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kvasek [131]
3 years ago
8

The Earth moving round the Sun in a circular orbit is acted upon by a

Physics
1 answer:
zalisa [80]3 years ago
3 0
No I do not agree. It is because work is done when force acting o a body displace or covers certain displacement in the direction of force applied .
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. One long wire carries a current of 30 A along the entire x axis. A second-long wire carries a current of 40 A perpendicular to
d1i1m1o1n [39]

Answer:

magnitude of net magnetic field at given point is

B = 5 \times 10^{-6} T

Explanation:

As we know that magnetic field due to a long current carrying wire is given as

B = \frac{\mu_o i}{2\pi r}

here we we will find the magnetic field due to wire which is along x axis is given as

i = 30 A

r = 2 m

now we have

B_1 = \frac{4\pi \times 10^{-7} (30)}{2\pi (2m)}

B_1 = 3\times 10^{-6} T into the plane

Now similarly magnetic field due to another wire which is perpendicular to xy plane is given as

i = 40 A

r = 2 m

now we have

B_2 = \frac{4\pi \times 10^{-7} (40)}{2\pi (2m)}

B_2 = 4\times 10^{-6} T along + x direction

Since the two magnetic field is perpendicular to each other

So here net magnetic field is given as

B = \sqrt{B_1^2 + B_2^2}

B = 5 \times 10^{-6} T

5 0
3 years ago
Currently, is our planet an open or closed system?
Dmitrij [34]
Our planet is closed system because there is a limit of how much matter could be exchanged.
8 0
3 years ago
Water, with a density of 1000 kg/m3, flows out of a spigot, through a hose, and out a nozzle into the air. The hose has an inner
stepan [7]

Answer:

   P₁ = 2.3506 10⁵ Pa

Explanation:

For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

          A₁ v₁ = A₂ v₂

Let's look for the areas

          r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm

          r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm

          A₁ = π r₁²

          A₁ = π 1.125²

          A₁ = 3,976 cm²

          A₂ = π r₂²

          A₂ = π 0.1²

          A₂ = 0.0452 cm²

Now with the continuity equation we can look for the speed of water inside the hose

           v₁ = v₂ A₂ / A₁

           v₁ = 11.2 0.0452 / 3.976

           v₁ = 0.1273 m / s

Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)

           P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂

          P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂

Let's calculate

           P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25

           P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴

           P₁ = 2.3506 10⁵ Pa

7 0
3 years ago
You have created a document that uses the term Frank L. Wright over 30 times. You want to change every occurrence of this term t
fgiga [73]
Find and Replace dialog box is probably the most useful.
3 0
4 years ago
Which of the following motions has a straight line?
skad [1K]
The answer would probably be B.
8 0
3 years ago
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