How does the density of water change when: (a) it is heated from 0o

A helicopter ambulance flew from one hospital to another in a straight line. The pilot had to change speed several times due to

Olenka [21]

Answer:

Explanation:

All the rest of the information is extraneous. The only 2 things you have to know are

d = 20 km

t = 8 minutes = 8/60 hours = 0.13333333

So the speed is s = d/t

s = 20/0.1333333 = 150 km/hour

Note: you have not specified what units the speed is. I suppose you could answer 20/8 = 2.5 km/min

4 0

3 years ago

When there's a hazard ahead, it's almost always quicker for you to _________ than to come to a full stop.

TEA [102]

When there's a hazard ahead, it's almost always quicker for you to steer away than to come to a full stop.

<h3>What is an hazard?</h3>

Hazard refers to any obstacle or other feature which causes risk or danger.

Living organisms respond to hazards via the production of adrenaline hormone. This hormone causes a flight response away from the hazard.

Therefore, when there's a hazard ahead, it's almost always quicker for you to steer away than to come to a full stop.

Learn more about hazards at: brainly.com/question/5338299

5 0

2 years ago

Which one of the following temperatures (in °C) is equivalent to 294 K?

vaieri [72.5K]

Answer:

21

Explanation: its actually 20.85 but i guess they round to 21

5 0

2 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

3 years ago

Two spectators at a soccer game see, and a moment later hear, the ball being kicked on the playing field. The time delay for the

WINSTONCH [101]

Answer:

a)188.65m

b)154.35m

c)243.7m

Explanation:

Given data:

$t_A=0.55s$

$t_B=0.45s$

(a) The distance from the kicker to each of the 2 spectators is given by:

$d_A=v \times t_A$

where,

v= speed of sound

$t_A$ =time taken for the sound waves to reach the ears

$d_A=343\times 0.55=188.65$ m

(b) $d_B=v \times t_B$

where,

v= speed of sound

$t_B$ =time taken for the sound waves to reach the ears

$d_B=343\times 0.45=154.35m$

(c)As the angle b/w slight lines from the two spectators to the player is right angle,

hypotenuse=the distance b/w 2 spectators

and, the slight lines are the other 2 lines

$D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m$

5 0

3 years ago