All categories

14 days ago

A beaker contain 200mL of water What is its volume in cm3 and m3

Physics

2 answers:

Sav [2.2K]14 days ago

7 0

The volumes are 200cm3 and 0.0002m3

Keith_Richards [2.2K]14 days ago

4 0

One mL corresponds to a cm^3, so it totals 200 cm^3. To convert from cm^3 to m^3, refer to the KHDBDCM conversion chart. You simply move the decimal point two spaces to the left, indicating that it equals 2 m^3.
In summary: 200 cm^3 and 2 m^3.

You might be interested in

3.113 A heat pump is under consideration for heating a research station located on Antarctica ice shelf. The interior of the sta

ValentinkaMS [2425]

Answer:

a. β = 8.23 K

b. β = 28.815 K

Explanation:

The performance of the heat pump can be calculated using the formula

β = TH / (TH - TC)

TH = 15 ° C + 273.15 K = 288.15 K

TC = - 20 ° C + 273.15 K = 253.15 K

β = 288.15 K / (288.15 K - 253.15 K)

β = 8.23 K

TH = 15 ° C + 273.15 K = 288.15 K

TC = 5 ° C + 273.15 K = 278.15 K

β = 288.15 K / (288.15 K - 278.15 K)

β = 28.815 K

6 0

14 days ago

A 5-kg concrete block is lowered with a downward acceleration of 2.8 m/s2 by means of a rope. The force of the block on the Eart

Maru [2355]

The moment the body impacts the ground, two types of Forces are produced: the gravitational pull and the Normal Force. This aligns with Newton's third law, indicating that every action has an equal and opposite reaction. If the downward force of gravity is directed toward the earth, the reactionary force from the block acts upwards, equivalent to its weight:

F = mg

Where,

m = mass

g = gravitational acceleration

F = 5*9.8

F = 49N

Consequently, the answer is E.

5 0

17 days ago

A wildlife researcher is tracking a flock of geese. The geese fly 4.0 km due west, then turn toward the north by 40° and fly ano

serg [2593]

Answer:

(a). The distance traveled is 7.06 km towards the west.

(b). Their displacement magnitude is 7.51 km.

Explanation:

The information given states that,

The geese initially move 4.0 km directly west, then alter course to the north at a 40° angle, covering an additional 4.0 km.

Based on the diagram,

(a). To determine the distance

We will apply the distance formula

$AD=AB+BD$

Insert the values into the equation

$D= 4+4.0\cos40^{\circ}$

$D=7.06\ km$

The resultant distance is 7.06 km westward.

(b). We will find the total displacement's magnitude

Using the displacement formula

$AC=\sqrt{(CD)^2+(AD)^2}$

Insert the values into the equation

$AC=\sqrt{(4.0\sin40)^2+(7.06)^2}$

$AC=7.51$

The total displacement magnitude is 7.51 km.

In conclusion, (a). The traveled distance is 7.06 km towards the west.

(b). The magnitude of their total displacement is 7.51 km.

7 0

16 days ago

When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (

inna [2205]

The answer is (C) 4 beats per second. The number of beats is computed as the difference between the frequencies of the two tuning forks. Plugging in the frequency values yields a result. Thus, the number of observable beats per second will be 4.

3 0

6 days ago

Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.

inna [2205]

The heat required to raise the temperature of a substance by $\Delta T$ is represented by
$Q=m C_p \Delta T$
where m stands for the mass of the substance and $C_p$ indicates the specific heat of the substance. In this situation, we possess $m=100~g=0.1~Kg$ and $C_p=4.19~KJ/(Kg K)$ , the specific heat of water.
Consequently, we can ascertain the temperature rise $\Delta T$ :
$\Delta T = \frac{Q}{m C_p}= \frac{50~KJ}{0.1~Kg cdot 4.19~KJ/(Kg K)}=119~K =119^{\circ}C$
Initially, the water's temperature was $25^{\circ}C$ , so the end temperature should be
$T_f = 25^{\circ}C+119^{\circ}C=144^{\circ}C$
Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
$Q=m C_L=0.1~Kg \cdot 2265~KJ/Kg=226.5~KJ$
where $C_L$ denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of $100^{\circ}C$ (the temperature at which the phase change begins)

4 0

25 days ago