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aivan3 [116]
3 years ago
8

An object is moving but we don't know its mass or velocity. A force of 20 newtons to the right

Physics
1 answer:
jok3333 [9.3K]3 years ago
8 0

The impulse is (force) x (time) = (20 N) x (20 sec) = 400 N-sec

When we grind through the units, we find that the [newton-second]
is exactly the same as the [kilogram-meter/sec] unit-wise, and once
we know that, it doesn't surprise us to learn that impulse is equivalent
to a change in momentum (mass x speed ... also kg-m/s).

So this impulse exerted on the moving object adds 400 kg-m/s of
linear momentum to its motion, directed to the right.  That may or
may not be the total change in its momentum during that 20-sec,
because our 20-N may not be the only force acting on it.


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The friends start by discussing what they know with respect to electric and magnetic fields. Alexa mentions several differences
Sveta_85 [38]

Answer:

A. "The electric force vector is along the direction of the electric field, whereas the magnetic force vector is perpendicular to the magnetic field."

D. "The kinetic energy of a charged particle moving in an electric field is not altered, whereas the kinetic energy of a charged particle moving in a magnetic field is either increased or decreased, depending on the direction of motion."

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8 0
3 years ago
If you and a friend are standing at opposites ends of a gymnasium and one of you claps, will the other hear the clap at the same
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3 0
3 years ago
A cylinder with a piston contains 0.200 mol of nitrogen at 1.50×105 Pa and 320 K . The nitrogen may be treated as an ideal gas.
Alja [10]

Answer:

Q = -105 J

Also we know that for cyclic process change in internal energy is always ZERO

Explanation:

First gas is compressed isobarically such that its volume is half of initial volume

So its temperature is also half

So heat given in this process is given as

Q = nC_p \Delta T

for diatomic gas we have

C_p = \frac{7}{2} R

so we will have

Q = 0.200(\frac{7}{2}R)(160 - 320)

Q = -930.7 J

Now in adiabatic process heat is not transferred

so in this process

Q = 0

so we have

T_1V_1^{1.4-1} = T_2V_2^{1.4-1}

(160)(\frac{V}{2})^{0.4} = T_2(V)^{0.4}

T_2 = 121.26 K

Now it is again reached to original pressure

so temperature will become initial temperature

so heat given in that part

Q_3 = nC_v\Delta T

here we know that

C_v = \frac{5}{2}R

Q_3 = (0.200)(\frac{5}{2}R)(320 - 121.26)

Q_3 = 825.76 J

So total heat given to the system is

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Q = -105 J

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3 0
3 years ago
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