I) Friction depends on the........ of surface.

Someone please help me!!! Please explain and I will mark brainliest!!!

BARSIC [14]

R=10+15+30

55 is the answer to the question

6 0

3 years ago

An electron (charge −e=−1.60×10−19C−e=−1.60×10−19C) is at rest at a distance 1.00×10−10 m 1.00×10−10m from the center of a nucle

shtirl [24]

Answer:

$-9.45\times10^{-16} \text{ J}$

Explanation:

According to Coulomb's law, the force of attraction between two point charges, $q_1$ and $q_2$ , separated by a distance $d$ is given by

$F = k\dfrac{q_1q_2}{d^2}$

$k$ is a constant with a value of $9\times10^9\text{ F/m}$ .

When we substitute the values from the question,

$F = (9\times10^9\text{ F/m})\dfrac{(-1.60\times10^{-19} \text{ C})\times(+1.31\times10^{-17}\text{ C})}{(1.00\times10^{-10}\text{ m})^2} = -1.89\times10^{-6} \text{ N}$

This value is negative because it is in a direction towards the positive charge.

The work done in moving the electron from the nucleus is

$W = F\times r$

$W = (-1.89\times 10^{-6} \text{ N})\times(5.00\times10^{-10}\text{ m}) = -9.45\times10^{-16} \text{ J}$

This is negative because work is done on the electron, not by it.

3 0

3 years ago

The development of new technology requires the evaluation of the overall benefit to cost ratio of the new technology.

Valentin [98]

D) Ride comfort, it won't let me answer unless there are 20 characters so I added this pointless sentence.

4 0

4 years ago

Read 2 more answers

A steel rod with a length of l = 1.55 m and a cross section of A = 4.89 cm2 is held fixed at the end points of the rod. What is

Vedmedyk [2.9K]

Answer:

The size of the force developing inside the steel rod is 32039.28 N

Explanation:

Given;

length of the steel rod, L = 1.55 m

cross sectional area of the steel, A = 4.89 cm²

temperature change, ∆T = 28.0 K

coefficient of linear expansion for steel, α = 1.17 × 10⁻⁵ 1/K

Young modulus of steel, E = 200.0 GPa.

Extension of the steel is given as;

α ∆T L = FL / AE

α ∆T = F/AE

F = AEα ∆T

F = ( 4.89 x 10⁻⁴)(200 x 10⁹)(1.17 × 10⁻⁵)(28.0 K)

F = 32039.28 N

Therefore, the size of the force developing inside the steel rod when its temperature is raised, is 32039.28 N

7 0

3 years ago

A number line goes from negative 5 to positive 5. Point D is at negative 4 and point E is at positive 5. A line is drawn from po

saul85 [17]

The location of the point F that partitions a line segment from D to E ( $\overline{DE}$ ), that goes from negative 4 to positive 5, into a 5:6 ratio is fifteen halves (option 4).