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4 years ago

The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.

1 answer:

5 0

The answer for this problem is clarified through this, the system is absorbing (+). And now see that it uses that the SURROUNDINGS are doing 84 KJ of work. Any time a system is overshadowing work done on it by the surroundings the sign will be +. So it's just 12.4 KJ + 4.2 = 16.6 KJ.

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1 point

Yuliya22 [10]

PE= 3kg x 10N/kg x 10m
= 300J

8 0

3 years ago

The distance between two consecutive nodesof a standing wave is 20.9cm.Thehandgen-erating the pulses moves up and down throughac

irina1246 [14]

Answer:

Velocity, v = 0.239 m/s

Explanation:

Given that,

The distance between two consecutive nodes of a standing wave is 20.9 cm = 0.209 m

The hand generating the pulses moves up and down through a complete cycle 2.57 times every 4.47 s.

For a standing wave, the distance between two consecutive nodes is equal to half of the wavelength.

$\dfrac{\lambda}{2}=0.209\ m\\\\\lambda=0.418\ m$

Frequency is number of cycles per unit time.

$f=\dfrac{2.57}{4.47}\\\\f=0.574\ Hz$

Now we can find the velocity of the wave.

Velocity = frequency × wavelength

v = 0.574 × 0.418

v = 0.239 m/s

So, the velocity of the wave is 0.239 m/s.

4 0

3 years ago

A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg

Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

$p_i=mv+0=mv$

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

$p_f=mv'+MV'$

Equating initial and the final momenta we get

$mv=mv'+MV'\\\\m(v-v')=MV'.....i$

Now since the surface is frictionless thus the energy is also conserved thus

$E_i=\frac{1}{2}mv^2$

Similarly the final energy becomes

$E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2$ \

Equating initial and final energies we get

$\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)$

Solving i and ii we get

$v+v'=V'$

Using this in equation i we get

$v'=\frac{v(m-M)}{(M-m)}=-v$

Thus putting v = -v' in equation i we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0

3 years ago

Which is younger, the rock layers or the fault?

jeka94

<h2>FAULT</h2>

The principle of cross-cutting relationships states that a fault or intrusion is younger than the rocks that it cuts. The fault labeled 'E' cuts through all three sedimentary rock layers (A, B, and C) and also cuts through the intrusion (D). So the fault must be the youngest formation that is seen.

<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em>

6 0

3 years ago

Read 2 more answers

Runner x starts 50 feet in front of runner z and moves with a velocity of 8.0 m/s. At what speed would runner z have to run to c

Mariulka [41]

Answer:

15.625

Explanation:

8 0

3 years ago