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What precautions should be taken on board a vessel during a lightning storm?

dusya [7]

The precautions which should be taken on board a vessel during a lightning storm are:

(i) All electrical equipments should be unplugged.

(ii) Keep yourself away from all metal parts such as engines, metal tanks, stoves etc.

(iii) Stay inside the boat.

7 0

3 years ago

Read 2 more answers

What happens to speed when wavelength decreases

Phantasy [73]

Speed = wavelength × frequency

giving that frequency is 0, wavelength and speed are directionally proportional. wavelength decrease = speed decrease

6 0

3 years ago

A box of unknown mass is sliding with an initial speed vi = 5.60 m/s across a horizontal frictionless warehouse floor when it en

Maksim231197 [3]

We know that the Delta E + W(Work done by non-conservative forces) = 0 (change of energy)

In here, the non-conservative force is the friction force where f = uN (u =kinetic friction coefficient)

W= f x d = uNd ; N=mg
Delta E = 1/2 mV^2 -1/2mVi^2

umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term)

This will then give us:

1/2Vi^2-ugd = 1/2V^2

V^2 = Vi^2 - 2ugd

So plugging in our values, will give us:

V= Sqrt (5.6^2 -2.3^2)

=sqrt (26.07)

= 5.11 m/s

6 0

3 years ago

How long would it take a machine to do 5.000

Gala2k [10]

Answer:

a

Explanation:

How long would it take a machine to do 5.000

joules of work if the power rating of the machine

is 100 watts?

6 0

3 years ago

Spring compressed 10cm by 100N force and held in place with Pin. Pin is pulled and block is pushed Up the incline. Uk(coefficien

otez555 [7]

The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s

The height at which the block stops rising is approximately 1.1415 m

The length of the incline is approximately 1.536 m

<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

$Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m} = \mathbf{ 1000\, N/m}$

Coefficient of kinetic friction, $\mu_k$ = 0.39

Therefore, we have;

Friction force = $\mathbf{\mu_k}$ ·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, W ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

$\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d$

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, W ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

The velocity of the block after the Spring extends 7 cm, v₂ ≈ 1.73 m/s

The height at which the block will stop moving, h, is given as follows;

$At \ the \ maximum \ height, \ h, \ we \ have ; \ \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x$

Which gives;

$Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{ 7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2 } \approx 1.536$

The distance up the inclined, the block rises, at maximum height is therefore;

$x_{max}$ ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

The height at which the block stops rising, h ≈ 1.1415 m

From the above solution for the height, the length of the incline is he

distance along the incline at maximum height which is therefore;

Length of the incline, $x_{max}$ = 1.536 m

Learn more about conservation of energy here:

brainly.com/question/7538238

5 0

2 years ago