At this rate, how long would it take for two continents 2900 km kilometers apart to collide?

Which of the following is an example of an irregular meter? A. Four beats per measure B. Five beats per measure C. Two beats per

Komok [63]

B. 5 beats per measure.

7 0

3 years ago

Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur

dimulka [17.4K]

Answer:

$-1.7908787542\times 10^{-9}\ C$

$3.4260289211\times 10^{-9}\ C$

$1.7908787542\times 10^{-9}\ C$

$1.6351501669\times 10^{-9}\ C$

Explanation:

r = Radius

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field is given by

$E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C$

The charge is $-1.7908787542\times 10^{-9}\ C$

$Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C$

The charge is $3.4260289211\times 10^{-9}\ C$

The charge inside will have the polarity changed

$q=+1.7908787542\times 10^{-9}\ C$

Outside the charge will be

$3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C$

3 0

3 years ago

Describe the climate and name the climate zone at alaska; portland, oregon; and key west, florida.

Tpy6a [65]

Alaska- Subartic Climate
Portland, Oregon- Marine West Coast Climate
Key West, Florida- Tropical Savannah Climate

3 0

3 years ago

An electron and a second particle both move in circles perpendicular to a uniformmagnetic field. The mass of the second particle

Katarina [22]

Answer:

The change on the second particle is $2.93\times 10^{-16}\ C$ .

Explanation:

The period of revolution of the particle in the magnetic field is given by the formula as follows :

$T=\dfrac{2\pi m}{Bq}$

It is given that the magnetic field is uniform. The mass of the second particle is the same as that of a proton but thecharge of this particle is different from that of a proton.

$m_s=m_p$

If both particles take the same amount of time to go once around their respective circles. So,

$T_e=T_s\\\\\dfrac{2\pi m_e}{Bq_e}=\dfrac{2\pi m_s}{Bq_s}\\\\\dfrac{m_e}{q_e}=\dfrac{m_p}{q_s}\\\\q_s=\dfrac{m_pq_e}{m_e}\\\\q_s=\dfrac{1.67\times 10^{-27}\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}\\\\q_s=2.93\times 10^{-16}\ C$

So, the change on the second particle is $2.93\times 10^{-16}\ C$ .

7 0

3 years ago

In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate th

Radda [10]

Answer:

a) Option D is correct.

The center of mass between the Eartg and the moon is inside the Earth.

Explanation:

Given,

Mass of the moon = (7.35×10²²) kg

Mass of the Earth = (6.00×10²⁴) kg

Mass of the Sun = (2.00×10³⁰) kg

Distance between the Earth and the moon = (3.80×10⁵) km

Distance between the Earth and the Sun = (1.50×10⁸) km

With the assumption that all.of the bodies being considered are on the same straight line on the x-axis,

Note that Centre of mass is given as

C.M = (Σmx)/(Σm)

For the Earth-moon system, let the earth be x=0, then the moon is at x = (3.80 × 10 5) km away.

C.M = (Σmx)/(Σm)

Σmx = (6.00×10²⁴)) × (0) + (7.35×10²²) × (3.80×10⁵) = (2.793 × 10²⁸) kg.km

Σm = (6.00×10²⁴) + (7.35×10²²) = (6.0735 × 10²⁴) kg

CM = (2.793 × 10²⁸) ÷ (6.0735 × 10²⁴)

CM = (4.60 × 10³) km = 4600 km

This means the centre of mass is 4600 km from the Earth.

The Earth's radius = 6378 km

Hence, the centre if mass is inside the Earth.

Hope this Helps!!!

8 0

3 years ago