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2 years ago

5.

Physics

1 answer:

iris [78.8K]2 years ago

7 0

Answer:

t = 1.659s

Explanation:

We can use the kinematics equations to solve this questions:

v = u + at

$v^{2} = u^{2} +2as$

where v = Final Velocity, u = initial velocity, a = acceleration, t = time, s = displacement

a) Given information from the question,

u = $\frac{70km}{h} =\frac{(70*1000)m}{(1*3600)s} = 19.444m/s$ (Convert km/h to m/s first)

a = $2m/s^{2}$

s = 35m

Now we can substitute these values into the 2nd kinematics equation to find v, final velocity.

$v^{2} =(19.444)^{2} +2(2)(35)\\v=\sqrt{(19.444)^{2} +2(2)(35)} \\v= 22.761m/s (5.sf)\\$

b) Now we have the final velocity, we can substitute the values into the first kinematics equation to find t , the time taken.

v = u + at

22.761 = 19.444 + 2t

2t = 22.761 - 19.444

t = $\frac{22.761-19.444}{2}$

t = 1.659s

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Answer:

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Explanation:

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Where:

$m_{i}$ - Mass of ice, in grams.

$m_{w}$ - Mass of the juice, in grams.

$c_{i}$ - Specific heat of ice, in joules per gram-degree Celsius.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in joules per gram.

$T_{1}$ - Initial temperature of ice, in degrees Celsius.

$T_{2}$ - Melting point of water, in degrees Celsius.

$T_{3}$ - Final temperature of the ice-juice system, in degrees Celsius.

$T_{4}$ - Initial temperature of the juice, in degrees Celsius.

If we know that $m_{i} = 100\,g$ , $c_{i} = 2.090\,\frac{J}{g\cdot ^{\circ}C}$ , $c_{w} = 4.18\,\frac{J}{g\cdot ^{\circ}C}$ , $L_{f} = 334\,\frac{J}{g}$ , $T_{1} = -10\,^{\circ}C$ , $T_{2} = 0\,^{\circ}C$ , $T_{3} = 10\,^{\circ}C$ and $T_{4} = 20\,^{\circ}C$ , then the mass of the juice is:

$m_{w} = \frac{m_{i}\cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})]}{c_{w} \cdot (T_{3}-T_{4})}$

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