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3 years ago

Your grandmother in Ireland sends you her favorite cookie recipe. Her instructions say to bake the cookies at

Physics

2 answers:

Anna [14]3 years ago

7 0

Answer:

350 Degrees Fahrenheit

Explanation:

Lapatulllka [165]3 years ago

3 0

$F=\frac{9}{5}C+32$

$F=\frac{9}{5}190.5C+32$

$F=342.9+32$

$F=374.9$

$374.9F=190.5C$

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Hii please help i’ll give brainliest if you give a correct answer please please hurry it’s timed

Yanka [14]

Answer:

Sorry I'm wrong The person above is correct.

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4 0

3 years ago

Read 2 more answers

Which law states that absolute zero cannot be reached?

padilas [110]

The third law of thermodynamics,the principle of temperature.

This law states that the entropy at 0 is always equel to 0.
This means that it is impossible to cool down a perfect 0 or absolute 0(-273.15 C)

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3 years ago

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A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,