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xenn [34]
3 years ago
10

Sail boats utilize energy from the wind. Wind energy is a form of

Physics
1 answer:
yarga [219]3 years ago
6 0
A. kinetic energy hope this helps
You might be interested in
A solenoid is designed to produce a magnetic field of 3.50×10^−2 T at its center. It has a radius of 1.80 cm and a length of 46.
Anna11 [10]

Answer:

a. 2143 turns/m

b. 111.5 m

Explanation:

a. The minimum number of turns per unit length (N/L) can be found using the following equation:

B = \frac{\mu_{0}NI}{L}

\frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m

Hence, the minimum number of turns per unit length is 2143 turns/m.

b. The total length of wire is the following:

N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns

Since each turn has length 2πr of wire, the total length is:

L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m

Therefore, the total length of wire required is 111.5 m.

I hope it helps you!

4 0
3 years ago
Which planet has a moon with many sulfur volcanoes?
Murljashka [212]

Jupiter i hope it is right answer

6 0
3 years ago
A force of 210 N is applied to an object and the object accelerates at 14m/s2. Determine the mass of the object in kg.
lisabon 2012 [21]

Answer:

15 kg

Explanation:

F = m*a

F/a = m

210/14 = 15 kg

7 0
3 years ago
What force is needed to accelerate a 1,000-kilogram car from a stop to 5 m/s/s?
solong [7]
That depends on how soon you want it to reach 5 m/s/s. Without friction, ANY force will accelerate the car, like a mosquito pushing on it, but a Space Shuttle booster will accelerate it at a greater rate.
3 0
3 years ago
If the earth's magnetic field has strength 0.50 gauss and makes an angle of 20.0 degrees with the garage floor, calculate the ch
lys-0071 [83]

Answer:

ΔΦ = -3.39*10^-6

Explanation:

Given:-

- The given magnetic field strength B = 0.50 gauss

- The angle between earth magnetic field and garage floor ∅ = 20 °

- The loop is rotated by 90 degree.

- The radius of the coil r = 19 cm

Find:

calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.

Solution:

- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.

- The strength of magnetic field B and the are of the loop A remains constant. So we have:

                         Φ = B*A*cos(θ)

                         ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )

- The initial angle θ_1 between the normal to the coil and B was:

                         θ_1  = 90° -  ∅

                         θ_1  = 90° -  20° = 70°

The angle θ_2 after rotation between the normal to the coil and B was:

                         θ_2  =  ∅

                         θ_2  = 20°

- Hence, the change in flux can be calculated:

                        ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )

                        ΔΦ = -3.39*10^-6

                       

8 0
3 years ago
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