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3 years ago

A 14-foot boat operating under oars after dark must show what color light to prevent a collision?

1 answer:

7 0

<span>A "White" colored light must be shown by a 14-foot boat that is operating under oars after it is dark, in order to prevent a collision.

Boats which are less than 23 feet cannot exceed a maximum speed of 7 knots. They need to exhibit an all-around white light when it is sailing in the dark.

The other practicable sidelights can also be used instead of the white lights, when they are sailing in international waters.
</span>

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Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it

d1i1m1o1n [39]

Answer:

Q = 165.95 cm³ / s, 1) v = $\sqrt{0.55^2 + 19.6 y}$ , 2) v = 2.05 m / s,

3) d₂ = 1.014 cm

Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

Q = v A

where Q is the flow rate, A is area and v is the velocity

the area of a circle is

A = π r²

radius and diameter are related

r = d / 2

substituting

A = π d²/4

Q = π/4 v d²

let's reduce the magnitudes

v = 0.55 m / s = 55 cm / s

let's calculate

Q = π/4 55 1.96²

Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

v = $\sqrt{0.55^2 + 2 \ 9.8\ y}$

v = $\sqrt{0.55^2 + 19.6 y}$

2) ask to calculate the velocity for y = 0.2 m

v = $\sqrt{0.55^2 + 19.6 \ 0.2}$

v = 2.05 m / s

3) We write the continuous equation for this point 2

Q = v₂ A₂

A₂ = Q / v₂

let us reduce to the same units of the SI system

Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

A₂ = 165.95 10⁻⁶ / 2.05

A₂ = 80,759 10⁻⁶ m²

area is

A₂ = π/4 d₂²

d₂ = $\sqrt{4 A_2 / \pi }$

d₂ = $\sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }$

d₂ = 10.14 10⁻³ m

d₂ = 1.014 cm

4 0

3 years ago

Electrical wire with a diameter of .5 cm is wound on a spool with a radius of 30 cm and a height of 24 cm.

kow [346]

Answer:

a) # lap = 301.59 rad , b) L = 90.48 m

Explanation:

a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm

# turns = 1 turn (24 cm / 0.5 cm)

# laps = 48 laps

Let's reduce to radians

# laps = 48 laps (2 round / 1 round)

# lap = 301.59 rad

b) Each lap gives a length equal to the length of the circle

L₀ = 2π R

L = # turns L₀

L = # turns 2π R

L = 48 2π 30

L = 9047.79 cm

L = 90.48 m

6 0

3 years ago

You observe that a negatively charged plastic pen repels a charged piece of magic tape. You then observe that the same piece of

hichkok12 [17]

Answer:

(2) The excess negative charge from the sphere spread out all over your body.

(7) After you touched it, the metal sphere was very nearly neutral.

Explanation:

Plastic pen repels magic tape so magic tape is also negatively charged . Further , magic tape repels small metal sphere that means small sphere also is negatively charged.

Now when small sphere is touched by a man insulated from ground , the charge is distributed between man and small sphere according to their capacitance .

Since human body will have greater capacitance ,it will acquire larger share of charge . Sphere being of very small size will retain very less charge and it will become almost neutral . Hence it will be attracted by charged tape .

5 0

3 years ago

7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If th

Ainat [17]

Answer:

Explanation:

Given

mass of children $m_1=20\ kg$

$m_2=30\ kg$

distance between two children $L=3\ m$

suppose small child is at a distance of x m from pivot point

so torque of small child and heavier child must be equal

$20\times (x)=30\times (3-x)$

$2x=9-3x$

$5x=9$

$x=1.8\ m$

7 0

3 years ago

A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t

MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2 m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0

3 years ago