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2 years ago

What were the three factors that affected density

Physics

1 answer:

Stolb23 [73]2 years ago

7 0

Answer:

Pressure, temperature and humidity all affect air density. And you can think of air density as the mass of air molecules in a given volume.

Explanation:

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An apple dropped from the branch of a tree hits the ground in 0.5 s. If the acceleration of the apple during its motion is 10 ms

Ipatiy [6.2K]

Given that,

Time = 0.5 s

Acceleration = 10 m/s²

(I). We need to calculate the speed of apple

Using equation of motion

$v=u+at$

Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

$v=0+10\times0.5$

$v=5\ m/s$

(III). We need to calculate the height of the branch of the tree from the ground

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times10\times(0.5)^2$

$s=1.25\ m$

(II). We need to calculate the average velocity during 0.5 sec

Using formula of average velocity

$v_{avg}=\dfrac{\Delta x}{\Delta t}$

$v_{avg}=\dfrac{x_{f}-x_{i}}{t_{f}-t_{0}}$

Where, $x_{f}$ = final position

$x_{i}$ = initial position

Put the value into the formula

$v_{avg}=\dfrac{1.25+0}{0.5}$

$v_{avg}=2.5\ m/s$

Hence, (I). The speed of apple is 5 m/s.

(II). The average velocity during 0.5 sec is 2.5 m/s

(III). The height of the branch of the tree from the ground is 1.25 m.

7 0

2 years ago

Why does the moon orbit the earth as the earth orbits the sun?<br><br>will give brainliest

AysviL [449]

As the Earth rotates, it also moves, or revolves, around the Sun. ... As the Earth orbits the Sun, the Moon orbits the Earth. The Moon's orbit lasts 27 1/2 days, but because the Earth keeps moving, it takes the Moon two extra days, 29 1/2, to come back to the same place in our sky.

6 0

2 years ago

If the Sun were the size of a small exercise ball (about 0.5 meters (m) in

suter [353]

Answer:

the size of a pea

Explanation:

4 0

2 years ago

A rain cloud contains 5.32 × 107 kg of water vapor. The acceleration of gravity is 9.81 m/s 2 . How long would it take for a 2.0

Alja [10]

Answer:

20.85 years

Explanation:

2.61 km = 2610 m

2.07 kW = 2070 W

First we need to calculate the potential energy required to take m = $5.32 * 10^7$ kg of rain cloud to an altitude of 2610 m is

$E = mgh = 5.32 * 10^7 * 9.81 * 2610 = 1.4*10^{12}J$

With a P = 2070 W power pump, this can be done within a time frame of

$t = E/P = \frac{1.4*10^{12}}{2070} = 658037739 s$

or 658037739/(60*60) = 182788 hours or 182788 / 24 = 7616 days or 7616 / 365.25 = 20.85 years

4 0

3 years ago

Some element can be either solid or liquid. At the melting point, the liquid has 8 × 10-22 J more enthalpy per atom than the sol

OlgaM077 [116]

Answer:

481.76 J/mol

133.33 K

Explanation:

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

Change in enthalpy is given by

$\Delta H=8\times 10^{-22}\times 6.022\times 10^{23}\\\Rightarrow \Delta H=481.76\ J/mol$

Entropy is given by

$\Delta S=6\times 10^{-24}\times 6.022\times 10^{23}\\\Rightarrow \Delta S=3.6132\ J/mol K$

Latent heat of fusion is given by

$L_f=\Delta H\\\Rightarrow L_f=481.76\ J/mol$

The latent heat of fusion is 481.76 J/mol

Melting point is given by

$T_m=\dfrac{L_f}{\Delta S}\\\Rightarrow T_m=\dfrac{8\times 10^{-22}\times 6.022\times 10^{23}}{6\times 10^{-24}\times 6.022\times 10^{23}}\\\Rightarrow T_f=133.33\ K$

Melting occurs at 133.33 K

3 0

2 years ago