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suter [353]
3 years ago
8

What type a medium is a ocean wave

Physics
2 answers:
andre [41]3 years ago
8 0
Ocean waves travel through the medium called
"ocean water" or "salt water" or simply "water".
Mama L [17]3 years ago
5 0

Answer:

ocean water i think

Explanation:

You might be interested in
Calculate the magnetic field and its direction at point P, which is 2.0 cm away from the top wire and 4.0 cm from the bottom wir
dusya [7]

Answer:

Explanation :  B = μ₀i / 2πr

B1 = (2×10^-7) × 1.5/0.02

       = 15 μT

B2 = (2×10^-7 ) ×  1.5/0.04

       = 7.5 μT

total field = 15 μT + 7.5 μT

                  = 22.5 μT

8 0
3 years ago
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts
Naddika [18.5K]

Answer:

Explanation:

75 N and 90 N are acting in opposite direction so net force = 90 - 75 = 15 N .

Friction force will act in the direction opposite to the direction of net force .

So friction force will act in the direction in which 75 N is acting .

Total force acting in the direction of 75 =  75 + 12 = 87 N

Net force acing on the third child = 90 - 87 = 3 N  

Its direction will be that in the direction of 90 N .

8 0
3 years ago
A tourist stands at the top of the Grand Canyon, holding a rock, overlooking the valley below. Find the final velocity, and disp
MrRa [10]

Answer:

a. -39.2 m/s; -78.4 m

b. -31.2 m/s; -46.4 m

c. -47.2 m/s; -110.4 m

Explanation:

<h2>Part (a)</h2>

We are given/can infer these variables:

  • t = 4.0 s
  • a = -9.8 m/s²
  • v_0 = 0 m/s

We want to find the displacement and the final velocity of the rock.

  • Δx = ?
  • v = ?

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = 0 + (-9.8)(4.0)
  • v = -9.8 * 4.0
  • v = -39.2 m/s

The final velocity of the rock is -39.2 m/s.

Now we can use this equation to find the displacement of the rock:

  • Δx = v_0 t + 1/2at²

Plug in the known variables into this equation.

  • Δx = 0 * 4.0 + 1/2(-9.8)(4.0)²
  • Δx = 1/2(-9.8)(4.0)²
  • Δx = -4.9 * 16
  • Δx = -78.4 m

The displacement of the rock is -78.4 m.

<h2>Part (b)</h2>

We are given/can infer these variables:

  • v_0 = 8.0 m/s
  • a = -9.8 m/s²
  • t = 4.0 s

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = 8.0 + (-9.8)(4.0)
  • v = 8.0 + -39.2
  • v = -31.2 m/s

The final velocity of the rock is -31.2 m/s.

We can use this equation to find the displacement:

  • Δx = v_0 t + 1/2at²

Plug in known variables:

  • Δx = 8.0(4.0) + 1/2(-9.8)(4.0)²
  • Δx = 32 - 4.9(16)
  • Δx = -46.4 m

The displacement of the rock is -46.4 m.

<h2>Part (c)</h2>

We are given/can infer these variables:

  • v_0 = -8.0 m/s
  • a = -9.8 m/s²
  • t = 4.0 s

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = -8.0 + (-9.8)(4.0)
  • v = -8.0 - 39.2
  • v = -47.2 m/s

The final velocity of the rock is -47.2 m/s.

We can use this equation to find the displacement:

  • Δx = v_0 t + 1/2at²

Plug in known variables:

  • Δx = -8.0(4.0) + 1/2(-9.8)(4.0)²
  • Δx = -32 - 4.9(16)
  • Δx = -110.4 m

The displacement of the rock is -110.4 m.

7 0
3 years ago
Read 2 more answers
If the balloon can barely lift an additional 3500 N of passengers, breakfast, and champagne when the outside air density is 1.23
Bingel [31]

The complete question is :

A hot-air balloon has a volume of 2100 m3 . The balloon fabric (the envelope) weighs 860 N . The basket with gear and full propane tanks weighs 1300 N .

If the balloon can barely lift an additional 3400 N of passengers, breakfast, and champagne when the outside air density is 1.23kg/m3, what is the average density of the heated gases in the envelope?

Solution :

Given volume of the hot air balloon $=2100 \ m^3$

The balloon fabric weights = 860 N

The weight of the basket with the gear and propane tank = 1300 N

Density of outside air $= 1.23 \ kg/m^3$

∴  Total pay load = Weight of the air displaced - weight of gas inside the balloon

Total pay load = 860 + 1300 + 3400

                        = 5560 N

Mass = density x volume

Weight  $= \text{mass} \times g$

Weight = volume x density $\times \text{ acceleration due to gravity (g)}$

Weight of the displaced air = 2100 x 1.23 x 9.8

                                             = 25313 N

Weight of the gas inside the balloon = density $\times \text{ acceleration due to gravity (g)}$ x volume

                                                             = density x 9.8 x 2100

                                                             = density x 20580 N

Therefore substituting the values, we get

⇒ 25313 - (density x 20580) = 5560

⇒ density $=\frac{19753}{20580}$

                 $= 0.96 \ kg/m^3$

So the density of the heated gas $= 0.96 \ kg/m^3$

8 0
3 years ago
A projectile is launched horizontally at a speed of 45.0 m/s from a
Free_Kalibri [48]

Answer:

answer will be c: 1.5 seconds :)

8 0
3 years ago
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