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3 years ago

Can I have the answer please

Physics

1 answer:

Margarita [4]3 years ago

8 0

Answer:

intial velocity, u=0 final velocity, v=0

Explanation:

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Define acceleration to show its scientific meaning

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According to Boyles’ law, PV = constant. If a graph is plotted with the pressure P against the volume V, the graph would be a(n)

Hunter-Best [27]

Answer:

C. hyperbola

Explanation:

From Boyle's law:

PV = k, where k is a constant

Solving for P:

P = k / V

At first glance, this equation doesn't fit any of the options. But when you graph it, you can see that it's actually a <em>rotated</em> hyperbola.

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Select the three words in the following sentence that indicate it is in the third person.

Triss [41]

Die meisten von ihnen haben die Möglichkeit zu den Anderen zu kommen oder die Möglichkeit für die Zeit der Arbeit mit dem Auto und der Wohnung zu

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2 years ago

Bar magnets placed on table may not point along N-S direction . Why? Give reasons.

denpristay [2]

Answer:

Hey!

Explanation:

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3 years ago

Suppose you apply a ﬂame and heat one liter of water, raising its temperature 10°C. If you transfer the same heat energy to two

polet [3.4K]

(a) the water density is $d=1000 kg/m^3$ , and 1 liter corresponds to a volume of $V=1 L=0.001 m^3$ . Therefore we can find the mass of the water in the first case:
$m=dV=(1000 kg/m^3)(0.001 m^3)=1 kg$

The amount of heat supplied to the water to raise its temperature by $\Delta T=10 ^{\circ} C$ is
$Q=m C_s \Delta T$ (1)
where
$C_s = 4.18 kJ/(kg ^{\circ} C}$ is the specific heat capacity of the water.
Using the data, we find
$Q=(1 kg)(4.18 kJ/(kg ^{\circ} C)(10^{\circ} C)=41.8 kJ$

We want to find the increase in temperature if we transfer the same amount of heat Q to 2 liters of water. The mass of 2 liters of water is
$m=dV=(1000 kg/m^3)(0.002 m^3)=2 kg$
And so by re-arranging equation (1) we can calculate the new increase of temperature:
$\Delta T_2 = \frac{Q}{m C_s} = \frac{41.8 kJ}{(2 kg)(4.18 kJ/(kg ^{\circ} C)}=5 ^{\circ} C$

(b) Now we have 3 liters of water. SImilarly to point (a), the mass is now
$m=dV=(1000 kg/m^3)(0.003 m^3)=3 kg$
And so, the increase in temperature if we use the same amount of heat as before is
$\Delta T_3= \frac{Q}{m C_s} = \frac{41.8 kJ}{(3 kg)(4.18 kJ/(kg ^{\circ} C)}=3.3 ^{\circ} C$

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3 years ago