Ask question

Ask question

All categories

11111nata11111 [884]

2 years ago

12

The density of a sample can be obtained by dividing its by its _____.

1 answer:

mixer [17]2 years ago

3 0

<span>Density can be calculated and found by dividing the sample's mass by its volume. D=m/v</span>

You might be interested in

How are electricity and magnets connected?

Stella [2.4K]

Answer:

Magnets are employed to generate electricity.

Explanation:

Magnets' characteristics are employed to generate electricity. Electrons are pulled and pushed by moving magnetic fields. When you move a magnet around a coil of wire, or a coil of wire around a magnet, the electrons in the wire are pushed out and an electrical current is created.

3 0

2 years ago

Read 2 more answers

A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 781 K, and r

andreev551 [17]

Answer:2.517 J/K

Explanation:

Given

Reservoir 1 Temperature $T_1=781 K$

Reservoir 2 Temperature $T_2=335 K$

Let Q is the amount of heat Flows i.e. $Q=1477 J$

thus change in Entropy is given by $\frac{\sum Q}{T}$

$\Delta S=\frac{\sum Q}{T}=-\frac{Q}{T_1}+\frac{Q}{T_2}$

$\Delta S=\frac{\sum Q}{T}=-\frac{1477}{781}+\frac{1477}{335}$

$\Delta S=\frac{\sum Q}{T}=-1.891+4.4089$

$\Delta S=\frac{\sum Q}{T}=2.517 J/K$

6 0

2 years ago

A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c

viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = $71\times 10^{- 9} C$

Q' = + 42 nC = $42\times 10^{- 9} C$

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

$\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E} = 708.03 N/C$

Now,

Electric field at the mid-point due to charge Q' is given by:

$\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E'} = 418.84 N/C$

Now,

The net Electric field is given by:

$\vec{E_{net}} = \vec{E} - \vec{E'}$

$\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C$

5 0

2 years ago

Ten narrow slits are equally spaced 3.00 mm apart and illuminated with indigo light of wavelength 444 nm. The width of bright fr

ankoles [38]

Answer:ask yo mama

Explanation:she finished school

8 0

2 years ago

Can u expert Anwser this I Would Like to Help U but Please Anwser This

Roman55 [17]

Streamline
Effect
Drag
Parachutes
Surface area
Friction
Air
Water

3 0

2 years ago