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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.

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Where is the Oort cloud located?

sergiy2304 [10]

Answer: C

Explanation: just did it

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What is the frequency of an x- ray if the wavelength is 4.5 E - 10m?

AnnZ [28]

V (speed) = F (frequency) x Wavelength
If we rearrange the formula, making frequency the subject;
F (frequency) = Speed ÷ Wavelength
F = 300,000 m\s x 4.5 e -10m
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3 years ago

An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength

Vladimir [108]

Answer:

3x10⁴v

Explanation:

Using

Wavelength= h/ √(2m.Ke)

880nm = 6.6E-34/√ 2.9.1E-31 x me

Ke= 6.6E-34/880nm x 18.2E -31.

5.6E-27/18.2E-31

= 3 x 10⁴ Volts

4 0

3 years ago

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

5 0

3 years ago

A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin

JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

$f_s = f(\dfrac{v}{v-(-v_s)})$

$f_s = f(\dfrac{v}{v+v_s})$

$v_s = v[\dfrac{f_s}{f_o}-1]$

$= 343[\dfrac{508}{490}-1]$

$v_s=12.6 m/s$

distance the tunning fork has fallen

$y_1=\dfrac{v^2}{2a_y}$

$=\dfrac{12.6^2}{2\times 9.8}$

=8.1 m

now, time required for the observed will be

$t = \dfrac{8.1}{343} = 0.023 s$

now, for the distance calculation

$y_2 = u\ t + \dfrac{1}{2}at^2$

$= 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2$

=0.293 m

total distance

= 8.1 + 0.293 = 8.392 m

3 0

3 years ago