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3 years ago

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Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while

Milada jumps horizontally with an initial speed of 95.0 cm/s. How far from the base of the cliff will Milada hit the ground? Ignore air resistance.

1 answer:

Vinvika [58]3 years ago

7 0

Answer:

Explanation:

Given

Time taken to reach ground is $t=2.7\ s$

Malda initial velocity $u=95\ cm/s$

Let h be the height of Cliff

using $h=ut+\frac{1}{2}at^2$

where, u=initial velocity

t=time

In first case chirpy drop downward thus u=0

$h=0+\frac{1}{2}(9.8)(2.7)^2$

$h=35.72\ m$

For Milada there is horizontal velocity u=95 cm/s=0.95 m/s[/tex]

time taken to reach the ground will be same so distance traveled in this time with 0.95 m/s horizontal velocity is given by

$R=u\times t$

$R=0.95\times 2.7=2.43\ m$

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A hot iron horseshoe (mass = 0.35 kg ), just forged, is dropped into 1.40 L of water in a 0.45 kg iron pot initially at 22.0°C.

olga_2 [115]

Answer:

$420$ °C

Explanation:

$m_{h}$ = mass of the horseshoe = 0.35 kg

$m_{w}$ = mass of the water = 1.40 L = 1.40 kg

$m_{i}$ = mass of the iron pot = 0.45 kg

$c_{i}$ = specific heat of iron = 450 J kg⁻¹ °C⁻¹

$c_{w}$ = specific heat of water = 4186 J kg⁻¹ °C⁻¹

$T_{hi}$ = initial temperature of the horseshoe = ?

$T_{wi}$ = initial temperature of the water = 22 °C

$T_{pi}$ = initial temperature of the iron pot = 22 °C

$T_{f}$ = final temperature = 32 °C

Using conservation of Heat

$m_{h}c_{i}(T_{hi} - T_{f}) = m_{w}c_{w}(T_{f} - T_{wi}) + m_{i}c_{i}(T_{f} - T_{pi})$

$(0.35)(450)(T_{hi} - 32) = (1.40)(4186)(32 - 22) + (0.45)(450)(32 - 22)$

$(157.5)(T_{hi} - 32) = 60629$

$T_{hi} - 32 = 384.95$

$T_{hi} = 420$ °C

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3 years ago

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The answer is

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Explanation:

Given;

mass of the first student, m₁ = 77kg

mass of the second student, m₂ = 59kg

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initial velocity of the second student, u₂ = 0

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3 years ago

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Answer:

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3 years ago