The correct answer is letter A. red. Young red stars would most likely be found in a giant molecular cloud. Red stars<span> are cooler than the sun, with surface temperatures of 3,500 K for a bright red star</span>

6 0

3 years ago

Data:

____ [38]

a. The speed of the tsunami at ocean depth of 8000 m is 1,019.66 km/h.

b. The speed of the tsunami at ocean depth of 3500 m is 660.75 km/h.

c. The speed of the tsunami at ocean depth of 70 m is 89.66 km/h.

d. The speed of the tsunami at ocean depth of 5 m is 30.625 km/h.

<h3>Speed of the tsunami at the given depths</h3>

The speed of the tsunami is calculated from method of extrapolation or interpolation as shown below;

<h3>Speed at 8000 meters</h3>

8000 m -------- ?

7000 m -------- 943 km/h

4000 m -------- 713 km/h

(8000 - 7000)/(7000 - 4000) = (? - 943)/(943 - 713)

0.3333 = (? - 943)/230

230(0.3333) = ? - 943

76.66 = ? - 943

? = 1,019.66 km/h

<h3>Speed at 3500 m</h3>

4000 m ---------- 713 km/h

3500 m ---------- ?

2000 m ----------- 504 km/h

(4000 - 3500)/(4000 - 2000) = (713 - ?) / (713 - 504)

0.25 = (713 - ?) /209

0.25(209) = 713 - ?

52.25 = 713 - ?

? = 660.75 km/h

<h3>Speed at 70 m</h3>

200 m ------ 159 km/h

70 m --------- ?

50 m ------- 79 km/h

(200 - 70)/(200 - 50) = (159 - ?)/(159 - 79)

0.8667 = 159 - ? / 80

80(0.8667) = 159 - ?

69.336 = 159 - ?

? = 89.66 km/h

<h3>Speed at 5 meters</h3>

50 m ------- 79 km/h

10 m --------- 36 km/h

5 m ---------- ?

(50 - 5)/(50 - 10) = (79 - ?)/(79 - 36)

1.125 = 79 - ?/43

43(1.125) = 79 - ?

48.375 = 79 - ?

? = 30.625 km/h

Learn more about speed of tsunami here: brainly.com/question/23136161

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8 0

1 year ago

A block is sliding down an inclined plane that makes an angle of 65o with the horizontal. If the coefficient of kinetic friction

Nezavi [6.7K]

Answer:

8.2 m/s²

Explanation:

m = mass of the block

μ = Coefficient of kinetic friction = 0.17

$F_{n}$ = Normal force on the block by the ramp

$f_{k}$ = kinetic frictional force

Force equation perpendicular to ramp surface is given as

$F_{n} = mg Cos65$

Kinetic frictional force is given as

$f_{k} = \mu F_{n}$

$f_{k} = \mu mg Cos65$

Force equation parallel to ramp surface is given as

$mg Sin65 - f_{k} = ma$

$mg Sin65 - \mu mg Cos65 = ma$

$g Sin65 - \mu g Cos65 = a$

$(9.8) Sin65 - (0.17) (9.8) Cos65 = a$

$a = 8.2$ m/s²

4 0

4 years ago

You need to calibrate your giant paintball bazooka, so you're fired at an angle of 60° and you find that the giant paintball spl

Alexxx [7]

Answer:

Vx = V0 cos 60 = V0 ./ 2 horizontal speed

Vy = V0 sin 69 = V0 * .866 = .866 V0 vertical speed

t = 107 / Vx = 214 / V0 time to travel 107 m

h = Vy t - 1/2 * g t*2 time to fall 1.7 m

-1.7 = .866 V0 * 214 / V0 - 4.9 * (214 / V0)^2 substituting

224400 / V0^2 = 187

V0 = (224400 / 187)^1/2 = 2=34.64 m/s

t = 214 / 34.64 = 6.178 stc

Check: h = Vy t - 1/2 g t^2

-1.7 = .866 * 34.64 * 6.178 - 4.9 * 6.178^2

-1.7 = -1.69

5 0

3 years ago