distance/liters of fuel = k
By variation,
distance,1/liters of fuel,1 = distance,2/liters of fuel,2, where 1 denotes situation 1 and 2 denotes situation 2. Therefore,
999999 km /<span>999 liters = x km /</span><span>121212 liters, where x is the unknown distance. We can now therefore find the value of x.
x = (999999*121212)/999
x = 121333212 kilometers</span>
Answer:
15,505 N
Explanation:
Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student
-ΔU = ΔK
-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²
-(0 - mgh) = 1/2mv² - 0
mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.
mgh = 1/2mv²
gh = 1/2v²
v² = 2gh
v = √(2gh)
v = √(2 × 9.8 m/s² × 1 m)
v = √(19.6 m²/s²)
v = 4.43 m/s
Upon impact on the ground and stopping, impulse I = Ft = m(v' - v) where F = force, t = time = 0.02 s, m =mass of student = 70 kg, v = initial velocity on impact = 4.43 m/s and v'= final velocity at stopping = 0 m/s
So Ft = m(v' - v)
F = m(v' - v)/t
substituting the values of the variables, we have
F = 70 kg(0 m/s - 4.43 m/s)/0.02 s
= 70 kg(- 4.43 m/s)/0.02 s
= -310.1 kgm/s ÷ 0.02 s
= -15,505 N
So, the force transmitted to her bones is 15,505 N
