The temperature in deep space is thought to be about 3K.what is 3K in degrees Celsius and in degrees Fahrenheit?

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t

marishachu [46]

Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete ( $\Delta t$ ), in seconds:

$\Delta t = t_{A}-t_{C}$ (1)

Where:

$t_{C}$ - Time spent by the sound in concrete, in seconds.

$t_{A}$ - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

$\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}$

$\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}} \right)$

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$ (2)

Where:

$v_{C}$ - Speed of the sound in concrete, in meters per second.

$v_{A}$ - Speed of the sound in the air, in meters per second.

$x$ - Distance traveled by the sound, in meters.

If we know that $\Delta t = 6.4\,s$ , $v_{C} = 3000\,\frac{m}{s}$ and $v_{A} = 343\,\frac{m}{s}$ , then the distance travelled by the sound is:

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$

$x = 2478.585\,m$

The impact occured at a distance of 2478.585 meters from the person.

7 0

2 years ago

A moving rope (parallel to the slope) is used to pull skiers up the mountain. If the slope of the hill is 37" and friction is ne

Charra [1.4K]

Since rope is parallel to the inclined plane so here we can say that net force parallel to the person which is pulling upwards must counterbalance the component of weight of the person.

Now here we will do the components of the weight of the person

given that weight of the person = 500 N

now its components are

$W_x = 500 cos37$