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Ipatiy [6.2K]
3 years ago
6

. Define 1 watt power​

Physics
2 answers:
iris [78.8K]3 years ago
6 0

Answer:

1 watt power is defined as the 1 Joule of energy consumed per second..

Art [367]3 years ago
5 0

Answer:

If a body does 1j work in 1 sec time then the power of that body is called 1watt power

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Please help me on this 1 assignment im giving 25 points GOOD LUCK!
kati45 [8]

Answer:

Explanation:

im sorry if this doesn't help but i think  its A

3 0
3 years ago
A long wave with a period of about 15 minutes will travel across the oceans at a speed of approximately _______.
Temka [501]

The speed of the wave  is mathematically given as

v=2266.66m/s

A long wave with a period of about 15 minutes will travel across the oceans at a speed of approximately v=2266.66m/s

<h3>Speed of the wave</h3>

Question Parameters:

A long wave with a period of about 15 minutes

Generally the equation for the Wave velocity  is mathematically given as

v=\lambda * frequency

Where

f=1/t

Therefore

v=\lambda * frequency

v=\lambda * 1/t

Therefore, with wavelenght of the ocean as 34km

v=34*1000*1/15

v=2266.66m/s

For more information on Speed

brainly.com/question/4931057

7 0
2 years ago
How can a magnetic field be used to generate an electric current?
Irina18 [472]
<span>When a magnet moves near a wire, it's changing field causes the electrons in the wire to flow as electric current.</span>
3 0
3 years ago
Water is pumped through a pipe of diameter 15.0 cm, from the Colorado River up to Grand Canyon Village, located on the rim of th
Aleksandr [31]

Answer:

p= 1.50289×10⁷ N/m²

Explanation:

Given

HA = (564 m)................(River Elevation)

HB = (2096 m).............(Village Elevation)

Area = A =(π/4){Diameter}² = (π/4){0.15 m}² = 0.017671 m²

ρ = (1 gram/cm³) = (1000 kg/m³)........(Water Density)

p(pressure)=?

Solution

p=PA - PB

p= ρ*g*HB - ρ*g*HA

p= (ρ*g)*(HB - HA)

p= (1000×9.81 )×{2096  - 564}  

p= 1.50289×10⁷ N/m²

8 0
3 years ago
A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper
Levart [38]

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

     ∑  F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

          cos 41 = T₁ₓ / T1

          sin 41 = T₁y / T1

          T₁ₓ = T₁  cos 41

          T₁y= T₁  sin 41

for cable gold

           cos 63 = T₂ / T₂

           sin 63 = T_{2y} / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

              T₃ - W = 0

              T₃ = W

              T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

   - T₁ₓ + T₂ₓ = 0

  T₁ₓ = T₂ₓ

   T1 cos 41 = T2 cos 63

   T1 = T2 cos 63 / cos 41                (1)

y Axis

      T_{1y} + T_{2y} - T3 = 0

       T₁ sin 41 + T₂ sin 63 = T₃          (2)

to solve the system we substitute equation 1 in 2

        T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

         T₂ (cos 63 tan 41 + sin 63) = W

         T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

          T₂ = 200 / (cos 63 tan 41 + sin 63)

          T₂ = 200 / 1,2856

           T₂ = 155.6 N

we substitute in 1

            T₁ = T₂ cos 63 / cos 41

             T₁ = 155.6 cos63 / cos 41

             T₁ = 93.6 N

therefore the tension in each cable is

            T₁ = 93.6 N

             T₂ = 155.6 N

             T₃ = 200 N

6 0
2 years ago
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