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scoundrel [369]

3 years ago

9

After dinner, Joanne did not put the leftover casserole away until after watching her favorite television show. What is the long

est time it should be left out at room temperature?

1 answer:

Sever21 [200]3 years ago

7 0

Answer:

At most 2 hours

Explanation:

According to my research of various health departments it is said that cooked food should not be left out in room temperature for more than 2 hours, assuming that room temperature is between 40°F and 140°F. This is because after 2 hours bacteria starts to grow at an increased rate and can have various health effects on the person consuming the food. Therefore food is best wrapped and placed in the refrigerator.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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(1) Explain : Market and its characteristics.

AlekseyPX

Answer:

market is a place where we sell or buy things ( goods)

it's characteristics are

buying and selling goods

perfect competitions

market doesn't refer only a fix place

6 0

3 years ago

How to measure the volume of a baseball bat ( need answers ASAP )

vaieri [72.5K]

<em>Measure the amount of water it displaces.</em>

This won't be easy, because the bat floats in water. But I think you can get around that little problem like this:

-- Get some kind of a tank or tub that's big enough to hold the whole bat under water.

-- Get a heavy weight, like a big wrench or a small rock.

-- Fill the tub almost to the tippy top with water.

-- Slip the heavy weight into the tub, slowly. Some water will run over the top and out of the tub. That's OK ... it's exactly what you want. If NO water runs over the top, pour some more in, until it runs out and then stops. You want the tub full to the brimmy rim with the rock at the bottom of it.

-- Take the heavy weight out of the tub.

-- Now set the tub into a bigger tub or a deep pan. The next time it overflows and some water runs out of it, you'll need to catch that water and measure it.

-- Get a short piece of heavy string. Tie the heavy weight to somewhere near the middle of the bat.

-- Slowly slide the bat into the water, with the rock tied to it. The bat needs to go complete underwater.

-- Some more water will run over the top and out of the tub, and INTO the lower tub. Wait until the overflow stops and everything settles down again.

-- Take the bat (tied to the weight) out of the tub. Slowly and carefully, so that your hand or your arm doesn't make any MORE water run over and out.

-- Lift the upper tub out of the lower tub.

-- Take the lower tub, with the overflow water in it. Using a kitchen measuring cup, or a saucepan or a bottle, or anything else with liquid amounts marked on it, measure how much water overflowed into the lower tub.

THAT amount is the volume of the bat.

You may have to do some units conversions. Like if you need the volume of the bat in cm³ and you used measuring vessels marked in fluid ounces. But you can find all those conversion factors with a search on Floogle.

8 0

3 years ago

How is a revolution related to a year

nikdorinn [45]

Revolution means orbiting around another body.
<span>A year is the time for a planet to complete one orbit around the Sun. 100% sure

</span>

4 0

3 years ago

Read 2 more answers

What is the correct answer?

inysia [295]

The answer to your question is metaphase

3 0

3 years ago

A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5º above the horizontal on a long flat firing range. (

Margarita [4]

Answer:

A) $h = 69.58 m$

D) $v = 58.12 \frac{m}{s}$ (Speed magnitude)

α = 22.49° (Speed direction above the horizontal)

Explanation:

Conceptual analysis:

To solve this problem we consider the following concepts:

1) The projectile in its movement describes a curved line called a parabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane.

The initial velocity (Vo) is tangent to the trajectory at the initial point and can be broken down into two components, one vertical (Voy) and one horizontal(Vox):

$V_{ox} = V_{o}Cos\alpha _{o}$ Formula (1)

$V_{oy} = V_{o}Sin\alpha _{o}$ Formula (2)

Where:

Vo: Initial velocity in m/s

$\alpha_{o}$ : Initial angle above the horizontal in grades

2) The formula to calculate its velocity at any vertical position(y) is as follows:

$v_{y}^{2} = v_{oy}^{2} -2gy$ Formula (3)

Where:

$v_{f}^{2}$ : Final speed component in vertical direction in m/s

$v_{oy}^{2}$ : Initial speed component in vertical direction in m/s

g: acceleration due to gravity in m/s2

3) The formulas to calculate the projectile velocity components at any time (t) are:

$v_{x} = v_{ox}$ Because the movement is uniform in the x direction (constant speed)

$v_{y} = v_{oy}-g*t$ Formula (4) Because the movement is uniformly accelerated in the y direction

Known information:

We know the following data:

$v_{o} = 65.2 \frac{m}{s}$

$\alpha _{o}$ = 34.5º above the horizontal

$g=9.8 \frac{m}{s^{2}}$

Development of the problem:

Initial speed components(Vox, Voy), (Formula (1), Formula (2)

$v_{ox} = 65.2*Cos34.5 = 53.7\frac{m}{s}$

$v_{oy} = 65.2*Sin34.5 = 36.93\frac{m}{s}$

A) Maximum height (h):

When the projectile reaches its maximum height (h) ,the speed component Vy = 0, then, we replace this value in the Formula (3):

$0=(36.93)^2-2*9.8*h$

$h=\frac{ (36.93)^2}{2*9.8} = 69.58 m$

D) Speed of the projectile 1.50s after firing

We replace t=1.5 s in the formula(4)

$v_{y} = 36.93-9.8*1.5 = 22.23 \frac{m}{s}$

$v_{x} = v_{ox} = 53.7 \frac{m}{s}$

$v = \sqrt{53.7^{2}+22.23^{2}} = 58.12 \frac{m}{s}$ (Speed magnitude)

$\alpha = tan^{-1} (\frac{v_{y}}{v_{x}}) = tan^{-1} (\frac{22.23}{53.7})$

α = 22.49° (Speed direction above the horizontal)

6 0

3 years ago