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trapecia [35]
2 years ago
12

The planet Gallifrey has 2 times the gravitational field strength and 2 times the radius of the Earth.

Physics
1 answer:
Alexeev081 [22]2 years ago
4 0

The mass of the planet Gallifrey is 8 times the mass of the Earth.

  • let the gravitational field of Earth = g
  • let the radius of the Earth = R
  • gravitational field of Gallifrey = 2g
  • radius of Gallifrey = 2R

<h3>What is gravitational potential energy?</h3>
  • This is the work done in moving an object to a certain distance against gravitational field.

The gravitational field strength of the Earth is given as follows;

g = \frac{GM}{r^2} \\\\G = \frac{gr^2}{M}

The gravitational field strength of the Planet Gallifrey is calculated as follows;

g_2 = \frac{GM_2}{r_2^2}

G = \frac{g_2r_2^2}{M_2} \\\\\frac{g_2r_2^2}{M_2} = \frac{gr^2}{M}\\\\M_2gr^2 = Mg_2r_2^2\\\\M_2 = \frac{Mg_2r_2^2}{gr^2} \\\\M_2 = \frac{M \times 2g \times (2r)^2}{gr^2} \\\\M_2 = \frac{M \times 2g \times 4r^2}{gr^2} \\\\M_2 = 8M

Thus, the mass of the planet Gallifrey is 8 times the mass of the Earth.

Learn more about gravitational field strength here:  brainly.com/question/14080810

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39 g H2O contains 1.3 ×1024molecules H2O.

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5 0
3 years ago
A) A mass spectrometer has a velocity selector that allows ions traveling at only one speed to pass with no deflection through s
KengaRu [80]

Answer:

(A). The speed of the ions is 1.2\times10^{6}\ m/s

(B). The radius of curvature of a singly charged lithium ion is 2.0\times10^{6}\ m

Explanation:

Given that,

Electric field = 60000 N/C

Magnetic field = 0.0500 T

(A). We need to calculate the velocity

For no deflection

F_{E}=F_{B}

Eq=Bqv

v = \dfrac{E}{B}

v=\dfrac{60000}{0.0500}

v=1.2\times10^{6}\ m/s

(B). We need to calculate the radius

Using magnetic force balance by centripetal force

Bqv=\dfrac{mv^2}{r}

r=\dfrac{mv^2}{Bqv}

Put the value into the formula

r=\dfrac{1.16\times10^{-26}\times(1.2\times10^{6})^2}{0.0500\times1.6\times10^{-19}}

r=2.0\times10^{6}\ m

Hence, (A). The speed of the ions is 1.2\times10^{6}\ m/s

(B). The radius of curvature of a singly charged lithium ion is 2.0\times10^{6}\ m

6 0
3 years ago
A cat falls from a table of height 1.3 m. What is the impact speed of the cat? ( i want the answer NOT THE FORMULA)
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The impact speed will be
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4 0
3 years ago
Read 2 more answers
Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying
agasfer [191]

Answer:

The charge density in the system is 4.25*10^4C/m

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C

We need to asume here the number of free electrons in a copper conductor, at which is generally of 8.5 *10^{28}m^{-3}

The equation to find the current is

I = VenA

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)

I= 22.11a

Now to find the linear charge density, we know that

I = \frac{Q}{t} \rightarrow Q = It

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

V_d = \frac{l}{t} \rightarrow l = V_d t

The charge density is defined as

\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}

Replacing our values

\lambda = \frac{22.11}{5.20*10{-4}}

\lambda= 4.25*10^4C/m

Therefore the charge density in the system is 4.25*10^4C/m

5 0
3 years ago
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