You have 100 grams of different substances, the most dense substance would have which volume? A. 1L B. 0.10 L. C. 100 L D. 10 L

1 answer:

6 0

Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. We do as follows:

A. 1L
100 g /1L = 100 g/L

B. 0.10 L.
100g/0.10L = 1000 g/L ------> MOST DENSE SUBSTANCE

C. 100 L
100g/100L = 1 g / L 

D. 10 L
100g/10L = 10 g/L

You might be interested in

Un lote de construcción rectangular mide 100 metros por 150 metros. Calcula el área de este lote en yardas cuadradas.

antoniya [11.8K]

Answer: 17939.74 yards

Explanation:

Given , A rectangular measures 100 meters by 150 meters

To find : Area of rectangle.

Formula :

Area of rectangle = Length x width

Here, let length = 100 meters and width = 150 meters

Then, Area of rectangle = 100 meters x 150 meters = 15,000 square meters

Also , 1 meter = 1.09361 yards

Then, Area of rectangle = 15,000 x 1.09361 x 1.09361 square yards

= 17939.7424815 square yards

≈ 17939.74 yards

Hence, the area of rectangle is 17939.74 yards .

8 0

3 years ago

A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

#SPJ4