Can anyone tell me the answer to this problem:)

Where in a river would you find fish that are adapted to low light and low oxygen levels?

JulijaS [17]

No place in the river is characterized like this

7 0

4 years ago

Distance Between Two Cars A car leaves an intersection traveling west. Its position 4 sec later is 19 ft from the intersection.

faltersainse [42]

Answer:

The answer to the question is;

The rate at which the distance between the two cars is changing is equal to 14.4 ft/sec.

Explanation:

We note that the distance traveled by each car after 4 seconds is

Car A = 19 ft in the west direction.

Car B = 26 ft in the north direction

The distance between the two cars is given by the length of the hypotenuse side of a right angled triangle with the north being the y coordinate and the west being the x coordinate.

Therefore, let the distance between the two cars be s

we have

s² = x² + y²

= (19 ft)² + (26 ft)² = 1037 ft²

s = $\sqrt{1037 ft^2}$ = 32.202 ft.

The rate of change of the distance from their location 4 seconds after they commenced their journeys is given by;

Since s² = x² + y² we have

$\frac{ds^{2} }{dt} = \frac{dx^{2} }{dt} + \frac{dy^{2} }{dt}$

→ $2s\frac{ds }{dt} = 2x\frac{dx}{dt} + 2y\frac{dy }{dt}$ which gives

$s\frac{ds }{dt} = x\frac{dx}{dt} + y\frac{dy }{dt}$

We note that the speeds of the cars were given as

Car B moving north = 12 ft/sec, which is the y direction and

Car A moving west = 8 ft/sec which is the x direction.

Therefore

$\frac{dy }{dt}$ = 12 ft/sec and

$\frac{dx}{dt}$ = 8 ft/sec

$s\frac{ds }{dt} = x\frac{dx}{dt} + y\frac{dy }{dt}$ becomes

$32.202 ft.\frac{ds }{dt} = 19 ft \times 8 \frac{ft}{sec} + 26ft\times 12\frac{ft}{sec}$ = 464 ft²/sec

$\frac{ds }{dt} = \frac{464\frac{ft^{2} }{sec} }{32.202 ft.}$ = 14.409 ft/sec ≈ 14.4 ft/sec to one place of decimal.

3 0

3 years ago

A boat that has a speed of 6km / h must cross a 200m wide river perpendicular to the current that carries a speed of 1m / s. Cal

ivanzaharov [21]

Answer:

a) 1.94 m/s

b) 120 m

Explanation:

Convert km/h to m/s:

6 km/h = 1.67 m/s

a) The final speed is found with Pythagorean theorem:

v = √((1.67 m/s)² + (1 m/s)²)

v = 1.94 m/s

b) The time it takes the boat to cross the river is:

t = (200 m) / (1.67 m/s)

t = 120 s

The displacement in the direction of the current is:

x = (1 m/s) (120 s)

x = 120 m

3 0

3 years ago

Which list places the layers of the sun in the correct order from outermost to innermost?

Valentin [98]

Answer - corona, chromosphere, photosphere

4 0

3 years ago

. A projectile is fired with an initial velocity of 113 m słatan angle of 60.0

bonufazy [111]

Answer:

a) 9.99 s

b) 538 m

c) 20.5 s

d) 1160 m

Explanation:

Given:

x₀ = 0 m

y₀ = 49.0 m

v₀ = 113 m/s

θ = 60.0°

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

a) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find t.

vᵧ = aᵧ t + v₀ᵧ

(0 m/s) = (-9.8 m/s²) t + (113 sin 60.0° m/s)

t ≈ 9.99 s

b) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find y.

vᵧ² = v₀ᵧ² + 2aᵧ (y − y₀)

(0 m/s)² = (113 sin 60° m/s)² + 2 (-9.8 m/s²) (y − 49.0 m)

y ≈ 538 m

c) When the projectile lands, y = 0 m. Find t.

y = y₀ + v₀ᵧ t + ½ aᵧ t²

(0 m) = (49.0 m) + (113 sin 60° m/s) t + ½ (-9.8 m/s²) t²

You'll need to solve using quadratic formula:

t ≈ -0.489, 20.5

Since negative time doesn't apply here, t ≈ 20.5 s.

d) When the projectile lands, y = 0 m. Find x. (Use answer from part c).

x = x₀ + v₀ₓ t + ½ aₓ t²

x = (0 m) + (113 cos 60° m/s) (20.5 s) + ½ (0 m/s²) (20.5 s)²

x ≈ 1160 m

4 0

3 years ago