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2 years ago

7

What are the benefits of the cool-down period following exercise?

1 answer:

kvv77 [185]2 years ago

5 0

The benefits of the cool down period are quite important, it allows your body to slow your heart rate at a nice healthy safe pace, if you stop right away it can cause breathing, heart, and muscle problems.

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Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 197

Leno4ka [110]

(a) 3.58 km 45° south of east

The total displacement is given by:

$d=vt$

where

v is the average velocity

t is the time

The average velocity is:

v = 3.53 m/s

While we need to convert the time from minutes to seconds:

$t=169 min \cdot 60 s/min = 10140 s$

Therefore, the magnitude of the displacement is

$d=(3.53)(10140)=35794 m = 3.58 km$

And the direction is the same as the velocity, therefore 45° south of east.

(b) 5.53 m/s 90° south of east

The velocity of the air relative to the ground is

$v_a = 2.00 m/s$

and the direction is exactly opposite to that of Allen, so it is 45° north of west. Allen's velocity relative to the ground is

v = 3.53 m/s

So this must be the resultant of Allen's velocity relative to the air (v') and the air's speed ( $v_a$ ). Since these two vectors are in opposite direction, we have

$v= v'-v_a$

Therefore we find v', Allen's velocity relative to the air:

$v'=v+v_a = 3.53 + 2.00 = 5.53 m/s$

The direction must be measured relative to the air's reference frame. In this reference frame, Allen is moving exactly backward, so his direction will be 90° south of east.

(c) 56.1 km at 90° south of east.

Since Allen's velocity relative to the air is

v' = 5.53 m/s

Then the displacement of Allen relative to the air will be given by

$d'=v't$

and substituting,

$d'=(5.53)(10140)=56074 m = 56.1 km$

And the direction is the same as that of the velocity, therefore will be 90° south of east.

3 0

3 years ago

Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us

nalin [4]

Answer:

$13 W/m^2$

Explanation:

The apparent brightness follows an inverse square law, therefore we can write:

$I \propto \frac{1}{r^2}$

where I is the apparent brightness and r is the distance from the Sun.

We can also rewrite the law as

$\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2}$ (1)

where in this problem, we have:

$I_1 = 1300 W/m^2$ apparent brightness at a distance $r_1$ , where

$r_1 = 150$ million km

We want to estimate the apparent brightness at $r_2$ , where $r_2$ is ten times $r_1$ , so

$r_2 = 10 r_1$

Re-arranging eq.(1), we find $I_2$ :

$I_2 = \frac{r_1^2}{r_2^2}I_1 = \frac{r_1^2}{(10r_1)^2}(1300)=\frac{1}{100}(1300)=13 W/m^2$

5 0

3 years ago

3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the

andreev551 [17]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

hence, W(output) = 4.84 x 10^6 J

The useful work done on the mass is 4.84 x 10^6 J

5 0

2 years ago

If it takes you 20 joules of work to move a couch 10 meters in 10 seconds, what is the power?

Debora [2.8K]

Answer:

Power = 2 j/s

Explanation:

Power = Work / Time

= 20/10

= 2 j/s

4 0

2 years ago

Can someone help me please? I’ve been trying to solve these questions all day.

qwelly [4]

#16

If we put a resistor in circuit it will slow the speed of current

Let's check ohms law

$\\ \rm\Rrightarrow \dfrac{V}{I}=R$

So if resistance is more current is less

#17

Again use ohms law

$\\ \rm\Rrightarrow V=IR$

$\\ \rm\Rrightarrow V\propto I$

Voltage must be increased

4 0

2 years ago