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3 years ago

7

Determine the force of gravitational attraction between the Earth and the moon. Their masses are 5.98 x 1024 kg and 7.26 x 1022

kg, respectively

1 answer:

monitta3 years ago

4 0

Answer:

$F=1.95\times 10^{20}\ N$

Explanation:

Mass of Earth, $m_e=5.98 \times 10^{24}\ kg$

Mass of Moon, $m_m=7.26\times 10^{22}\ kg$

The distance between Earth and the Moon is, $d=384,400\ km$

We need to find the force of gravitational attraction between the Earth and the moon. The force of gravity is given by :

$F=G\dfrac{m_em_m}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{5.98 \times 10^{24}\times 7.26\times 10^{22}}{(384400 \times 10^3)^2}\\\\F=1.95\times 10^{20}\ N$

So, the required force is $1.95\times 10^{20}\ N$ .

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Describe the electron transfers that occur in the formation of calcium fluoride from elemental calcium and elemental fluorine.

boyakko [2]

Answer:

Check explanation

Explanation:

In the formation of calcium fluoride we take calcium and fluorine.

in elemental form calcium exist in solid form and fluorine in gaseous form.

formation of compound takes place to complete their octet, in case of calcium need to remove two electron and need to add one elecron in fluorine to complete their octet so two electron will ransferred from calcium to two fluorine atom.

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3 years ago

A metaphysical poet is a writer whose

EastWind [94]

Not really sure but...

<span>A metaphysical poet is a writer whose </span>focus is on universal human experiences.
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8 0

3 years ago

a fan acquires a speed of 180 rpm in 4s, starting from rest. calculate the speed of the fan at the end of the 5th second startin

KengaRu [80]

Answer:

225 rpm

Explanation:

The angular acceleration of the fan is given by:

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_f$ is the final angular speed

$\omega_i$ is the initial angular speed

$\Delta t$ is the time interval

For the fan in this problem,

$\omega_i = 0\\\omega_f = 180 rpm\\\Delta t=4 s$

Substituting,

$\alpha = \frac{180-0}{4}=45 rpm/s$

Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:

$\omega' = \omega_i + \alpha t$

where

$\omega_i = 0\\\alpha = 45 rpm/s\\t = 5 s$

Substituting,

$\omega' = 0 + (45)(5)=225 rpm$

7 0

3 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

3 years ago

A model rocket has an acceleration of 12 m/s2. A net force of 18 N is acting on the rocket. What is the mass of the rocket? its

larisa86 [58]

Answer:

C.) 1.5 kg

Explanation:

Start with the equation:

$F_n_e_t=ma$

Plug in what you know, and solve:

$18=m(12)\\m=1.5kg$

Find matching soluation:

C.) 1.5 kg

5 0

2 years ago