What’s the awnser to 1 and 2

A charge per unit length given by l(x) = bx, where b = 12 nc/m2, is distributed along the x axis from x = +9.0 cm to x = +16 cm.

Pavel [41]

Hopefully this will help you.

4 0

2 years ago

Which type of radiation can reach the furthest? A. Beta B. Gamma C. Alpha

Mashutka [201]

Answer:

C. Alpha

Explanation:

Gamma radiation, unlike alpha or beta, does not consist of any particles, instead consisting of a photon of energy being emitted from an unstable nucleus. Having no mass or charge, gamma radiation can travel much farther through air than alpha or beta, losing (on average) half its energy for every 500 feet.

7 0

3 years ago

Read 2 more answers

PLEASE HELP 100 POINTS! Please fill in the scale distance from sun and diversion factor, listing off the numbers works

steposvetlana [31]

Solved your another question same like this with scaling to Cm this time we go with metre(m)

Scale factor

1km=10³m
1m=10^{-3}km

Mercury

58000m

Ven us

108000m

Earth

150000m

Mars

228000m

Jupiter

778000m

Saturn

1430000m

Uranus

2870000m

Neptune

4500000m

7 0

2 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is: