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Serga [27]
3 years ago
8

The mass of the Earth is approximately 5.98 × 1024 kilograms and the mass of the Moon is approximately 7.35 × 1022 kilograms. Ap

proximately how much greater is the mass of the Earth than the mass of the Moon?
Physics
1 answer:
soldier1979 [14.2K]3 years ago
8 0

It is a question that leads us to know the proportion of one element to another. We have initially that the mass of the earth is of

m_E = 5.98*10^{24}kg

While the mass of the Moon is of

m_M = 7.35*10^{22}kg

The ratio  between the two is given by the function

\frac{m_E}{m_M} = \frac{5.98*10^{24}kg}{7.35*10^{22}kg}

\frac{m_E}{m_M} = 81.36

This means that the mass of the earth is 81.36 times larger than that of the Moon.

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Which statements describe the results of the effect of angle of insolation on absorption experiment? Check all that apply.
lilavasa [31]

Answer:

At 6 minutes, the soil temperature at a 45° angle of insolation is higher than at 0°.

At 15 minutes, the soil temperature at a 90° angle of insolation is higher than at 45°

Explanation:

5 0
3 years ago
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A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

\omega=-\alpha t

\alpha=-\dfrac{\omega}{t}

\alpha=-\dfrac{50.0}{20.0}

\alpha=-2.5\ rad/s^2

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

\theta=\omega_{0}t+\dfrac{1}{2}\alpha t

Put the value into the formula

\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2

\theta=500\ rad

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

\vec{\tau}=\vec{r}\times\vec{f}

\tau=r\times f\sin\theta

Put the value into the formula

\tau=2.5\times4\times 9.8\sin60

\tau=84.87\ N-m

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0
3 years ago
The movement of protons through atp synthase occurs from the
Lera25 [3.4K]

Answer:

cytoplasm and channel gates

Explanation:

The movement originates from the cytoplasm. This is the fluid medium through which ions are shuttle from one place to another. However, though simple as it might appear to be, the movement requires carrier proteins. The are proteins that facilitate in the movement of the ions. These proteins have specially controlled gates called channel proteins. These are regulated proteins that open and close based on hydrogen ion concentration. These proteins are able to facilitate the movement of ATP molecules.

5 0
3 years ago
An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of
Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, E=390\ keV=390\times 10^3\ eV

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2meE}}

\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}

\lambda=1.96\times 10^{-12}\ m

(b) According to Bragg's law,

n\lambda=2d\ sin\theta

n = 1

For nickel, d=0.092\times 10^{-9}\ m

\theta=sin^{-1}(\dfrac{\lambda}{2d})

\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})

\theta=0.010^{\circ}

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0
3 years ago
A 150kg person stands on a compression spring with spring constant 10, 000 N/m and norminal.Length of 0.50.What is the togal len
Eva8 [605]

Answer:

Total length of spring 0.647 m

Explanation:

We have given mass of the person m = 150 kg

Acceleration due to gravity g=9.8m/sec^2

Spring constant k = 10000 N/m

Nominal length of spring = 0.50

According to hook's law

mg=kx

150\times 9.8=10000\times x

x = 0.147 m

So total length of spring = 0.50+0.147 = 0.647 m

4 0
3 years ago
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