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riadik2000 [5.3K]
3 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]3 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]3 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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Explanation:

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3 years ago
our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three
Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The torque produced by the pile of rocks is \tau = 35.63\ N \cdot m  

b

The distance of the single for equilibrium to occur is r_s =1.62 \ m

Explanation:

From the question we are told that

     The mass of the left rock is  m_s = 2.25 \ kg

     The mass of the rock on the right m_p = 10.1 kg

    The distance from  fulcrum to the center of the pile of rocks is  r_p = 0.360 \ m

   

Generally the torque produced by the pile of rock is mathematically represented as

           \tau = m_p * g * r_p

Substituting values

         \tau = 10.1 * 9.8  * 0.360                  

          \tau = 35.63\ N \cdot m      

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

   The torque due to the single rock is

           \tau = m_s  * g * r_s

At equilibrium the both torque are equal

            35.63 = m_s * r_s * g

Making r_s the subject of the formula

             r_s = \frac{35.63 }{m_s * g}

Substituting values

            r_s = \frac{35.63 }{2.25 * 9.8}

            r_s =1.62 \ m

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Explanation:

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pav-90 [236]
I attached the missing picture.
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For point A we have:
F_a=F_cf-F_g
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At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
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timurjin [86]

Answer:

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Explanation:

Given;

mass of the sports car, m = 1100 kg

initial velocity of the sports car, u = 0 m/s

final velocity of the sports car, v = 32 m/s

time of motion, t = 10 s

The kinetic energy of the car is given by;

K.E = ¹/₂m(v² - u²)

K.E = ¹/₂mv²

K.E = ¹/₂ x 1100 x 32²

K.E = 563200 J

The average power of the engine of the sports car is given by;

Pavg = Energy / time

Pavg = 563200 / 10

Pavg = 56320 W

Pavg = 56.32 kW

Therefore, the average power of the engine of the sports car is 56.32 kW

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