All categories

4 years ago

A compound is made up of two or more of what?

Physics

2 answers:

dlinn [17]4 years ago

7 0

A compound is made up of two or more elements~
Hope this helps~

jolli1 [7]4 years ago

5 0

A compound is made up of two or more of atoms that are bonded together with each other via a chemical bonding. It is possible to break the compounds into separate smaller components by the process of chemical reaction. One very simple example of compound is water as it is made of two atoms of hydrogen and one atom of oxygen. Whenever atoms are joined together to create a compound, the atoms definitely loose their individual atomic properties. Other very commonly found compound is carbon dioxide. It is made up of one atom of carbon and two atoms of oxygen.

You might be interested in

What energy transformations take place when you eat a taco and then go for a jog

sesenic [268]

Answer:

Chemical to kinetic and thermal.

Explanation:

You would eat the food (chemical) than you would jog and move around (kinetic). While running your body would also give off heat (thermal).Than your body would sweat to cool itself down.

6 0

3 years ago

Given: G = 6.672 × 10−11 N · m2 /kg2 Io, a satellite of Jupiter, has an orbital period of 1.24 days and an orbital radius of 4.1

Dahasolnce [82]

Answer:

Mass of Jupiter = 4.173×10^15kg

Explanation:

Using Kepler's 3rd law, it states that the orbital period T is related to the distance,r as:

T^2 = GM/4 pi × r^3

Where G = universal gravitational constant

r = radius

M = masd of jupiter

Rearranging the formular to make M the subject of formular

T^2 × 4 pi = G M × r^3

(T^2 × 4 pi) / (G× r^3) = M

(1.24^2 × 4 × 3.142) /(6.672×10^-11)(4.11×10^8)^3

M = 19.32 /6.672×10^-11)(4.11×10^8)^3

M = 19.32 / 4.63 ×10^15

M = 4.173×10^15kg

6 0

3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

3 years ago

Reactants → products

Nat2105 [25]

The answer is A

Explanation: the conservation of matter means that the mass stays the same

4 0

3 years ago

Read 2 more answers