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Gelneren [198K]
4 years ago
15

2 A kovács a meleg vasat a hideg vízbe teszi. A vas

Physics
1 answer:
Anton [14]4 years ago
4 0

Answer:

300 kJ  

Explanation:

There are two energy flows in this process.

Heat lost by iron + heat gained by water =      0

            q₁             +                    q₂              =      0

       -300 kJ        +                     q₂              =      0  

                                                   q₂              = 300 kJ  

The water gained 300 kJ.

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C) A sample of substance of volume 10 cm3 was brought back to Earth from the Moon.
postnew [5]

Answer:

8125

Explanation:

P=0,13 N\\a=1,6 N/kg\\m=P/a=0,13/1,6=0,08125\\V=10(cm)^{3} =10^{-5} meters^{3}  \\p=m/V=0,08125/10^{-5} =0,08125*100000=8125kg/meters^{3}

3 0
4 years ago
Two technicians are discussing fuse testing. Technician A says that a test light should light on both test points of the fuse if
Murljashka [212]

Answer:

C. Both technicians A and B

Explanation:

Both technicians are absolutely correct because a functional test light is meant to light on both test point if the fuse is working fine which implies that, if the test light doesn't light on both sides then there must be a fault with the fuse. So, both technicians A and B are very correct.

4 0
3 years ago
Two protons are 1 × 10−10 m apart (about one atomic radius). Which interaction between two protons is stronger, the gravitationa
anastassius [24]

Answer: The electric repulsion between the two protons is stronger than the gravitational attraction.

Explanation: Please see the attachments below

4 0
4 years ago
Find velocity vx(t) and coordinate x(t) for a particle of mass m which is subject to the force given by: Fx = F0 e −kt , where F
muminat

Answer:

v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})

x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)

Explanation:

Given that the force of the particle is,

F_{x}=F_{0}e^{-kt}

Now it can be further written as

m\frac{dv}{dt}= F_{0}e^{-kt}\\\frac{dv}{dt}=\frac{ F_{0}}{m} e^{-kt}\\dv=\frac{ F_{0}}{m}e^{-kt}dt\\ v=\frac{ F_{0}}{-km}e^{-kt}+C

Now the initial conditions are v=1 at t=0.

So,

1=\frac{ F_{0}}{-km}e^{0}+C\\C=1+\frac{ F_{0}}{km}

Now the velocity will become.

v_{x} (t)=\frac{ F_{0}}{-km}e^{-kt}+1+\frac{ F_{0}}{km}\\v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})

And,

\frac{dx}{dt} =1+\frac{ F_{0}}{km}(1-e^{-kt})\\dx=(1+\frac{ F_{0}}{km}(1-e^{-kt}))dt\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+C\\

And, another initial condition is x=0 at t=0

0=0+\frac{ F_{0}0}{km}+\frac{ F_{0}t}{k^{2} m}e^{0}+C\\C=-\frac{ F_{0}t}{k^{2} m}

Now,

x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+-\frac{ F_{0}t}{k^{2} m}\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)

5 0
3 years ago
You don;t run across elastic collision in real life
pentagon [3]

Answer:

no comment sorry bro

Explanation:

4 0
3 years ago
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