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vampirchik [111]

3 years ago

12

2. SHOCKING BLANCHED VEGETABLES MAKES THEM, WHAT?

1 answer:

JulijaS [17]3 years ago

7 0

GET ACID STOP COOKING

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Alecsey [184]

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kipiarov [429]

Answer:

This can be part of your paragraph.

Explanation:

From the cornea, the light passes through the pupil. The iris, or the colored part of your eye, controls the amount of light passing through. From there, it then hits the lens. This is the clear structure inside the eye that focuses light rays onto the retina.

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2 years ago

A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translat

Talja [164]

Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

$m g =\dfrac{mv^2}{r}$

$9.81 =\dfrac{v^2}{0.675}$

v² = 6.622

v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

where, $I =\dfrac{2}{5}mr^2$

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

$0.7 u^2 + g H = 0.7 v^2 + g(2R)$

$0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)$

0.7 u² = 7.85092

u² = 11.2156

u = 3.35 m/s

the initial speed is 3.35 m/s

8 0

2 years ago

You throw a glove straight upward to celebrate a victory. Its initial kinetic energy is K and it reaches a maximum height h. Wha

Rasek [7]

Answer:

K/2

Explanation:

The law of conservation of mechanical energy states that the sum of the kinetic and potential energies is a constant at any point.

At maximum height, the glove has purely potential energy but at the bottom, it has purely kinetic energy.

The potential energy at the top = kinetic energy at the bottom. The potential energy is given by

$PE = mgh$

At half height, this potential energy is

$PE = \frac{1}{2}mgh$

At this height, PE + KE = Constant = KE at bottom or PE at maximum height.

$mgh = \frac{1}{2}mgh +KE$

$KE = \frac{1}{2}mgh = K/2$

5 0

2 years ago