What is so remarkable about the migratory monarch butterflies?

A bullet with a mass m b = 11.5 g is fired into a block of wood at velocity v b = 249 m/s. The block is attached to a spring tha

Ostrovityanka [42]

Answer:

0.358Kg

Explanation:

The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system

0.5Ke^2 = 0.5Mv^2

0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2

Using conservation of momentum between the bullet and the block

0.0115(265) = (M + 0.0115)v

3.0475 = (M + 0.0115)v

v = 3.0475/(M + 0.0115)

plugging into Energy equation

12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2

12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )

12.56 = 0.5 × 9.2872/ M + 0.0115

12.56 = 4.6436/ M + 0.0115

12.56 ( M + 0.0115 ) = 4.6436

12.56M + 0.1444 = 4.6436

12.56M = 4.6436 - 0.1444

12.56 M = 4.4992

M = 4.4992÷12.56

M = 0.358 Kg

4 0

3 years ago

A brown bear runs at a speed of 9.0\,\dfrac{\text m}{\text s}9.0 s m 9, point, 0, start fraction, start text, m, end text, div

KengaRu [80]

Explanation:

Below is an attachment containing the solution.

7 0

3 years ago

Read 2 more answers

If you lived on the Moon, would you see Earth go through phases? If so, would the sequence of phases be the same as those of the

marissa [1.9K]

Answer:

yes; yes

Explanation:

Phases of the moon refers to the shapes of the moon due to the lit part of it visible from the Earth. On a new moon day, the moon comes between the sun and the earth such that the lit portion is not visible from the Earth. On a full moon day, the earth comes between the sun and the moon and the whole lit part is visible.

When one would view the earth from the moon, the earth would also be visible as going through the phases. The order would be reversed. Understand this with the following example, On a new moon day, the Earth would be visible completely lit from the moon. So it will be full Earth day on the moon. On a full moon day, the lit side of the Earth would be completely away and hence, from the moon, new earth would be there.

8 0

3 years ago

A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide w

barxatty [35]

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

$Moles = \frac{Mass\ taken}{Molar\ mass}$