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Vsevolod [243]
3 years ago
6

If a bottlel filled with water 25c and freezes at -5c the waters tempreture drop what is the sign of the systems q,w,ΔE

Physics
1 answer:
laila [671]3 years ago
6 0
Triangle because h2o and celsious have yhree silibyles and the triangle has three sides
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The sun is bigger, but has less mass than the earth
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3 years ago
which statements about velocity are true? check all that apply. a. for velocity, you must have a number, a unit, and a direction
satela [25.4K]
a). for velocity, you must have a number, a unit, and a direction.
Yes.  This one isn't bad.  The 'number' and the 'unit' are the speed.

b). the si units for velocity are miles per hour.
No.  That's silly. 
'miles' is not an SI unit, and 'miles per hour'
is only a speed, not a velocity. 

c). the symbol for velocity is .
You can use any symbol you want for velocity, as long as
you make its meaning very clear, so that everybody knows
what symbol you're using for velocity.
But this choice-c is still wrong, because either it's incomplete,
or else it's using 'space' for velocity, which is a very poor symbol.

d). to calculate velocity, divide the displacement by time.
Yes, that's OK, but you have to remember that the displacement
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3 0
3 years ago
What is the gravitational potential energy of a 3 kg ball kicked into the air at a height of 5 meters?
sladkih [1.3K]

formula for gravitational P.E =mgh

Solution:-mass=3kg height=5metre and gravity=9.8 or 10m/sec² so P.E=mgh , 3×9.8×5=147kgm²/sec²

7 0
3 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
Students had two batteries and two different resistors. During four trials, they build four different circuits and measure the c
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Answer:

bhi jo bhi of gp oh oh gi IG 7u to uff do if goo td to yd do FP ae rt 7g hi pic vo icon

Explanation:

bh hi h bhi vc di oh x At jb jo iv hp of di of dr hi o hc x gh ki vc hi jo

5 0
2 years ago
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