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1 month ago

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

Physics

2 answers:

Sav [3.1K]1 month ago

7 0

The volumes are 200cm3 and 0.0002m3

Keith_Richards [3.2K]1 month 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.

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An object weighs 7.84 N when it is in air and 6.86 N when it is immersed in water of density 1000 kg/m3. What is the density of

Sav [3153]

The object's density is 8000 kg/m^3. The object's weight in air is 7.84 N while it measures 6.86 N when submerged in water, where the density of water is 1000 kg/m^3. According to Archimedes' principle, an immersed object experiences an upward buoyant force equivalent to its loss of weight in the fluid. By calculating the weight difference (7.84 - 6.86 = 0.98 N) and employing the standard equations relating density and volume, we find that 10^-4 m^3 corresponds to a density of 8000 kg/m^3.

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1 month ago

In broad daylight, the size of your pupil is typically 3 mm. In dark situations, it expands to about 7 mm. How much more light c

Sav [3153]

Answer:

31.4 mm²

Explanation:

The ability of a telescope or eye to gather light can be expressed by the formula,

$GDP=\pi } \frac{d^{2} }{4}$

where d signifies the diameter of the pupil.

In bright daylight, the usual size of the pupil is 3 mm.

$GDP_{b} =\pi \frac{3^{2} }{4}$

Conversely, in darkness, the diameter typically enlarges to 7 mm.

$GDP_{b} =\pi \frac{7^{2} }{4}$

This indicates an increase in light-gathering capacity.

$Increase=\pi \frac{49}{4} -\pi \frac{9}{4} \\Increase=31.4 mm^{2}$

Thus, the amount of light the eye can capture is 31.4 mm².

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2 months ago

A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel

kicyunya [3294]

The rocket's acceleration is described here as

$a_y = 2.60* t$

now recognizing that

$\frac{dv}{dt} = 2.60t$

we integrate both sides

$\int dv = \int 2.60t dt$

$v = 2.60\frac{t^2}{2}$

$v = 1.30 t^2$

given that the rocket is accelerating for a duration of t = 10 s

thus, we have

$v = 1.30 * 10^2$

$v = 130 m/s$

consequently, after t = 10 s, the rocket will achieve a speed of 130 m/s in an upward direction

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1 month ago

A 45.0-kg person steps on a scale in an elevator. The scale reads 460 n. What is the magnitude of the acceleration of the elevat

ValentinkaMS [3465]

The elevator's acceleration is 0.422 m/s². To clarify the solution: By applying Newton's Law, the net forces in the motion's direction equal the mass multiplied by the acceleration. The forces comprise 460 N in the motion's direction and the person's weight acting in the opposite direction... The weight is determined by the mass and gravity's acceleration (W = mg). Here m = 45 kg and g = 9.8 m/s², leading to W = 441 N. With the scale indicating 460 N, we apply F - W = ma, yielding 19 = 45 a. Dividing both sides by 45 gives a = 0.422 m/s².

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1 month ago

What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?

serg [3582]

Response:

U = 12,205.5 J

Clarification:

To determine the internal energy of an ideal gas, use the following equation:

$U=\frac{3}{2}nRT$ (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: the temperature of the gas = 100K

Substituting the parameter values into equation (1):

$U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J$

The overall internal energy for 10 moles of Oxygen at 100K is 12,205.5 J

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1 month ago