Ask question

Ask question

All categories

2 years ago

14

Which measurement tool is used to find the volume of a small piece of granite? Select one: a. thermometer b. metric ruler c. gra

duated cylinder d. triple beam balance

1 answer:

Natalka [10]2 years ago

3 0

Answer:

Explanation:

c. graduated cylinder

You might be interested in

A 23.0 kg iron weightlifting plate has a volume of 2920 cm3 . what is the density of the iron plate in g/cm3?

yanalaym [24]

The first thing you should know for this case is that density is defined as the quotient between mass and volume:
D = M / V
In addition, you should keep in mind the following conversion:
1Kg = 1000g
Substituting the values we have:
D = (23.0 * 1000) / (2920) = 7.88 g / cm ^ 3
answer
the density of the iron plate is 7.88 g / cm ^ 3

8 0

2 years ago

What would be the final temperature if you mixed a liter of 40C water with 2 liters of 20C water?

riadik2000 [5.3K]

Answer:

I found

33 ∘ C

Explanation:

i hope this helps :)

5 0

2 years ago

Power is measured in unit of Joules per second or

Rasek [7]

D watts yw .yea yww

5 0

2 years ago

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

Why do we use microwaves to communicate berween earth and satellites

nlexa [21]

It’s frequency is high and microwaves can pass through the atmosphere of the Earth.

7 0

3 years ago

Read 2 more answers