In act 3, who says the following lines to Bottom?

Why should you rotate food when thawing it 360 training?

Anni [7]

Rotation during thawing is necessary to ensure that your food reaches the right temperature in a secure and efficient manner. This prevents the ice from becoming stuck under some of the food. By rotating it, you can reveal any trapped, covered, or chunked ice.

<h3>What do you mean by Microwave oven?</h3>

A microwave oven, often known as a microwave, is an electric oven that uses electromagnetic radiation with a microwave frequency range to cook and heat food.
As a result, the food is heated via a process known as dielectric heating, which causes the polar molecules there to rotate and produce heat energy.
Because excitation is rather homogenous in the exterior 25–38 mm (1–1.5 inches) of a homogeneous, high water content food item, microwave ovens heat food fast and effectively.

<h3>What are the uses of a microwave oven?</h3>

A microwave oven, sometimes known as an electronic oven, is a device that uses microwaves, which are high-frequency electromagnetic waves, to cook food.
A microwave oven is a compact, box-shaped oven that increases food's temperature by exposing it to an electromagnetic field with a high frequency.

Learn more about microwave oven here:

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5 0

1 year ago

In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. the dish is on a shelf above the poi

Georgia [21]

components of the speed of the coin is given as

$v_x = v cos60$

$v_x = 6.4 cos60 = 3.2 m/s$

$v_y = vsin60$

$v_y = 6.4 sin60 = 5.54 m/s$

now the time taken by the coin to reach the plate is given by

$t = \frac{\delta x}{v_x}$

$t = \frac{2.1}{3.2}$

$t = 0.656 s$

now in order to find the height

$h = vy * t + \frac{1}{2} at^2$

$h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2$

$h = 1.52 m$

so it is placed at 1.52 m height

3 0

3 years ago

If an object moving at 12 m/s has a kinetic energy of 72 j what is the mass

EleoNora [17]

Simply, apply the formula $E = \frac{1}{2}mv^{2}$ and insert the values of m = mass, v = velocity and E = Energy.
The result will be $72J = \frac{1}{2}m(12m/s)^{2}$ , m = 1 kg

5 0

3 years ago

If you push any floating object down from equilibrium and release it, it bobs up and down. That looks like an oscillation, so le

GarryVolchara [31]

Answer:

$F_{y}$ = ( ρ_fluid g A) y

Explanation:

This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force

for the first part, let's write Newton's equilibrium equation

B₀ - W = 0

B₀ = W

ρ_fluid g V_fluid = W

the volume of the fluid is the area of the cube times the height it is submerged

V_fluid = A y

For the second part, the body introduces a quantity and below this equilibrium point, the equation is

B - W = m a

ρ_fluid g A (y₀ + y) - W = m a

ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a

ρ_fluid g A y + (B₀-W) = ma

the part in parentheses is zero since it is the force when it is in equilibrium

ρ_fluid g A y = m a

this equation the net force is

$F_{y}$ = ( ρ_fluid g A) y

we can see that this force varies linearly the distance and measured from the equilibrium position

8 0

3 years ago

Pls how to solve this problem. help would be appreciated

Bingel [31]

Answer:

80 m/s

Explanation:

Given:

a = -5 m/s²

v = 0 m/s

Δx = 640 m

Find: v₀

v² = v₀² + 2a(x − x₀)

(0 m/s)² = v₀² + 2(-5 m/s²) (640 m)

v₀ = 80 m/s

7 0

3 years ago