Question 23 of 32

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

2 years ago

Boxes A and B are in contact on a horizontal, frictionless surface. Box A has mass 21.0 kg and box B has mass 8.0 kg. A horizont

vfiekz [6]

Answer:

2.75 N

Explanation:

Given that,

Box A has a mass 21.0 kg and box B has a mass 8.0 kg.

A horizontal force of 100N is exerted on box A.

Let a be the acceleration of the system. Using second law of motion,

$F=(m_A+m_B)a\\\\a=\dfrac{F}{(m_A+m_B)}\\\\a=\dfrac{10}{(21+8)}\\\\a=0.344\ m/s^2$

Now applying Newton's second law to box B. So,

$F_A=m_Ba\\\\=8\times 0.344\\\\=2.75\ N$

So, 2.75 N is the force that box A exerts on box B.

8 0

3 years ago

An object is dropped from a platform100 ft high. Ignoring wind resistance, what will its speed be when it reaches the ground?

sweet-ann [11.9K]

Answer:

80 ft/s

Explanation:

Given:

Δy = 100 ft

v₀ = 0 ft/s

a = 32.2 ft/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 ft/s)² + 2 (32.2 ft/s²) (100 ft)

v = 80.2 ft/s

Rounded, the speed when it reaches the ground is 80 ft/s.

4 0

3 years ago

Greek engineers had the unenviable task of moving large columns from the quarries to the city. One engineer, Chersiphron, tried

nexus9112 [7]

I believe this should be everything you need:)

7 0

3 years ago

A 1.0 kg ball at the end of a 2.0 m string swings in a vertical plane. At its lowest point the ball is moving with a speed of 10

Trava [24]

Answer:

a) 4.65m/s

b) 59.8 N , 1.01125 N

Explanation:

a)

m = mass of the ball = 1 kg

r = length of the string = 2.0 m

h = height gained by the ball as it moves from lowest to topmost position = 2r = 2 x 2 = 4 m

v = speed at the lowest position = 10 m/s

v' = speed at the topmost position = ?

Using conservation of energy

Kinetic energy at topmost position + Potential energy at topmost position = Kinetic energy at lowest position

(0.5) m v'² + m g h = (0.5) m v²

(0.5) v'² + g h = (0.5) v²

(0.5) v'² + (9.8 x 4) = (0.5) (10)²

v' = 4.65m/s

b)

T' = Tension force in the string when the ball is at topmost position

T = Tension force in the string when the ball is at lowest position

At the topmost position:

force equation is given as

$mg + T' = \frac{m v'^{2}}{r}$

$(1)(9.8) + T' = \frac{(1) (4.65)^{2}}{2}$

T' = 1.01125 N

At the lowest position:

force equation is given as

$T - mg = \frac{m v^{2}}{r}$

$T - (1) (9.8) = \frac{(1) (10)^{2}}{2}$

T = 59.8 N

8 0

4 years ago