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Which type of wave has a longer wavelength: AM radio waves (with frequencies in the kilohertz range) or FM radio waves (with fre

faltersainse [42]

Answer:

AM has longer wavelength

Explanation:

The relation between the wavelength and teh frequency is given by

v = f x λ

Where, f is the frequency and λ be the wavelength.

It shows that the wavelength is inversely proportional to the frequency.

So, higher the frequency, smaller be the wavelength.

So, FM has high frequency than AM, thus, FM has lower wavelength as compared to AM.

6 0

2 years ago

Listed in the Item Bank are some important labels for sections of the image below. To find out more information about labels, so

erica [24]

1 - cytoplasm

2 - DNA

3 - rybosome

4 - cell membrane

5 - nucleus

5 0

2 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

2 years ago

Organism undergo constant chemical changes as they maintain an internal balance known as?

Aleks04 [339]

The answer is homeostasis.

3 0

3 years ago

Read 2 more answers

A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time int

Inessa05 [86]

Let F be the magnitude of the force applied to the cart, m the mass of the cart, and a the acceleration it undergoes. After time t, the cart accelerates from rest v₀ = 0 to a final velocity v. By Newton's second law, the first push applies an acceleration of

F = m a → a = F / m

so that the cart's final speed is

v = v₀ + a t

v = (F / m) t

If we force is halved, so is the accleration:

a = F / m → a/2 = F / (2m)

So, in order to get the cart up to the same speed v as before, you need to double the time interval t to 2t, since that would give

(F / (2m)) (2t) = (F / m) t = v

3 0

2 years ago