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Vlad1618 [11]
3 years ago
13

Why does the dime transfer less momentum?

Chemistry
1 answer:
Pepsi [2]3 years ago
6 0

Answer:

This because the mass of the dime is very less = 2.268 grams only

Explanation:

Momentum :<em><u> It is product of mass and velocity of the objec</u></em>t.The affect of momentum can be observed by the force exerted <em><u>. Greater the change in momentum, greater the force exerted by the object</u></em>.

The formula of momentum(p) is :

p=m\times v

Momentum depends upon :

1. Mass = Greater the mass of the object , greater the momentum . For example : you get more hurt by a heavy stone then a light ball when both hit you with the same velocity.

2. Velocity = Momentum of the object with more velocity is more. This application is used by the cricketers. While catching the ball , the fielder pushes his hand away. By doing this he decreases the velocity of the object. Hence the momentum decrease and the ball comes to catcher's hand easily.

The mass of dime = 2.26 gram So it has less momentum because of less mass.

Hence the force exerted or the momentum transferred by the Dime is very low .

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La densidad del óxido de magnesio. MgO, es de 3.581 g/cm3 El MgO, es de 3.581 g/cm3 El MgO cristaliza con ordenamiento cúbico co
Jobisdone [24]

Answer:

a=4.213cm

r=1.490x10^{-8}cm

Explanation:

Hola.

En este caso, para calcular la longitud (a) de una cara de celda unitaria, consideramos la siguiente ecuación:

\rho =\frac{#at*M}{a^3N_A}

En la que consideramos el número de átomos por celda (4 para FCC), la masa molar (40.3 g/mol para MgO) y el número de avogadro para obtener:

3.581g/mol = \frac{4atom/celda*40.3g/mol}{a^3*6.02x10^{23}atom/mol}

Despejando para a, obtenemos:

a^3 = \frac{4atom/celda*40.3g/mol}{3.581g/cm^3*6.02x10^{23}atom/mol}\\\\a=\sqrt[3]{7.478cm^3} \\\\a=4.213cm

Finalmente, el radio lo calculamos como:

r=\frac{\sqrt{2}*a}{4}=\frac{\sqrt{2}*4.213x10^{-8}cm}{4}\\\\r=1.490x10^{-8}cm

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