Most rocks contain more than one type of

Give three factors which are responsible for the vanishing forest

slava [35]

Answer:

1. Huge wildfires

2. Deforestation

3. Reduced amount of aforestation, etc

5 0

3 years ago

Which electromagnetic wave has the lowest frequencies (less than 3×109 hertz)?

ICE Princess25 [194]

Answer:

Radio waves

Explanation:

The electromagnetic spectrum includes all different types of waves, which are usually classified depending on their frequency. Ordering them from the highest frequency to the lowest frequency, they are:

- Gamma rays

- X-rays

- Ultraviolet

- Visible light

- Infrared radiation

- Microwaves

- Radio waves

Radio waves are the electromagnetic waves with lowest frequency, their frequency is lower than 300 GHz ( $3\cdot 10^{11} Hz$ ) and therefore they are the electromagnetic waves with lowest energy (in fact, the energy of an electromagnetic wave is proportional to its frequency). They are generally used for radio and telecommunications since this type of waves can travel up to long distances.

5 0

3 years ago

Read 2 more answers

How are the energy conversions of machines similar to the energy conversions in your body

sleet_krkn [62]

Answer:

Explained below

Explanation:

When we eat food, our body gets chemical energy from it. Now, this chemical energy from the food is changed into some different energy forms that is useful to it. They include:

-Chemical to mechanical energy to aid in movement of muscles

- chemical to thermal energy to aid in regulating the body temperature.

- chemical to electrical energy to aid the brain in thinking.

Thus is similar to how a machine converts energy because machines also generate energy after being powered and convert to other forms of energy. For example, an alarm clock converts electrical energy to sound energy, hair dryer converts electrical energy to thermal/heat energy.

6 0

3 years ago

An electron is accelerated by a 3.6 kv potential difference. the charge on an electron is 1.60218 × 10−19 c and its mass is 9.10

katen-ka-za [31]

By definition, the potential energy is:
U = qV
Where,
q: load
V: voltage.
Then, the kinetic energy is:
K = mv ^ 2/2
Where,
m: mass
v: speed.
As the power energy is converted into kinetic energy, we have then:
U = K
Equating equations:
qV = mv ^ 2/2
From here, we clear the speed:
v = root (2qV / m)
Substituting values we have:
v = root ((2 * (1.60218 × 10 ^ -19) * 3600) /9.10939×10^-31))
v = 3.56 × 10 ^ 7 m / s
Then, the centripetal force is:
Fc = Fm
mv ^ 2 / r = qvB
By clearing the magnetic field we have:
B = mv / qr
Substituting values:
B = (9.10939 × 10 ^ -31) * (3.56 × 10 ^ 7) / (1.60218 × 10 ^ -19) * 0.059
B = 3.43 × 10 ^ -3 T
Answer:
A magnetic field that must be experienced by the electron is:
B = 3.43 × 10 ^ -3 T

6 0

3 years ago

To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl

Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁ + 2.43 x 0 = v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂² = 0.07032

v₂² = 0.07032 / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

5 0

3 years ago