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3 years ago

12

Which part of an Adam is actively exchanged or shared in a chemical bond?

1 answer:

Lemur [1.5K]3 years ago

5 0

Atom* the particles are (Electrons)

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How to test the pH level of a soap?

olchik [2.2K]

Just test the soap book

8 0

3 years ago

Read 2 more answers

A wave with a frequency of 500 Hz is traveling at a speed of 200 m/s. What is the wavelength?

pshichka [43]

Answer:0.4 is what i got

Explanation:

200÷500

6 0

3 years ago

When the temperature of the air is 25 degress C, the velocity of a sound wave traveling through the air is approximately

dusya [7]

Assuming an ideal gas, the speed of sound depends on temperature
only. Air is almost an ideal gas.

Assuming the temperature of 25°C in a "standard atmosphere", the
density of air is 1.1644 kg/m3, and the speed of sound is 346.13 m/s.

The velocity can't be specified, since the question gives no information
regarding the direction of the sound.

6 0

3 years ago

A pendulum consists of a large balanced mass hanging on the end of a long wire. At the point where a 28-kg pendulum has the grea

Ray Of Light [21]

Answer:

The length of the wire is approximately 67.1 m

Explanation:

The parameters of the pendulum are;

The mass of the pendulum, m = 28 kg

The angle between the pendulum weight and the wire, θ = 89°

The magnitude of the torque exerted by the pendulum's weight, τ = 1.84 × 10⁴ N·m

We have;

Torque, τ = F·L·sinθ = m·g·l·sinθ

Where;

F = The applies force = The weight of the pendulum = m·g

g = The acceleration due to gravity ≈ 9.8 m/s²

l = The length of the wire

Plugging in the values of the variables gives;

1.84 × 10⁴ N·m = 28 kg × 9.8 m/s² × l × sin(89°)

Therefore;

l = 1.84 × 10⁴ N·m/(28 kg × 9.8 m/s² × sin(89°)) = 67.0656080029 m ≈ 67.1 m

The length of the wire, l ≈ 67.1 m

6 0

3 years ago

In an engine governor, the two spheres (total mass of 1.0kg) are at 0.05m and rotating at 37rad/s If the engine increases the an

kirill115 [55]

Here we can say that there is no external torque on this system

So here we can say that angular momentum is conserved

so here we will have

$I_1\omega_1 = I_2\omega_2$

now we have

$I_1 = mr^2$

$I_1 = (1kg)(0.05^2)$

$I_1 = 25\times 10^{-4} kg m^2$

similarly let the final distance is "r"

so now we have

$I_2 = mr^2$

$I_2 = 1r^2$

now from above equation we have

$(25\times 10^{-4})37 = (r^2)(58)$

$r = 0.04 m$

so final distance is 0.04 m between them

8 0

3 years ago