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Oksana_A [137]
3 years ago
15

UN BLOQUE TIENE DE DIMENSIONES AB = 4 cm, BC = 2.5 cm, BF = 6 cm Y UNA MASA DE 300 gr  DETERMINE SU VOLUMEN  HALLE SU DENSIDAD

 COMPARE LA DENSIDAD HALLADA, CON LAS DENSIDADES DE SOLIDOS QUE APARECEN EN LAS TABLAS Y DIGA A QUE MATERIAL CORRESPONDE O NO
Physics
1 answer:
torisob [31]3 years ago
6 0

Responder:

- Volumen del bloque = 60cm³

- Densidad = 5g / cm³

Explicación:

Dada la dimensión del bloque

AB = 4 cm, BC = 2.5 cm y BF = 6 cm

Volumen del bloque = Largo × Ancho × Alto

Volumen del bloque = AB × BC × BF

Volumen del bloque = 4 × 2.5 × 6

Volumen del bloque = 60cm³

Dada la masa del bloque = 300 gramos

Densidad = Masa / Volumen

Densidad = 300g / 60cm³

Densidad = 5g / cm³

Algunos materiales que tienen una densidad de 5g / cm³ o cercanos son radio, germanio y europio con densidades de 5g / cm³, 5.3g / cm³ y 5.24g / cm³ respectivamente.

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Mass of a body is 210g and its density is 7.981g/cm^3 what will be its volume with regard to significant figures
aleksandrvk [35]

Density is mass per unit volume,

\rho=\dfrac mV

so to get the volume, divide the mass by the density,

V=\dfrac m\rho

So the volume is

\dfrac{210\,\mathrm g}{7.981\frac{\rm g}{\mathrm{cm}^3}}\approx26.312\,\mathrm{cm}^3\approx\boxed{26\,\mathrm{cm}^3}

3 0
3 years ago
Ben franklin showed that 1 teaspoon of oil would cover about 0.50 acre of still water. if you know that 1.0 x 104m2 = 2.47 acres
Dmitry_Shevchenko [17]

To solve for the thickness of the latter of oil, we can simply divide the volume by the total surface area:

Thickness = Volume / Surface Area

We are both given the volume and the surface area, all we have to do now is to convert the units in like terms so that we can cancel those out.

Surface Area = 0.50 acres (1.0 x 10^4 m^2 / 2.47 acres)

Surface Area = 2,024.29 m^2

Further converting this into cm:

Surface Area = 2,024.29 m^2 (100 cm / m)^2

Surface Area = 20,242,914.98 cm^2

 

Therefore the thickness is:

Thickness = 5.0 cm^3 / 20,242,914.98 cm^2

Thickness = 2.47 x 10^-7 cm

or

<span>Thickness = 2.47 x 10^-9 m = 2.47 nm = 24.7 A</span>

3 0
3 years ago
Soft ball,kick ball,wiffle ball??
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6 0
2 years ago
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A light with a second-order bright band forms a diffraction angle of 30. 0°. The diffraction grating has 250. 0 lines per mm. Wh
Luden [163]

The distance between two successive troughs or crests is known as the wavelength. The wavelength of the light will be 1000 nm.

How do you define wavelength?

The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.

The wavelength is also defined as the distance between two locations in a wave that have the same oscillation phase.

Diffraction angle= 30⁰

Diffraction grating per mm= 250

wavelength = ?

Mathematically the equation of bright band is given by

\rm \lambda= \frac{sin\theta}{nN}

\rm \lambda= \frac{sin23^0}{250\times 2}

\rm \lambda= 0.000001 m

\rm \lambda= 1000 nm

Hence the wavelength of the light will be 1000 nm.

To learn more about the wavelength refer to the link;

brainly.com/question/7143261

8 0
2 years ago
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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
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