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2 years ago

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What value should be reported for the total mass of three samples of iron weighing 117.0 g, 19.43 g, and 6.1043 g?

1 answer:

AURORKA [14]2 years ago

5 0

The value that should be reported for the total mass of three samples of iron will be 0.143 Kg or 143 g

<h3>What is Mass and Weight ?</h3>

Mass is the quantity of matter. While weight is a gravitational pull on an object. Mass is measured in Kg while weight is measured in Newton.

What value should be reported for the total mass of three samples of iron weighing 117.0 g, 19.43 g, and 6.1043 g?

The total mass will be the sum of the three masses.

The total mass = 117 + 19.43 + 6.1043

The total mass = 142.5343 g

Convert gram to kilogram by dividing the answer by 1000

The total mass = 142.5343/1000

The total mass = 0.1425343 Kg

Therefore, the value that should be reported for the total mass of three samples of iron will be 0.143 Kg or 143 g approximately

Learn more about Mass and Weight here: brainly.com/question/1384116

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If the temperature of an iron sphere is increased A)its density will decrease .B)its volume will decrease. C)its mass will decre

babymother [125]

Answer:

A) Its density will decrease

Explanation:

When an object is heated, its volume increases. This is due to the fact that the particles in the medium vibrate more (if it is a solid) or they move more (if it is a liquid or a gas), therefore they tend to occupy a larger space.

At the same time, the mass of the object does not change, because the mass just represents the amount of matter contained in the object, so it does not increase/decrease at different temperatures.

The density of an object is defined as the ratio between the mass (m) and the volume (V):

$d=\frac{m}{V}$

We said that the mass remains unchanged while the volume increases: since the density is inversely proportional to the volume, this means that the density decreases.

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3 years ago

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Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.

Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

$f=\frac{R}{2}$

For this mirror, R = 20 cm, so its focal length is

$f=\frac{20}{2}=+10 cm$ (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm$

b)

We can solve this part by using the magnification equation:

$M=-\frac{y'}{y}=\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

$y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm$

and the negative sign means that the image is inverted.

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3 years ago

Antonina throws a coin straight up from a height of

vichka [17]

Answer:

$s=vt-\frac{1}{2}gt^2$

Explanation:

We could use the following suvat equation:

$s=vt-\frac{1}{2}gt^2$

where

s is the vertical displacement of the coin

v is its final velocity, when it hits the water

t is the time

g is the acceleration of gravity

Taking upward as positive direction, in this problem we have:

s = -1.2 m

$g=-9.8 m/s^2$

And the coin reaches the water when

t = 1.3 s

Substituting these data, we can find v:

$v=\frac{s}{t}+\frac{1}{2}gt=-\frac{1.2}{1.3}+\frac{1}{2}(-9.8)(1.3)=-7.3 m/s$

where the negative sign means the direction is downward.

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3 years ago

The instrument used to measure the diameter of a thin wire is

Pie

micrometer is used to measure the diameter of a thin wire

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what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three

grin007 [14]

Answer:

Approximately $5.11 \times 10^{-19}\; {\rm J}$ .

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: $R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}$ .

Look up the speed of light in vacuum: $c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}$ .

Look up Planck's constant: $h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}$ .

Apply the Rydberg formula to find the wavelength $\lambda$ (in vacuum) of the photon in question:

$\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}$ .

The frequency of that photon would be:

$\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}$ .

Combine this expression with the Rydberg formula to find the frequency of this photon:

$\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}$ .

Apply the Einstein-Planck equation to find the energy of this photon:

$\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}$ .

(Rounded to three significant figures.)

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