In complete agreement

An object, with mass 83 kg and speed 15 m/s relative to an observer, explodes into two pieces, one 4 times as massive as the oth

Agata [3.3K]

Answer:

The amount of Kinetic energy added to the system is 2334.3J

Solution:

As per the question:

Mass of object, m = 83 kg

Relative velocity of the object, $v_{mb} = v_{m} - v_{o} = 15 m/s$

After the explosion,

mass of the fragment is M and the other fragment, M' = 4M

Velocity of the lighter fragment after collision, v = 0 m/s

Now,

Mass of heavier fragment, M' = $\frac{4}{5}m$

Mass of lighter fragment, M' = $\frac{1}{5}m$

Let the velocity of the heavier fragment be v'.

Therefore by the law of conservation of momentum, we have:

Momentum of the object before collision = Momentum of the object after collision

$mv_{mo} = Mv + M'v'$

$mv_{mo} = M.0 + \frac{4}{5}mv'$

$v' = \frac{5}{4}\times 15 = 18.75 m/s$

Now, the change in Kinetic Energy gives the amount of Kinetic energy added to the system:

$\Delta KE = KE_{final} - KE_{initial}$

$\Delta KE = \frac{1}{2}Mv'^{2} - \frac{1}{2}mv_{mo}^{2}$

Since, the lighter particle stops, it won't have any kinetic energy.

$\Delta KE = \frac{1}{2}\times \frac{4}{5}times 83\times {18.75}^{2} - \frac{1}{2}\times 83\times 15^{2}$

$\Delta KE = 11671.8 - 9337.5 = 2334.3 J$

7 0

3 years ago

Is this object showing acceleration for the first 2 seconds? explain your answer.

iVinArrow [24]

Yes. The line is increasing. The flat line at the top of the graph is where there is not acceleration and the decreasing line is deceleration.

3 0

3 years ago

Cuál fue la primera derrota que sufrieron los fascistas

valentinak56 [21]

its c so componets arr r\very fragle im aying this because it needs more then 20 letters

8 0

3 years ago

Physics question, answer completely with work

Alenkasestr [34]

The force is 2.0 N east

Explanation:

The impulse exerted by a force is defined as the product between the force itself and the time interval during which the force is applied. Mathematically, it is equal to the change in momentum experienced by the object on which the force is acting:

$I=F\Delta t = \Delta p$

Where

I is the impulse

F is the force

$Delta t$ is the time interval during which the force is applied

$\Delta p$ is the change in momentum

In this problem,

$\Delta t = 3.0 s$ is the time interval

$I=6.0 N\cdot s$ (east) is the impulse

Therefore, the magnitude of the force is

$F=\frac{I}{\Delta t}=\frac{6.0}{3.0}=2.0 N$

And the direction is the same as the impulse (east).

Learn more about impulse and change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

7 0

3 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between m₁ and m₂ has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between m₁ and m₃ has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for x :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

3 years ago

In complete agreement​

In complete agreement