The attraction between two oppositely charged Atoms or groups of Atoms is which type of bond

Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter

kaheart [24]

The moon clock is A) (9.8/1.6)h compared to 1 hour on Earth

Explanation:

The period of a simple pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

In this problem, we want to compare the period of the pendulum on Earth with its period on the Moon. The period of the pendulum on Earth is

$T_e=2\pi \sqrt{\frac{L}{g_e}}$

where

$g_e = 9.8 m/s^2$ is the acceleration of gravity on Earth

The period of the pendulum on the Moon is

$T_m=2\pi \sqrt{\frac{L}{g_m}}$

where

$g_m = 1.6 m/s^2$ is the acceleration of gravity on the Moon

Calculating the ratio of the period on the Moon to the period on the Earth, we find

$\frac{T_m}{T_e}=\frac{g_e}{g_m}=\frac{9.8}{1.6}$

Therefore, for every hour interval on Earth, the Moon clock will display a time of

A) (9.8/1.6)h

#LearnwithBrainly

6 0

3 years ago

A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,

Rus_ich [418]

Answer:

$\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\$

Explanation:

Given that

$Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\$

As both charges are negative so there exist force of repulsion in direction as shown in figure.

$F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N$

Angle at which force F12 is acting is

$\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o$

$F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\$

$\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=$

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

$F_{21}=-5.63\times 10^{-11}N$

$\vec{F}_{21}=$

7 0

3 years ago

Light is an electromagnetic wave and travels at a speed of 3.00 × 108 m/s. The human eye is most sensitive to yellow-green light

Nimfa-mama [501]

Answer:

$f = 5.95 \times 10^{14} Hz$

Explanation:

As we know that the frequency of the wave is given as

$f = \frac{c}{\lambda}$

here we know that

$c = 3 \times 10^8 m/s$

also we know that

$\lambda = 5.04 \times 10^{-7} m$

now we have

$f = \frac{3 \times 10^8}{5.04 \times 10^{-7}}$

$f = 5.95 \times 10^{14} Hz$

6 0

3 years ago

A long, straight, cylindrical wire of radius R carries a current uniformly distributed over its cross section.

Mice21 [21]

Answer:

Explanation:

We shall solve this question with the help of Ampere's circuital law.

Ampere's ,law

∫ B dl = μ₀ I , B is magnetic field at distance x from the axis within wire

we shall find magnetic field at distance x . current enclosed in the area of circle of radius x

= I x π x² / π R²

= I x² / R²

B x 2π x = μ₀ x current enclosed

B x 2π x = μ₀ x I x² / R²

B = μ₀ I x / 2π R²

Maximum magnetic B₀ field will be when x = R

B₀ = μ₀I / 2π R

Given

B = B₀ / 3

μ₀ I x / 2π R² = μ₀I / 2π R x 3

x = R / 3

b ) The largest value of magnetic field is on the surface of wire

B₀ = μ₀I / 2π R

At distance x outside , let magnetic field be B

Applying Ampere's circuital law

∫ B dl = μ₀ I

B x 2π x = μ₀ I

B = μ₀ I / 2π x

Given B = B₀ / 3

μ₀ I / 2π x = μ₀I / 2π R x 3

x = 3R .

3 0

3 years ago

Given that the atomic weight of hydrogen is approximately 1 gram per mole, use the method above to estimate the average molecula

denpristay [2]

Answer:

2

Explanation:

6 0

3 years ago