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3 years ago

how can there be so many different substances in the world if there are only a few elements that are common?

1 answer:

3 0

-- If there are only 10 elements in the universe that can make compound molecules, and a compound molecule can be formed by combining 1, 2, 3, or 4 different elements, then that's already the possibility of at least 400 different molecules.

-- There are many more than 10 elements that can combine to form compound molecules.

-- Every single "organic" molecule, of which there are thousands, is the combination of carbon with other elements.

-- Most all of the substances that can be distilled out of oil, including the paraffin waxes, the alcohols, gasoline, kerosene, butane, propane, octane, and natural gas, are made of just carbon, hydrogen, and oxygen, only with different numbers of each one.

-- Plastics, drugs, rubber, and DNA are examples of molecules that are made of hundreds of atoms.

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A block of mass m moving due east at speed v collides with and sticks to a block of mass 2m that is moving at the same speed vv

lapo4ka [179]

Answer:

30.36°

Explanation:

By using linear momentum; linear momentum can be expressed by the relation:

$mv_xi + mv_yj$

where ;

m= mass

$v_x$ = velocity of components in the x direction

$v_y$ = velocity of components in the y direction

If we consider the east as the positive x and north as positive y which is synonymous to what we usually have on a graph.

Then;

Initial momentum = $mvi + 2mvcos 45i + 2mvsin45 j$

= $(mv+2mvcos45)i + (2mvsin45)j$

However, the masses stick together after collision and move with a common velocity: $V_xi +V_yj$

∴ Final momentum = $3mv (V_xi +V_yj)$

= $3mV_xi + 3mV_yj$

From the foregoing ;

initial momentum = final momentum

$3mV_xi + 3mV_yj = (mv+2mvcos45)i+(2mvsin45j)$

So;

$3mV_x = mv + 2mv cos 45 \\\\3mV_y = 2mV sin 45$

$V_x = \frac{mv+2mvcos45 }{3m}\\\\V_x = \frac{v+2vcos45}{3}$

$V_y = \frac{2mvsin45}{3m} \\\\V_y = \frac{2vsin45}{3}$

Finally;

The required angle θ = $tan^{-1} = \frac{V_y}{V_x}$

θ = $tan^{-1} = \frac{\frac{2vsin45}{3}}{\frac{v+2v.cos45}{3}}$

θ = $tan^{-1} = \frac{2sin 45}{1+2cos45}\\\\$

θ = 30.36°

7 0

3 years ago

A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on

kari74 [83]

Answer:

$9.16rad/s^2$

Explanation:

We are given that

Mass, $m_1=3.3 kg$

Radius,r=0.8 m

$m_2=4.9 kg$

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

$4.9g-T=4.9a$

$Tr=I\alpha$

Where

$\alpha=\frac{a}{r}$

$Tr=\frac{1}{2}m_1ra$

$T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a$

Substitute the value

$4.9g-\frac{1}{2}(3.3a)=4.9a$

$4.9\times 9.8=4.9a+\frac{3.3a}{2}$

Where $g=9.8 m/s^2$

$48.02=a(4.9+1.65)=6.55a$

$a=\frac{48.02}{6.55}=7.33m/s^2$

Angular acceleration, $\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2$

7 0

3 years ago

Name one thing that causes domains of a magnets atoms to lose alignment

kogti [31]

Heat cause domains of magnets atoms to lose alignment

7 0

3 years ago

Read 2 more answers

A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler

Romashka [77]

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

$W=K_f - K_i$

Since the cart was initially at rest, $K_i = 0$ , so

$W=K_f = \frac{1}{2}mv^2$ (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, $F_g = 99.5 N$ :

$m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg$

So solving eq.(1) for v, we find the final speed of the cart:

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s$

2) $2.51\cdot 10^7 J$

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

$F=4.28 \cdot 10^5 N$

$d=586 m$

So the work done is

$W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J$

3) $2.51\cdot 10^7 J$

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

$W=\Delta K = K_f - K_i$

where

W is the work done

$\Delta K$ is the change in kinetic energy

Therefore, the change in kinetic energy is

$\Delta K = W = 2.51\cdot 10^7 J$

4) 37.2 m/s

According to the work-energy theorem,

$W=\Delta K = K_f - K_i$

where

$K_f$ is the final kinetic energy of the train

$K_i = 0$ is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

$W=K_f = \frac{1}{2}mv^2$

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s$

7 0

4 years ago

the earth's moon has a gravitational field strength of about 1.6 n/kg near its surface. the moon has a mass of 7.35x10^22 kg. wh

Andrew [12]

Answer:

1750km

Explanation:

8 0

3 years ago