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3 years ago

Please help me - i beg u

Physics

1 answer:

oksano4ka [1.4K]3 years ago

8 0

Answer:

516526.863 m

Explanation:

From the question given above, the following data were obtained:

Acceleration due to gravity (g) = 2.24 m/s²

Mass (M) = 8.96×10²¹ Kg

Gravitational constant (G) = 6.67×10¯¹¹Nm²/Kg²

Radius (r) =?

The radius of the planet can be obtained as follow:

g = GM/r²

2.24 = 6.67×10¯¹¹ × 8.96×10²¹ / r²

2.24 = 5.97632×10¹¹ / r²

Cross multiply

2.24 × r² = 5.97632×10¹¹

Divide both side by 2.24

r² = 5.97632×10¹¹ / 2.24

r² = 2.668×10¹¹

Take the square root of both side

r = √2.668×10¹¹

r = 516526.863 m

Thus, the radius of the planet is 516526.863 m

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A cube that is 20 nanometer on an edge contains 399,500 silicon atoms, and each silicon atom has 14 electrons and 14 protons. In

Sergeu [11.5K]

Answer:

Total 3 holes are available for conduction of current at 300K.

Explanation:

In order to develop a semiconductor, two type of impurities can be added as given below:

N-type Impurities: Pentavalent impurities e.g. Phosphorous, Arsenic are added to have an additional electron in the structure. Thus a pentavalent impurity creates 1 additional electron.
P-type Impurities: Trivalent impurities e.g. Boron, Aluminium are added to have a positive "hole" in the structure. Thus a trivalent impurity creates 1 hole.

Now for estimation of extra electrons in the impured structure is as

$N_{electrons-free}=n_{pentavalent \, atoms}\\N_{electrons-free}=4\\$

Now for estimation of "holes" in the impured structure is as

$N_{holes}=n_{trivalent \, atoms}\\N_{holes}=7\\$

Now when the free electrons and "holes" are available in the structure ,the "holes" will be filled by the free electrons therefore

$N_{holes-net}=N_{holes}-N_{electrons-free}\\N_{holes-net}=7-4\\N_{holes-net}=3$

So total 3 "holes" are available for conduction of current at 300K.

6 0

3 years ago

How much work do you do on a 15 N book in lifting it straight up for a distance of<br> 0.40 meters?

astra-53 [7]

Answer:

Work done, W = 6 J

Explanation:

It is given that,

Force of gravity acting on the book, weight of the book is 15 N

We need to find the work done in lifting the book straight up for a distance of 0.4 meters.

The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

$W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J$

So, the magnitude of work done in lifting the book is 6 joules.

7 0

3 years ago

A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

3 years ago

A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb

NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

$\sum \tau = F*d$

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

$F*(27-6)= 6*600$

$F = \frac{6*600}{21}$

$F= 171.42 lb$

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 0

3 years ago

A cat, walking along the window ledge of a new york apartment, knocks off a flower pot, which falls to the street 280 feet below

Harlamova29_29 [7]

H = 280 ft, the height of the flower pot.
g = 32 ft/s²

Neglect air resistance.
Note that 1 ft/s = 15/22 mi/h

The initial vertical velocity is zero.
Let v = the velocity with which the flower pot hits the ground.
Then
v² = 2gh
= 2*(32 ft/s²)*(280 ft)
= 17920 (ft/s)²
v = 133.866 ft/s

Also,
v = (133.866 ft/s)*(15/22 (mi/h)/(ft/s)) = 91.272 mi/h

Answer: 133.9 ft/s or 91.3 mi/h

5 0

3 years ago