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3 years ago

7

Gravity anything with mass as gravity we know the earth has gravity because you and I are standing on earth and not floating off

into space when we jump we come right back to earth which model best represents the force of gravity pulling us all down to earth?

1 answer:

Paraphin [41]3 years ago

8 0

Answer:

$F=mg$

Explanation:

Close to Earth's surface, the force of gravity that pulls an object towards the ground is

$F=mg$ (2)

where

m is the mass of the object

g is the acceleration due to gravity, which is $9.81 m/s^2$ close to Earth's surface

This is an approximation of the general formula of gravity valid only close to Earth's surface. The more general formula is

$F=G\frac{Mm}{r^2}$ (1)

where

G is the gravitational constant

M is the Earth's mass

m is the object's mass

r is the distance of the object from Earth's center

At the Earth's surface,

r = R (Earth's radius), and by calling the following factor

$g=\frac{GM}{R^2}$

we see that eq.(1) becomes eq.(2).

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How many photons will be required to raise the temperature of 1.8 g of water by 2.5 k ?'?

tatyana61 [14]

Missing part in the text of the problem:
"<span>Water is exposed to infrared radiation of wavelength 3.0×10^−6 m"</span>

First we can calculate the amount of energy needed to raise the temperature of the water, which is given by
$Q=m C_s \Delta T$
where
m=1.8 g is the mass of the water
$C_s = 4.18 J/(g K)$ is the specific heat capacity of the water
$\Delta T=2.5 K$ is the increase in temperature.

Substituting the data, we find
$Q=(1.8 g)(4.18 J/(gK))(2.5 K)=18.8 J=E$

We know that each photon carries an energy of
$E_1 = hf$
where h is the Planck constant and f the frequency of the photon. Using the wavelength, we can find the photon frequency:
$\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{3 \cdot 10^{-6} m}=1 \cdot 10^{14}Hz$

So, the energy of a single photon of this frequency is
$E_1 = hf =(6.6 \cdot 10^{-34} J)(1 \cdot 10^{14} Hz)=6.6 \cdot 10^{-20} J$

and the number of photons needed is the total energy needed divided by the energy of a single photon:
$N= \frac{E}{E_1}= \frac{18.8 J}{6.6 \cdot 10^{-20} J} =2.84 \cdot 10^{20} photons$

4 0

3 years ago

A 55 kg cheerleader uses an oil-filled hydraulic lift to hold four 110 kg football players at a height of 1.0 m. If her piston i

AVprozaik [17]

Answer:

$D = 55.2 cm$

Explanation:

As we know that the total mass of the all four players is given as

$M = 4\times 110$

$M = 440 kg$

diameter of the piston of cheer leader is given as

$d_1 = 16 cm$

are of cross-section is given as

$A_1 = \pi r^2$

$A_1 = \pi(0.08)^2 = 0.02 m^2$

mass of the cheer leader is given as

m = 55 kg

so the pressure due to cheer leader is given as

$P_{in} = \frac{mg}{A_1}$

$P_{in} = \frac{55 \times 9.81}{0.02}$

$P_{in} = 26835 Pa$

Now on the other side pressure must be same

so we have

$\frac{Mg}{A} + \rho gH = P_{in}$

$\frac{440 \times 9.8}{A} + (900)(9.8)(1) = 26835$

$A = 0.24 m^2$

$\pi r^2 = 0.24$

$r = 0.276 m$

so diameter on the other side is given as

$D = 2 r$

$D = 55.2 cm$

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Rainbow can appear at night,they are called Moonbow?

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Yes, they are also known as white rainbows or lunar rainbows.

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4 years ago

Which of these equations will you used to find the final velocity if the initial

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Answer:

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blsea [12.9K]

Answer:

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Explanation:

2 x 76 = 152

bruh its not that hard

6 0

3 years ago

Read 2 more answers