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3 years ago

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The mass of Planet X is one-tenth that of the earth, and its diameter is one-half that of the earth. the acceleration due to gra

vity at the surface of Planet X is most nearly...

1 answer:

guapka [62]3 years ago

5 0

Fg=(G*M1*M2)/(r^2)

(1/10)/(1/2^2)=4/10=2/5

Planet X is approx 2/5 of earth’s gravity

9.8m/s^2 * 2/5 = 3.92m/s^2

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Equations E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 are approximations of the magnitude of the electric field of an electric dipole,

tamaranim1 [39]

Answer:

The ratio of $E_{app}$ and $E_{act}$ is 0.9754

Explanation:

Given that,

Distance z = 4.50 d

First equation is

$E_{act}=\dfrac{qd}{2\pi\epsilon_{0}\times z^3}$

$E_{act}=\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}$

Second equation is

$E_{app}=\dfrac{P}{2\pi\epsilon_{0}\times z^3}$

We need to calculate the ratio of $E_{act}$ and $E_{app}$

Using formula

$\dfrac{E_{app}}{E_{act}}=\dfrac{\dfrac{P}{2\pi\epsilon_{0}\times z^3}}{\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}}$

$\dfrac{E_{app}}{E_{act}}=\dfrac{(z^2-\dfrac{d^2}{4})^2}{z^3(z)}$

Put the value into the formula

$\dfrac{E_{app}}{E_{act}}=\dfrac{((4.50d)^2-\dfrac{d^2}{4})^2}{(4.50d)^3\times4.50d}$

$\dfrac{E_{app}}{E_{act}}=0.9754$

Hence, The ratio of $E_{app}$ and $E_{act}$ is 0.9754

8 0

3 years ago

Suppose that a simple pendulum consists of a small 81 g bob at the end of a cord of negligible mass. If the angle θ between the

MariettaO [177]

Answer:

(a) 0.115 m

(b) 2.08 x 10^-5 J

Explanation:

mass of bob, m = 81 g = 0.081 kg

The equation of oscillation is given by

θ = 0.068 Cos {9.2 t + Ф}

Now by comparison

The angular velocity

ω = 9.2 rad/s

(a) $\omega^{2} =\frac{g}{L}$

where, L be the length of the pendulum

$L =\frac{g}{\omega ^{2}}$

$L =\frac{9.8}{9.2 \times 9.2}$

L = 0.115 m

(b) A = L Sinθ

A = 0.115 x Sin 0.068

A = 7.8 x 10^-3 m

Maximum kinetic energy

K = 0.5 x mω²A²

K = 0.5 x 0.081 x 9.2 x 9.2 x 7.8 x 7.8 x 10^-6

K = 2.08 x 10^-5 J

8 0

3 years ago

While walking to work in Boston shortly after sunrise you notice that the water level in the bay is exceptionally low. Based on

Andrew [12]

While walking to work in Boston shortly after sunrise you notice that the water level in the bay is exceptionally low. Based on this very low low-tide and the dark cloudless sky last night, the current phase of the moon is a waxing crescent. Among the 8 phases of the Moon, the waxing crescent follows the new-moon phase.

The phases of Moon happen because of the presence of moon's orbit between the Earth and the Sun. The surface of the Moon reflects sunlight and this light reaches the Earth at different angles depending on the location of the Moon in its orbit. This gives us the impression of a waxing and a waning Moon.

5 0

3 years ago

Can you help me please? I'll mark brainest

DanielleElmas [232]

A. 33 because it’s a reflection the angle is the same. correct me if i’m wrong, but i’m pretty sure it’s 33 because it’s a reflection not a rotation or any sort.

5 0

3 years ago

6. Thorium-234 decays by giving off one particle, and the decay can be represented by the following nuclear equation:

Nady [450]

<span><span>23892</span>U→<span>l<span>23490</span></span>Th<span>+<span>42</span></span>He</span>

Explanation:

Uranium-238 produces thorium-234 by alpha decay.

An α-particle is a helium nucleus. It contains 2 protons and 2 neutrons, for a mass number of 4.

During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.

Thus, uranium-238 decays through α-particle emission to form thorium-234 according to the equation:

<span><span>23892</span>U→<span>l<span>23490</span></span>Th<span>+<span>42</span></span>He</span>

Note that the sum of the subscripts (atomic numbers or charges) is the same on each side of the equation.

Also, the sum of the superscripts (masses) is the same on each side of the equation.

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3 years ago