When a muscle is working, it _________, or tightens, thus making itself stronger.

A pilot flies in a straight path for 1 hour 30 minutes. She then makes a course correction, heading 45 degrees to the right of h

mrs_skeptik [129]

Answer:

1199 miles

Explanation:

1 hour 30 minutes = 1 + 30/60 = 1.5 hours

2 hours 15 minutes = 2 + 15/60 = 2.25 hours

The distance she flew in the 1st segment is:

1.5*345 = 517.5 miles

The distance she flew in the 2nd segment is:

2.25 * 345 = 776.25 miles

Since the 2nd segment is 45 degree with respect to the 1st segment, this means that she has flown

776.25 * cos(45) = 549 miles in-line with the 1st segment and

776.25* sin(45) = 549 miles perpendicular to the 1st segment:

So the distance from the end to her starting position is

$\sqrt{(517.5 + 549)^2 + 549^2} = 1199 miles$

4 0

3 years ago

Two elements in Period 4 that are most similar to Cobalt?

lisabon 2012 [21]

Iron and nickel are the elements most similar to cobalt

5 0

3 years ago

Do it in order. <br> from smallest to largest

Sholpan [36]

Answer:

The earth, The sun, the solar system and the milky way.

7 0

2 years ago

Which object will have greater acceleration? Why?

dexar [7]

Answer:

Object D

Explanation:

Use Newton's Second Law to determine the acceleration that each object has.

F = ma

The force applied in both cases is 50 N, but the mass for object C and object D is different.

Let's start with object C first:

F = ma
50 N = 10 kg · a
50 = 10a
5 = a

The acceleration object C undergoes is 5 m/s².

Now let's calculate object D next:

F = ma
50 N = 2 kg * a
50 = 2a
25 = a

The acceleration object D undergoes is 25 m/s².

Object D has greater acceleration because it has a smaller mass. The object with a smaller mass will accelerate more in order to satisfy Newton's 2nd Law.

7 0

3 years ago

Suppose a car manufacturer tested its cars for front-en4 collisions by hauling them up on a crane and dropping then; from a cert

Brrunno [24]

Answer:

a

Generally from third equation of motion we have that

$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

u is the initial speed of the car which is zero

$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

$v^2 = 0 + 2g[H - 0]$

=> $v = \sqrt{ 2 g H}$

b

$H = 9.86 \ m$

Explanation:

Generally from third equation of motion we have that

$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

u is the initial speed of the car which is zero

$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

$v^2 = 0 + 2g[H - 0]$

=> $v = \sqrt{ 2 g H}$

When $v = 50 \ km/h = \frac{50 *1000}{3600} = 13.9 \ m/s$ we have that

$13.9 = \sqrt{ 2 g H}$

=> $H = \frac{13.9^2}{2 * 9.8}$

=> $H = 9.86 \ m$

6 0

3 years ago