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fgiga [73]
3 years ago
13

A rocket explodes into two fragments, one 25 times heavier than the other. The magnitude of the momentum change of the lighter f

ragment is A) 25 times as great as the momentum change of the heavier fragment. B) The same as the momentum change of the heavier fragment. C) 1/25 as great as the momentum change of the heavier fragment. D) 5 times as great as the momentum change of the heavier fragment. E) 1/4 as great as the momentum change of the heavier fragment.
Physics
1 answer:
V125BC [204]3 years ago
4 0

Answer:

B) The same as the momentum change of the heavier fragment.

Explanation:

Since the initial momentum of the system is zero, we have

0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.

0 = p + p'

p = -p'

Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0  = p

The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'

Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'

<u>So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment.  </u>

So, option B is the answer

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Tasya [4]

Answer:

Coefficient of friction = 0.836

Explanation:

If v be the speed after one quarter of the circular path

v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8

tangential acceleration = 5.8 m/s²

radial acceleration = v² /r = 5.8

total acceleration = √2 x 5.8

m x√2 x 5.8 = m x g xμ

μ = √2 x 5.8 / 9.8 = 0.836

7 0
3 years ago
The animation shows a ball which has been kicked upward at an angle. Run the animation to watch the motion of the ball. Click in
sergejj [24]

Answer:

The acceleration of the ball is  a_y =  - 0.3672 \ m/s^2

Explanation:

From the question we are told that

       The maximum height the ball reachs is H_{max} =  42.24 \ m

       The horizontal component of the initial velocity of the ball is v_{ix} = 5.57 \ m/s

       The vertical component of the initial velocity of the ball is v_{iy} = = 16.18 m/s

The vertically motion of the ball can be mathematically represented as

       v_{fy}^2  =  v_{iy} ^2 + 2 a_{y} H_{max}

Here the final velocity at the maximum height is zero so v_{fy} = 0 \ m/s

Making the acceleration a_y the subject we have

        a_y =  \frac{v_{iy} ^2}{2H_{max}}

substituting values

      a_y =  - \frac{5.57^2}{2* 42.24}

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The negative sign shows that the direction of the acceleration is in the negative y-axis

6 0
3 years ago
The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f
zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

                 584 = Ha + Lg

an also this mean that torques acting on the body is balanced

         So,   0.410 Ha  = 0.840 Lg

    Making Lg the subject of formula in the equation above we

   Lg = 0.4881 Ha

 Considering the first equation and replacing Lg with this recent equation we have

                      584 = Ha + 0.4881 Ha

          Therefore Ha = 392.44 N

This value obtained is  for both hands for each hand we divide by 2

Therefore we have for each hand = 392.44/2 =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

             Lg = 0.4881 ×  392.44

                  = 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

                  191.22/2 = 95.8 N.

8 0
4 years ago
If the system is operated on mars, through what distance would the 18.0-kg mass have to fall to give the same amount of kinetic
-Dominant- [34]
The previous part of the exercise says:
"<span>Engineers are designing a system by which a falling mass m imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum. There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on Earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 m/s². In the Earth tests, when m is set to 18.0 kg and allowed to fall through 5.50 m, it gives 300.0 J of kinetic energy to the drum."

Since Kearth = Kmars, we have, for conservation of energy, that also the potential energies must be equal:
Uearth = Umars

which means:
m </span>· gearth · hearth = m · gmars <span>· hmars

we can solve for hmars:
hmars = (gearth / gmars) </span>· hearth
           = (9.8 / 3.71) · 5.50
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Therefore, the correct answer will be: the mass would have to fall from an height of 14.53m.

5 0
3 years ago
All of the following would be questions that could be scientifically investigated except:
ra1l [238]
All of the following would be questions that could be scientifically investigated except A.
That is an opinion and cannot become a fact.
6 0
3 years ago
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