what gas is most abundant greenhouse gas ? a) ozone b) chlorofluorocarbon c) carbon dioxide d) methane e) water vapor

A wave has a wavelength of 4. 9 m and a velocity of 9. 8 m/s. The medium through which this wave is traveling is then heated so

tatiyna

The wavelength of a wave is obtained by taking the ratio of wave speed and frequency.

The wavelength of the heated wave is 9.8 m. Hence, option (b) is correct.

What is frequency of a wave?

The number of oscillations completed by a wave in one second is known as the frequency of a wave. It is expressed as the ratio of the velocity of the wave to its wavelength.

Given data-

The wavelength of the wave is, $\lambda = 4.9 \;\rm m$ .

The velocity of the wave is, v = 9.8 m/s.

The mathematical expression for the frequency of the wave is,

$f = \dfrac{v}{\lambda}$

Solving as,

$f = \dfrac{9.9}{4.9}\\\\f =2 \;\rm Hz$

Now, with constant frequency and the double magnitude of velocity (v' = 2 × 9.8 = 19.6 m/s). The wavelength of the heated wave is calculated as,

$f = \dfrac{v'}{\lambda'}$

here,

$\lambda'$ is the wavelength of the heated wave.

Solving as,

$2 = \dfrac{19.6}{\lambda'}\\\\\lambda' = \dfrac{19.6}{2}\\\\\lambda' = 9.8 \;\rm m$

Thus, we can conclude that the wavelength of the heated wave is 9.8 m. Hence, option (b) is correct.

Learn more about the frequency of wave here:

brainly.com/question/1324797

3 0

2 years ago

As more and more resistors are added in parallel to a circuit, the equivalent resistance of the circuit ____________ (increases,

Anestetic [448]

Answer:

Decreases, Increases

Explanation:

Resistance is parallel can be calculated using

1/Req = 1/R1 + 1/R2 + 1/R3 +....

Then, as more resistor is added in parallel the equivalent resistance is reduced.

Let use a simple sample

Let all the resistor have equal resistances

Let say R = R1 = R2 = R3 =...Rn

Then, 1/Req = 1/R1 + 1/R2 + 1/R3 +....

1/Req = 1/R + 1/R + 1/R +.... 1/Rn

Req = R/n

Check attachment on how I got that.

This implies that, the equivalent resistance will always be less than the original resistance, since n>1

So, as n increases (I.e. as the number of resistance increases), the equivalent resistance reduces.

B. Now, to know if the current reduces or increases

Using Ohms law

V = iR

Then, I = V/R

So, let assume the voltage is constant, then, the current is inversely proportional to the resistance, so as we know that the resistance is reducing, then the current will be increasing.

So current increase as we add more resistor in parallel to a circuit

3 0

3 years ago

Which has more kinetic energy, a 4.0 kg bowling ball moving at 1.0 m/s or a 1.0 kg

jolli1 [7]

Answer:

The kinetic energy of bocce ball is more.

Explanation:

Given that,

Mass of a bowling ball, m₁ = 4 kg

Speed of the bowling ball, v₁ = 1 m/s

Mass of bocce ball, m₂ = 1 kg

Speed of bocce ball, v₂ = 4 m/s

We need to say which has more kinetic energy.

The kinetic energy of an object is given by :

$E=\dfrac{1}{2}mv^2$

Kinetic energy of the bowling ball,

$E_1=\dfrac{1}{2}m_1v_1^2\\\\E_1=\dfrac{1}{2}\times 4\times (1)^2\\\\E_1=2\ J$

The kinetic energy of the bocce ball,

$E_2=\dfrac{1}{2}m_2v_2^2\\\\E_2=\dfrac{1}{2}\times 1\times (4)^2\\\\E_2=8\ J$

So, the kinetic energy of bocce ball is more than that of bowling ball.

5 0

2 years ago

How does the nitrogen enter the food web?

nikklg [1K]

When bacteria in the soil takes nitrogen from the air it becomes nitrates it can finally move through the food chain in this form.

6 0

3 years ago

A thin rod of length 0.75 m and mass 0.42 kg is suspended

MrRissso [65]

Answer:

a) K = 0.63 J, b) h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

w² = $\frac{m g d}{I}$

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

I = ⅓ m L²

we substitute

w = $\sqrt{\frac{mgL}{2} \ \frac{1}{\frac{1}{3} mL^2} }$

w = $\sqrt{\frac{3}{2} \ \frac{g}{L} }$

w = $\sqrt{ \frac{3}{2} \ \frac{9.8}{0.75} }$

w = 4.427 rad / s

an oscillatory system is described by the expression

θ = θ₀ cos (wt + Φ)

the angular velocity is

w = dθ /dt

w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

w = 4.0 rad / s

the kinetic energy is

K = ½ I w²

K = ½ (⅓ m L²) w²

K = 1/6 m L² w²

K = 1/6 0.42 0.75² 4.0²

K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

Em₀ = K = ½ I w²

final point. Highest point

Em_f = U = m g h

energy is conserved

Em₀ = Em_f

½ I w² = m g h

½ (⅓ m L²) w² = m g h

h = 1/6 L² w² / g

h = 1/6 0.75² 4.0² / 9.8

h = 0.153 m

5 0

2 years ago