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Natalka [10]
3 years ago
9

Which attributes does the exosphere have?

Physics
1 answer:
coldgirl [10]3 years ago
8 0
The exosphere has high temperatures and low gravity
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Explain Archimedes Principle
rusak2 [61]

Answer:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces. Archimedes' principle is a law of physics fundamental to fluid mechanics. It was formulated by Archimedes of Syracuse.

Explanation:

6 0
3 years ago
A force of 100 newtons Is applled to a box at an angle of 36° with the horizontal. If the mass of the box Is 25 kilograms, what
faust18 [17]
Horizontal component of force = 100cos(36)= 80.9 N
5 0
3 years ago
A proton that has a mass m and is moving at +164 m/s undergoes a head-on elastic collision with a stationary carbon nucleus of m
Irina18 [472]
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:

m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,

m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂'  --> equation 1

The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is

(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2

Solving equations 1 and 2 simultaneously, v₁' =  -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
7 0
3 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0
3 years ago
The main purpose of an air bag is to stop a passenger during a car accident in a greater amount of time than if the air bag were
Simora [160]

Answer:

a) 45571 N  

b) 22786 N

c) 4557 N

Explanation:

  • Since the goal of the airbag is helping the person to stop after the collision in a greater time, this means that the change in momentum must finish when this is just zero.
  • In other words, the change in momentum, must be equal to the initial one, but with opposite sign.

       \Delta p = - p_{o} = -m*v = -55 kg*29m/s = -1595 kgm/s (1)

  • Now, just applying the original form of  Newton's 2nd Law, we know that this change in momentum must be equal to the impulse needed to stop the person:

       \Delta p = F* \Delta t  (2)

  • So, as we know the magnitude of Δp from (1) and we have different Δt as givens, we can get the different values of F (in magnitude) required to stop the person for each one of them, as follows:

       F_{1} = \frac{\Delta p}{\Delta t_{1}} = \frac{1595kgm/s}{0.035s} = 45571 N (3)

       F_{2} = \frac{\Delta p}{\Delta t_{2}} = \frac{1595kgm/s}{0.07s} = 22786 N (4)

       F_{3} = \frac{\Delta p}{\Delta t_{3}} = \frac{1595kgm/s}{0.35s} = 4557 N (5)

4 0
3 years ago
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