All categories

3 years ago

In terms of volume,how do ml & cm3 relate to one another?

Physics

2 answers:

kicyunya [14]3 years ago

5 0

1 millilitre = 1 cubic cm

Keith_Richards [23]3 years ago

3 0

1 milliliter = 1 cubic centimeter (cm^3)

You might be interested in

Visible light with a wavelength of 480 nm appears Question 33 options:

AlladinOne [14]

It would appear as blue

5 0

3 years ago

Consider the circuit below, which is powered by a 6-v battery. switch s is opened at t = 0 after having been closed for a long t

Semmy [17]

The answer should be B

3 0

3 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

8 0

3 years ago

a mass of 1.00 kg of water at temperature T is poured from a height of 0.100 km into a vessel containing water of the same tempe

Mariana [72]

Answer:

1.34352 kg

Explanation:

$m_w$ = Mass of water falling = 1 kg

h = Height of fall = 0.1 km

$\Delta T$ = Change in temperature = 0.1

c = Specific heat of water = 4186 J/kg K

g = Acceleration due to gravity = 9.81 m/s²

$m_v$ = Mass of water in the vessel

Here the potential energy will balance the internal energy

$m_wgh=m_wc\Delta T+m_vc\Delta T\\\Rightarrow m_v=\dfrac{m_wgh-m_wc\Delta T}{c\Delta T}\\\Rightarrow m_v=\dfrac{m_wgh}{c\Delta T}-m_w\\\Rightarrow m_v=\dfrac{1\times 9.81\times 100}{4186\times 0.1}-1\\\Rightarrow m_v=1.34352\ kg$

Mass of the water in the vessel is 1.34352 kg

6 0

3 years ago

Which statement is true of cooling down after physical activity? A cool-down is a period of semistrenuous activity after physica

Xelga [282]

The statement that is true of cooling down after physical activity is that you should cool down for about 5 to 10 minutes after being physically active.

8 0

3 years ago

Read 2 more answers