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3 years ago

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An eagle is flying horizontally 16.4 meters above a lake at a speed of 9.3 m/s, carrying a small pumpkin in its talons. The pump

kin slips free. How far horizontally will the pumpkin travel after it slips from the eagle until it hits the ground?

1 answer:

Dima020 [189]3 years ago

4 0

Answer:

The horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m

Explanation:

Given;

height above the ground, h = 16.4 m

speed of the eagle, v = 9.3 m/s

The time it will take the pumpkin to fall at the given height is calculated as;

$t = \sqrt{\frac{2h}{g} }\\\\t = \sqrt{\frac{2*16.4}{9.8} }\\\\t = 1.83 \ s$

The horizontal distance traveled at this time is given by;

x = vt

x = (9.3)(1.83)

x = 17.02 m

Therefore, the horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m

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3 years ago

the chef was careless and put the aluminum cookie sheet directly on the hot stove, which melted 0.05 kg of the aluminum. how muc

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Heat required to melt 0.05 kg of aluminum is 28.7 kJ.

<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>

The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.

The formula to be used is given below:

Heat required = mass * heat capacity * temperature change

Assuming the aluminum sheet was at room temperature initially.;

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Melting point of aluminum = 660.3 °C

Temperature difference = (660.3 - 25) = 635.3 903

Heat capacity of aluminum = 903 J/kg/903

Heat required = 0.05 * 903 * 635.3

Heat required = 28.7 kJ

In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.

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3 0

2 years ago

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

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The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

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3 years ago

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Paraphin [41]

Answer:

v = 3200 m/s

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As we know that the frequency of the sound wave is given as

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so we will have

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4 0

3 years ago

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marusya05 [52]

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That's <em>B </em>on Earth.

It would be some other number on other bodies.

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3 years ago