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3 years ago

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1. Calculate the velocity of a 0.8 kg kitten with a forward momentum of 5 kg*m/s.

1 answer:

V125BC [204]3 years ago

3 0

Q.1

Answer . Momentum = mass x velocity

Momentum/mass = velocity

5/0.8 = velocity

6.25 m/s = velocity

Q.2

Answer . F = ma => a = f/m => a = 7/3.2 => a = 2.1875m/s²

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Ryan places 0.150 kg of boiling water in a thermos bottle. How many kgs of ice at –12.0 °C must Ryan add to the thermos so that

Greeley [361]

Answer:

The value is $m_i = 0.0234 \ kg$

Explanation:

Generally from the calorimetry principle we have that

$Heat \ loss\ by\ ice = heat\ gained\ by\ water\$

So here heat gained water is mathematically represented as i.e

$Q_w = m_w * c_w * (T_w - T )$

substituting 0.150 kg for $m_w$ , 4200 J/kg.°C for $c_w$ , 100°C for $T_w$ and 75°C for T

We have

$Q_w = 0.150 * 4200 * (100 - 75 )$

$Q_w =15750 \ J$

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$Q_i = Q_1 + Q_2 + Q_3$

Here $Q_1$ is the energy to move the ice to its melting point which is evaluated as

$Q_1 = m_i * c_i * ( T_o -T_i)$

Here $m_i$ is the mass of ice

$c_i$ is the specific heat of ice with value $2.05 * 10^3 J/kg.°C$

$T_o$ temperature of ice at melting point with value 0°C

$T_i$ is the temperature of ice with value -12°C

$Q_2$ is the energy to move the ice from its its melting point to liquid which is evaluated as

$Q_2 = m_i * L$

Here L is the Latent heat of melting of ice with value $334 * 10^3 J/kg$

$Q_3$ is the energy to move the ice from liquid to the equilibrium temperature which is evaluated as

$Q_1 = m_i * c_w * ( T -T_o)$

So

$Q_i = m_i [ c_i * ( T_o -T_i) + L + c_w * ( T -T_o) ]$

=> $Q_i = m_i [ 2.05 * 10^3 * ( 0 -(-12)) + 334 * 10^3 + 4200 * ( 75 - 0) ]$

From

$Heat \ loss\ by\ ice = heat\ gained\ by\ water\$

We have that

$m_i * 673600 =15750$

=> $m_i = \frac{15750}{673600}$

=> $m_i = 0.0234 \ kg$

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