During a warm-up, it's important to control your

OLEGan [10]

Answer:

up swelling moves water horizontally

3 0

3 years ago

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

Jet001 [13]

<h2>The impact is occurred at a distance 348.55 m far from person.</h2>

Explanation:

Let s be the distance to the impact position and t be the time when he hears the sound though concrete after impact.

Time when he hears impact through air = t + 0.90

The distance traveled by sound wave in both concrete and air is s.

Speed of sound in air = 343 m/s

Speed of sound in concrete = 3000 m/s

Distance traveled in air = Distance traveled in concrete

343 x Time when he hears impact through air = 3000 x Time when he hears impact through concrete

343 x (t + 0.90) = 3000 x t

3000t - 343 t = 343 x 0.90

2657t = 308.7

t = 0.116 s

Distance traveled in concrete = 3000 x t = 3000 x 0.116 = 348.55m

So the impact is occurred at a distance 348.55 m far from person.

4 0

3 years ago

(1 point) A gun has a muzzle speed of 90 meters per second. What angle of elevation θ, where 0≤θ≤π4, should be used to hit an ob

Likurg_2 [28]

Answer:

0.122 radians

Explanation:

Given that the muzzle speed is, $u=90m/s$

And the distance of the object is, $s=200 m$

And acceleration due to gravity is $g=9.8 m/s^{2}$

Therefore negative acceleration due to gravity for upward motion.

Now,

The x component of velocity will be, $u_{x}=ucos\theta$

The y component of velocity will be, $u_{y}=usin\theta$

And the horizontal displacement is given. Now horizontal displacement is,

$s=u_{x}\times t$

Here, $u_{x}$ is horizontal velocity and t is time.

Substitute all the values in above equation, we get

$s=ucos\theta\times t\\200=90cos\theta\times t\\t=\dfrac{200}{90cos\theta}\\t=\dfrac{20}{9cos\theta}$

Now, in y direction we can calculate displacement and in y direction displacement is zero.

$s_{y}=u_{y}t+\frac{1}{2}at^{2}$

Substitute all the variables.

$0=usin\theta t+\frac{1}{2}at^{2}\\0=90sin\theta (\frac{20}{9cos\theta})+\frac{1}{2}(-9.8 m/s^{2})\times (\frac{20}{9cos\theta})^{2} \\200tan\theta=4.9\times \frac{400}{81}\times \frac{1}{cos^{2}\theta}\\200sin\theta=\frac{24.1975}{cos\theta}\\ 2sin\theta cos\theta=0.241975\\sin2\theta=0.241975\\2\theta=sin^{-1}(0.241975)\\ \theta=0.122 radians$

Therefore the required angle of elevation is 0.122 radians.

8 0

4 years ago

While sleeping, a 15.5 kg dog exerts a downward force on the ground. What is the magnitude of the upward force being exerted on

IrinaK [193]

He's right halos slid D also s dos wmpwH do a s
Wks eke wen

5 0

3 years ago

If cooling occurred at the bottom of a lake instead of its surface, the lake

AlladinOne [14]

No because the top has more oxygen to it which will break the particles up and have them freeze easier.

so if something froze at the bottom of the lake it would float to the top and that's why water always freezes from the top down :))

5 0

3 years ago

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