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riadik2000 [5.3K]
2 years ago
8

The landing of a spacecraft is cushioned with the help of airbags. During its landing on Mars, the velocity of downward fall is

16 meters/second. Immediately after the impact, the velocity is reduced to 1.2 meters/second. If the spacecraft has a mass of 11.5 × 104 kilograms, what is your estimate for the impulse if the time of impact is 0.8 seconds?
Physics
2 answers:
Svetach [21]2 years ago
7 0

Answer:

-1.7\cdot 10^6 kg m/s

Explanation:

The impulse is equal to the product between the force (F) and the time of impact (\Delta t):

I=F \Delta t

However, the impulse is also equal to the change in momementum of the spacecraft:

I = \Delta p= m (v_f - v_i)

where

m=11.5 \cdot 10^4 kg is the mass of the spacecraft

v_f = 1.2 m/s is the final velocity

v_i = 16 m/s is the initial velocity

Substituting these numbers into the formula, we find

I=(11.5\cdot 10^4 kg)(1.2 m/s-16 m/s)=-1.7\cdot 10^6 kg m/s

where the negative sign simply means that the impulse is in the opposite direction to the motion of the spacecraft (in fact, it makes it slowing down).

maria [59]2 years ago
5 0
You can use the impulse momentum theorem and just subtract the two momenta.
P1 - P2 = (16-1.2)(11.5e4)=1702000Ns
If you first worked out the force and integrated it over time the result is the same
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A hot iron ball of mass 200 g is cooled to a temperature of 22°C. 6.9 kJ of heat is lost to the surroundings during the process.
qaws [65]
Heat lost or gained, H = mc(θ₂ - θ₁) 
Where m = mass, c = Specific heat capacity, θ₂= final temperature, θ₁ = initial temperature

m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C  (Since it was cooled).

H = 6.9 kj = 6.9 *1000J = 6900 J

6900 = 200*0.444* (θ₂ - 22)

6900/(200*0.444)  =  θ₂ - 22

77.70 = θ₂ - 22

θ₂ - 22 = 77.7

θ₂      =  77.7 + 22 = 99.7

So initial temperature before cooling ≈ 100°C .  Option C.


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3 years ago
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1. An astronaut in a spacesuit has a mass of 80 kilograms. What is the weight of this astronaut on the surface of the Moon where
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A ball is dropped from rest from the top of a cliff that is 15.0 m high. From ground level, a second ball is thrown straight upw
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7 0
3 years ago
If a rock is thrown upward on the planet Mars with a velocity of 15 m/s, its height above the ground (in meters) after t seconds
s2008m [1.1K]

(a) The velocity (in m/s) of the rock after 1 second is 11.28 m/s.

(b) The velocity of the rock after 2 seconds is 7.56 m/s.

(c) The time for the block to hit the surface is 4.03.

(d) The velocity of the block at the maximum height is 0.

<h3>Velocity of the rock</h3>

The velocity of the rock is determined as shown below;

Height of the rock after 1 second; H(t) = 15(1) - 1.86(1)² = 13.14 m

v² = u² - 2gh

where;

  • g is acceleration due to gravity in mars = 3.72 m/s²

v² = (15)² - 2(3.72)(13.14)

v² = 127.23

v = √127.23

v = 11.28 m/s

<h3>Velocity of the rock when t = 2 second</h3>

v = dh/dt

v = 15 - 3.72t

v(2) = 15 - 3.72(2)

v(2) = 7.56 m/s

<h3>Time for the rock to reach maximum height</h3>

dh/dt = 0

15 - 3.72t = 0

t = 4.03 s

<h3>Velocity of the rock when it hits the surface</h3>

v = u - gt

v = 15 - 3.72(4.03)

v = 0

Learn more about velocity at maximum height here: brainly.com/question/14638187

8 0
2 years ago
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