If heat is transferred away from a gas, what could happen?

Mrs. Paul and Dr. Mykannen ride a tandem bike on the beach. They ride along the beach for about 455 meters. Their final velocity

algol [13]

Answer:

Time = 80.91 seconds

Explanation:

Given the following data;

Velocity = 5.50 m/s.

Distance = 445 meters

To find the time;

Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.

Mathematically, velocity is given by the equation;

$Velocity = \frac{distance}{time}$

Substituting into the formula, we have;

5.5 = 445/time

Time = 445/5.5

Time = 80.91 seconds

5 0

2 years ago

A "moving sidewalk" in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5

pishuonlain [190]

Answer:

a.14 s

b.70 s

Explanation:

a.Let the sidewalk moving in positive x- direction.

Speed of sidewalk relative to ground= $v_s=1m/s$

Speed of women relative to sidewalk=v=1.5m/s

The speed of women relative to the ground

$v_w=v_s+v=1+1.5=2.5m/s$

Distance=35 m

Time= $\frac{distance}{speed}$

Using the formula

Time taken by women to reach the opposite end if she walks in the same direction the sidewalk is moving= $\frac{35}{v_w}=\frac{35}{2.5}=14s$

b.If she gets on at the end opposite the end in part (a)

Then, we take displacement negative.

Speed of sidewalk relative to ground= $v_s=1m/s$

Speed of women relative to sidewalk=v=-1.5 m/s

The speed of women relative to the ground= $v_w=v_s+v=1-1.5=-0.5m/s$

Time= $\frac{-35}{-0.5}=70 s$

Hence, the women takes 70 s to reach the opposite end if she walks in the opposite direction the sidewalk is moving.

3 0

3 years ago

a wave in a rope is traveling at 6 m/s and at a frequency of 2 Hz. what is the wavelength of the wave producced

Tems11 [23]

Answer:

3

Explanation:

4 0

3 years ago

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both

Nuetrik [128]

Answer:

Part a)

Width of the slit is

$a = 580 nm$

Part b)

Ratio of intensity is given as

$\frac{I}{I_o} = 0.81$

Explanation:

Part a)

As we know by the formula of diffraction we will have

$a sin\theta = \lambda$

so we have

$\theta = 90$

$\lambda = 580 nm$

so we will have

$a sin90 = 580 nm$

$a = 580 nm$

Part b)

As we know that the intensity in diffraction pattern is given as

$I = I_o (\frac{sin\theta}{\theta})^2$

$\frac{I}{I_o} = (\frac{sin\theta}{\theta})^2$

so for angle 45 degree

$\frac{I}{I_o} = (\frac{sin45}{\pi/4})^2$

$\frac{I}{I_o} = 0.81$

7 0

3 years ago

Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par

taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ = 5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

$t_1 = \frac{72v_2}{v_1-v_2}\\\\t_1 = \frac{72*5.5}{8.5-5.5}\\\\t_1 = 132 \ s$

The distance traveled is given by;

Δx = v₁t₁

Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

4 0

3 years ago