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3 years ago

Can you change your mass without changing your weight?

1 answer:

6 0

No, because mass is the amount of matter in something and weight is the pull of gravity on a object, for example you would weigh 65 pounds.. and have the mass of 1058..if you were in space your weight would change but your mass will always stay the same

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Which of the following correctly describe electric field lines?

Xelga [282]

-- Electric field lines DO never cross. (A)

-- Electric field lines that are close together DO indicate a stronger electric field. (B)

-- Electric field lines DO not affect the charge that created them. (C)

-- Electric field lines DON'T begin on north poles and end on south poles. North and South "poles" are the way we talk about magnets, not electric charges.

6 0

3 years ago

Read 2 more answers

A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

$P=\rho \times g \times h$

Here

P is the pressure which is given as 68 kPa.
ρ is the density of the oil whose SG is 0.92. It is calculated as

$\rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\$

g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

So the height of column is 7.54m

a-3

By the relation of volume and density

$M=\rho \times V$

Here

ρ is the density of the oil which is 920 kg/m^3
V is the volume of cylinder with diameter 16m calculated as follows

$V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3$

Mass is given as

$M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg$

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

$\Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\$

Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

The original height of column is 5.98m

4 0

3 years ago

if quasars often resemble little blue stars, what was it about them that so surprised astronomers when they were discovered?

Anna11 [10]

Quasar is famous for being an intergalactic object which is billions of years away from the earth yet can still be seen, unlike the other star body, unlike giant galaxies.

Hence, the fact that quasars can be detected from distances where even the biggest and most luminous galaxies cannot be seen means that "they must be intrinsically far more luminous than the brightest galaxies."

This condition, including other related evidence gotten in recent years concerning our galaxy, has shown that quasars are probably the central nuclei of very distant, very active galaxies.

The surprising thing was that quasars and active galaxies have a lot of mass in the center of the very small volume of the space.

Therefore, the surprising thing about quasars was that due to this mass and energy they are 100 times more luminous than Milky Way which means they have high recession velocity and a very large amount of red-shifting.

To learn more about quasars, refer: brainly.com/question/9965257

#SPJ4

8 0

1 year ago

(25 pts) Estimate how much collector area and storage capacity would be required for an active solar hot-water system designed t

Aneli [31]

Answer:

The required total area is 1.48 m²

Explanation:

Given that,

Latitude = 44+° N

New Mexico,

Latitude= 35+° N

Heat capacity = 4200 J/Kg°C

Temperature = 60°C

Let us assume the input temperature 22°C

Estimate volume of water 100 ltr for 4 person.

We need to calculate the heat

Using formula of heat

$H=mc_{p}\Delta T$

$H=mc_{p}(T_{f}-T_{i})$

Put the value into the formula

$H=100\times4200\times(60-22)$

$H=15960\ KJ$ ...(I)

Let solar radiation for 6 hours/day.

We need to calculate the total energy per unit area

Using formula of energy

$E=1000\times6\times3600\ J/m^2$

$E=21600\ KJ/m^2$

Let the efficiency of collector is 50 %

Then, the total energy per unit area will be

$E=21600\times\dfrac{50}{100}$

$E=10800\ KJ/m^2$ ....(II)

We need to calculate the required total area

Using equation (I) and (II)

$A=\dfrac{H}{E}$

Where, H = heat

E = total energy

Put the value into the formula

$A=\dfrac{15960}{10800}$

$A=1.48\ m^2$

Hence, The required total area is 1.48 m²

6 0

3 years ago

In a recent experiment, anti-bacterial hand sanitizer was compared to hand soap. Students washed their hands using the sanitizer

Elden [556K]

B) Both soap brands out perform hand sanitizer in killing bacteria.

Looking at the data, the conclusion is pretty obvious, the two different types of soap resulted in about half the bacteria cultures as either the hand sanitizer, or just using warm water. In fact, the hand sanitizer looks like it's even worse than simply using warm water. So looking at the four available options, the best choice is "B) Both soap brands out perform hand sanitizer in killing bacteria."

7 0

3 years ago

Read 2 more answers