All categories

3 years ago

What does wire gauze help you do

Physics

2 answers:

Nookie1986 [14]3 years ago

5 0

Answer:

Explanation:

it helps beakers,flask and other glassware

Sladkaya [172]3 years ago

4 0

Answer:

Image result for what does wire gauze help you do

A wire gauze is a sheet of thin metal that has net-like patterns or a wire mesh. Wire gauze is placed on the support ring that is attached to the retort stand between the Bunsen burner and the glassware to support the beakers, flasks, or other glassware during heating.

Explanation:

Wire gauze is an important piece of supporting equipment in a laboratory because beakers or any type of glassware cannot be heated directly with a flame from the Bunsen burner, and the wire gauze will help protect the glassware.

You might be interested in

When water freezes, it expands. what does this say about the density of ice compared with the density of water?

masya89 [10]

<span>When water freezes to form ice, its volume expands. However, we know from conservation of mass that the mass of the ice is the same as the mass of the water. Since density is defined fundamentally as mass / volume, and we have an expanding volume at a constant mass, the denominator of the equation grows, and thus the density of ice is lower than that of liquid water.</span>

4 0

4 years ago

A metal sphere of radius 11 cm has a net charge of 2.8 × 10–8 C. (a) What is the electric field at the sphere's surface? (b) If

harina [27]

Answer:

Therefore,

a) $E=20.82\ kN/C$

b) $V= 2290\ Volt$

c) distance from the sphere's surface = 1.8 cm

Explanation:

Given:

Radius, r = 11 cm = 0.11 m

Charge, $Q = 2.8\times 10^{-8}\ C$

To Find:

a) electric field at the sphere's surface = ?

b) If V = 0 at infinity, what is the electric potential at the sphere's surface = ?

c) At what distance from the sphere's surface has the electric potential decreased by 320 V = ?

Solution:

Electric field at the sphere's surface is given as,

$E=\dfrac{k\times Q}{r^{2}}$

Where,

E = Electric Field,

$k = Coulomb's\ constant = 9\times 10^{9}$

Q = Charge

r = Radius

Substituting the values we get

$E=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{(0.11)^{2}}=20.82\ kN/C$

Now, Electric Potential at point surface is given as,

$V=\dfrac{k\times Q}{r}$

Substituting the values we get

$V=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{0.11}=2.29\ kV$

For distance from the sphere's surface has the electric potential decreased by 320 V,

So V becomes,

V = 2290 - 320 = 1970 Volt, then r =?

$V=\dfrac{k\times Q}{r}$

Substituting the values we get

$r=\dfrac{k\times Q}{V}\\\\r=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{1970}=0.128\ m$

Therefore the distance from the sphere's surface,

$d = 12.8 - 11 = 1.8\ cm$

Therefore,

a) $E=20.82\ kN/C$

b) $V= 2290\ Volt$

c) distance from the sphere's surface = 1.8 cm

7 0

4 years ago

A distance of 2.00 mm separates two objects of equal mass. If it gravitational force between them is 0.0104 N, find the mass of

Radda [10]

Put values in this formula, you will get the answer.

F=Gm^2/r^2

G=6.67X10^-11

8 0

4 years ago

A 5 kg brick is dropped from a height of 12m on a spring with a spring constant 8 kN/m. If the spring has unstretched length of

Ivan

Answer:

0.1164 m

0.49387 m

Explanation:

m = Mass = 5 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height from which brick falls = 12 m

k = Spring constant = 8 kN/m

a) Potential energy of the brick

$PE=mgh\\\Rightarrow PE=5\times 9.81\times 12\\\Rightarrow PE=588.6\ J$

Potential energy in spring

$PE=\frac{1}{2}\times kx^2\\\Rightarrow x=\sqrt{\frac{PE\times 2}{k}}\\\Rightarrow x=\sqrt{\frac{588.6\times 2}{8000}}\\\Rightarrow x=0.3836\ m$

The compression of spring = 0.5-0.3836 = 0.1164 m

b) Weight of the brick

$F=5\times 9.81\\\Rightarrow F=49.05\ N$

$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{49.05}{8000}\\\Rightarrow x=0.00613\ m$

The final length of the spring = 0.5-0.00613 = 0.49387 m

4 0

3 years ago

How far will u go if your wheel' diameter 34in?

Lapatulllka [165]

Technically you can go forever on and on, but maybe your question was like how many rotations in a certain distance?

4 0

3 years ago