How much of a kilogram is a gram?

What are some examples of irreversible processes in nature?

castortr0y [4]

Answer:

Burning. When you burn something, it turns into ash you can't make that thing turn into what it was before.

8 0

3 years ago

A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t

Oduvanchick [21]

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

$T = \frac{2u Sin\theta }{g}$

Final horizontal component of velocity, vx = ux = u Cos 30

Let vy is teh final vertical component of velocity.

Use first equation of motion

vy = uy - gT

$v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}$

$v_{y}=u Sin 30 - 2u Sin 30$

vy = - u Sin 30

The magnitude of final velocity is given by

$v = \sqrt{v_{x}^{2}+v_{y}^{2}}$

$v = \sqrt{\left (uCos 30 \right )^{2}+\left (uSin 30 \right )^{2}}$

v = u

Thus, the velocity is same as it just reaches the ground.

6 0

3 years ago

A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o

Crank

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

$\delta = \frac{PL}{AE}$

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

$P = 68*10^3 N$

E = 200GPa

$A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2$

Replacing we have,

$\delta = \frac{PL}{AE}$

$\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}$

$\delta = 0.001932m$

$\delta = 1.93mm$

Therefore the change in length is 1.93mm

7 0

2 years ago

What is true about pluto?

Lapatulllka [165]

Answer:

Explanation:

Pluto is a dwarf planet that lies in the Kuiper Belt, an area full of icy bodies and other dwarf planets out past Neptune. Pluto is very small, only about half the width of the United States and its biggest moon Charon is about half the size of Pluto

8 0

2 years ago

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In which of these cases do we have enough information to say that the atom is electrically neutral?

Luba_88 [7]

<span>So we want to know when do we have enough information about the atom that we can say that it is electrically neutral. So the information we need to know is the number of protons, whic are positively charged particles and the number of electrons which are negatively charged particles. Number of protons has to be equal to number of electrons for the atom to have neutral charge. Neutrons are electrically neutral so their number doesn't change the charge of the atom. So the correct answer is B. </span>

5 0

3 years ago

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