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Serga [27]
1 year ago
10

The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the averag

e depth of the water in kilometers. What is the approximate depth of water for a tsunami traveling at 200 kilometers per hour?.
Physics
1 answer:
Goshia [24]1 year ago
7 0

0.32 km

How to reach the answer:

The formula below provides the tsunami's speed in response to the query.

s = 356√d

when the tsunami's speed is 200 km/h, the equation's "s" is changed to "200," and the equation of speed is then expressed as follows.

200 = 356√d

The equation can be changed to compute the depth of the water as,

√d = 200÷356

√d = 0.562

Now on squaring both sides:

(√d)² = (0.562)²

d = (0.562)² = 0.316 or 0.32

Therefore the approximate depth of water for a tsunami travelling at 200km/h is 0.32 km

Know more about speed here:

brainly.com/question/19930939

#SPJ4

​

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A car interior is heated to 48 C by the sun What type of energy transfer?
Aneli [31]

Answer:

Radiation heat energy transfer

Explanation:

The type of heat transfer from the Sun is radiation heat transfer, which is the transfer of heat through electromagnetic radiation

The distance of the Sun to the Earth is several million kilometers away, with the space between being composes of vacuum and the nuclear reaction in the Sun's core generates vast amount of electromagnetic radiation that is transferred all across the universe and reaches the Earth as visible light and radiant energy at the speed of light

The radiant energy transferred from the Sun heats up the Earth, including the car's interior.

7 0
3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
2 years ago
A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is th
Soloha48 [4]

Answer:

The velocity of mass 2m is  v_B = 0.67 m/s

Explanation:

From the question w are told that

     The mass of the billiard ball A is =m

     The initial speed  of the billiard ball A = v_1 =1 m/s

    The mass of the billiard ball B is = 2 m

    The initial speed  of the billiard ball  B = 0

Let the final speed  of the billiard ball A  = v_A

Let The finial speed  of the billiard ball  B = v_B

      According to the law of conservation of Energy

                 \frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2

              Substituting values  

                \frac{1}{2} m (1)^2  = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2

Multiplying through by \frac{1}{2}m

                1 =v_A^2 + 2 v_B ^2 ---(1)

    According to the law of conservation of Momentum

            mv_1 + 2m(0) = mv_A + 2m v_B

    Substituting values

            m(1)  = mv_A + 2mv_B

Multiplying through by m

           1 = v_A + 2v_B ---(2)

making v_A subject of the equation 2

            v_A = 1 - 2v_B

Substituting this into equation 1

         (1 -2v_B)^2 + 2v_B^2 = 1

         1 - 4v_B + 4v_B^2 + 2v_B^2 =1

          6v_B^2  -4v_B +1 =1

          6v_B^2 -4v_B =0

Multiplying through by \frac{1}{v_B}

          6v_B -4 = 0

            v_B = \frac{4}{6}

            v_B = 0.67 m/s

4 0
3 years ago
1. The planet Jupiter completes a revolution of the sun in 11.5 years. Express it in seconds. Given that one year= 3.154 × 10^7
xenn [34]

Answer:

The planet Jupiter completes one revolution of the sun in 362710000 seconds. Long time, right?

Explanation:

3.154x10^7=3.154x10000000=31540000

11.5x31540000=362710000

7 0
1 year ago
Which statement accurately describes the illustration?
kvv77 [185]
The answer is B. I hope this helps! :)
3 0
3 years ago
Read 2 more answers
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