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3 years ago

Which is a property that mendeleev predicted for gallium

Physics

2 answers:

seropon [69]3 years ago

8 0

A; Metal This is because of maths

earnstyle [38]3 years ago

5 0

A. He predicted it would be a metal because it was his idea to arrange the elements according to their chemical and physical properties.

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Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3

sweet [91]

Given that,

Rank in order from largest to smallest the magnitude of the electric field at block dot.

Electric field :

Electric field is proportional to the charge divided by square of distance.

In mathematically,

$E\propto\dfrac{q}{r^2}$

Where, q = charge

r = distance

If the charge is greater then electric field will be greater.

If the distance is greater then electric field will be smaller.

We need to find the electric field at black dot

According to figure,

(I). The electric field at black dot due to positive charge point q to left. the distance is r.

The electric field will be

$E=\dfrac{kq}{r^2}$

The electric field will be largest.

(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.

Then, the electric field will be

$E=\dfrac{k2q}{(2r)^2}$

$E=\dfrac{kq}{2r^2}$

The electric field will be smallest.

(III). The electric field at black dot due to positive charge point 2q to left. The distance is r.

Then, the electric field will be

$E=\dfrac{k2q}{(r)^2}$

The electric field will be very largest.

(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.

Then, the electric field will be

$E=\dfrac{kq}{(2r)^2}$

$E=\dfrac{kq}{4r^2}$

The electric field will be very smallest.

So, The electric field from largest to smallest will be

$E_{3}>E_{1}>E_{2}>E_{4}$

Hence, The ranking will be 3, 1, 2, 4.

(D) is correct option.

4 0

3 years ago

A simulation that shows how the space shuttle docks with another craft would be which type of model?

My name is Ann [436]

That would be a. Computer

8 0

2 years ago

A satellite orbits the moon for 3685 s. It has an orbital radius of 2008177 m. What is the acceleration of the satellite?

Anna11 [10]

A satellite orbits the moon for 3685 s. It has an orbital radius of 2008177 m. then the acceleration of the satellite would be 5.832 m/s²

<h3 /><h3>What is a uniform circular motion?</h3>

It is defined as motion when the object is moving in a circle with a constant speed and its velocity is changing with every moment because of the change of direction but the speed of the object is constant in a uniform circular motion

The acceleration during the uniform circular motion is given by the formula

a = v²/r

where a is the acceleration

v is the velocity of the object

r is the radius of the orbit

As given in the problem a satellite orbits the moon for 3685 s. It has an orbital radius of 2008177 m.

velocity of the satellite

v = 2πr/t

v= (2×3.14×2008177)/3685

v = 3422.34 m/s

acceleration of the satellite

a = v²/r

a = 3422.34²/2008177

a = 5.832 m/s²

Thus, the acceleration of the satellite would be 5.832 m/s²

Learn more about uniform circular motion from here

brainly.com/question/2285236

#SPJ1

7 0

2 years ago

A car starts with an initial speed of 12 m/s and accelerates at 3 m/s/s for 5 seconds

Nostrana [21]

Explanation:

Vi = 12 m/s

a = 3 m/s^2

t = 2 s

Vf = Vi + a × t = 12 + 3 ×2 = 18 m/s

3 0

3 years ago

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

r = distance of corner of the square from it center

$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

$V = 0.815 V$

now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

$r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}$

$r_2 = 11.2 nm$

now potential is given as

$V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}$

$V = -0.576 - 0.257 = -0.833 V$

now final potential energy is given as

$U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J$

Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

8 0

3 years ago