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2 years ago

Sb-26 what information is most important when passing near a lighthouse

1 answer:

6 0

Most important information when passing near a lighthouse is the water depth.
Today there are many complex navigation instruments,but in the past the lighthouse were the only navigation sailors had. However, the lighthouse still is very important in the navigation. There is an information chart on every lighthouse that gives information about the location of the lighthouses and the depth of the water. With these information an experienced sailor can navigate even without modern instruments.

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One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned

liq [111]

Answer:

$T = 3183 N$

Explanation:

When the screw is turned by two turns then change in the length of the wire is given as

$\Delta L = 2(2\pi r)$

$\Delta L = 4\pi r$

$\Delta L = 4(\pi)(1.0 mm)$

$\Delta L = 12.56 mm$

now we know by the formula of Young's modulus

$Y = \frac{T/A}{\Delta L/L}$

so we have

$T = \frac{AY \Delta L}{L}$

$T = \frac{\pi (0.55 \times 10^{-3})^2(2.0 \times 10^{11})(12.56 \times 10^{-3})}{0.75}$

$T = 3183 N$

3 0

2 years ago

Which situation would create a field like the one shown here? A. an interaction between two magnetic south poles B. an interacti

alexgriva [62]

Answer:

you have no picture

Explanation:

if it shows the arrows going towards eachother it is north and south. if it is not it will be either north and north or south and south

4 0

2 years ago

Read 2 more answers

A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, t

LenKa [72]

Answer:

a) W = - 318.26 J, b) W = 0 , c) W = 318.275 J , d) W = 318.275 J , e) W = 0

Explanation:

The work is defined by

W = F .ds = F ds cos θ

Bold indicate vectors

We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces

Let's use trigonometry to break down weight

sin θ = Wₓ / W

Wₓ = W sin 60

cos θ = Wy / W

Wy = W cos 60

X axis

How the body is going at constant speed

fr - Wₓ = 0

fr = mg sin 60

fr = 15 9.8 sin 60

fr = 127.31 N

Y Axis

N - Wy = 0

N = mg cos 60

N = 15 9.8 cos 60

N = 73.5 N

Let's calculate the different jobs

a) The work of the force of gravity is

W = mg L cos θ

Where the angles are between the weight and the displacement is

θ = 60 + 90 = 150

W = 15 9.8 2.50 cos 150

W = - 318.26 J

b) The work of the normal force

From Newton's equations

N = Wy = W cos 60

N = mg cos 60

W = N L cos 90

W = 0

c) The work of the friction force

W = fr L cos 0

W = 127.31 2.50

W = 318.275 J

d) as the body is going at constant speed the force of the tape is equal to the force of friction

W = F L cos 0

W = 127.31 2.50

W = 318.275 J

e) the net force

F ’= fr - Wx = 0

W = F ’L cos 0

W = 0

4 0

2 years ago

The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a

navik [9.2K]

Answer:

230

Explanation:

$\omega$ = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

$K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J$

Energy used in one acceleration

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J$

Number of accelerations would be given by

$n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4$

So the number of complete accelerations is 230

8 0

2 years ago

Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f

stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2$

When s = Db, t = 2t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2$

Dividing the two equations

$\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db$

Hence, Da=(1/4)Db

3 0

2 years ago