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3 years ago

What type of boat requires navigation lights?

1 answer:

4 0

All the boats that are operating from sunset to sunrise must have navigation lights.
Navigation lights are also for our safety, and it should be powered on after sunset to sunrise. Because at night due to darkness, if it is on, it is an indication that boat is there. Some of the boats which should have navigation lights are sailboats, powerboats, rowboats etc.

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rectangular plate, whose streamwise dimension (or chord c) is 0.2 m and whose width (or span b) is 1.8 m, is mounted in a wind t

Andrei [34K]

Answer:

See explaination for the details of the answer.

Explanation:

Density is a measurement that compares the amount of matter an object has to its volume. An object with much matter in a certain volume has high density An object with little matter in the same amount of volume has a low density. Density is found by dividing the mass of an object by its volume.

The Calculate of the chordwise distribution of the skin friction coefficient and the displacement thickness is done and represented in the attachment.

Please kindly check attachment for the step by step answer.

8 0

4 years ago

A horizontal spring-mass system has low friction, spring stiffness 205 N/m, and mass 0.6 kg. The system is released with an init

Andreyy89

Answer:

(a). The maximum stretch during the motion is 20.7 cm

(b). The maximum speed during the motion is 3.84 m/s.

(c). The energy is 0.060 Watt.

Explanation:

Given that,

Spring stiffness = 205 N/m

Mass = 0.6 kg

Compression of spring = 13 cm

Initial speed = 3 m/s

(a). We need to calculate the maximum stretch during the motion

Using conservation of energy

$E_{initial}=E_{final}$

$\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_{m}^2$

Put the value into the formula

$\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times205\times x_{m}^2$

$x_{m}=\sqrt{\dfrac{4.43\times2}{205}}$

$x_{m}=20.7\ cm$

(b). Maximum speed comes when stretch is zero.

We need to calculate the maximum speed during the motion

Using conservation of energy

$E_{initial}=E_{final}$

$\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}mv'^2$

Put the value into the formula

$\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times0.6\times v'^2$

$v'=\sqrt{\dfrac{4.43\times2}{0.6}}$

$v'=3.84\ m/s$

(c). Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system

We need to calculate the time period

Using formula of time period

$T=2\pi\sqrt{\dfrac{m}{k}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{0.6}{205}}$

$T=0.33\ sec$

We need to calculate the energy

Using formula of energy

$E=\dfrac{P}{t}$

Put the value into the formula

$E=\dfrac{0.02}{0.33}$

$E=0.060\ Watt$

Hence, (a). The maximum stretch during the motion is 20.7 cm

(b). The maximum speed during the motion is 3.84 m/s.

(c). The energy is 0.060 Watt.

5 0

3 years ago

What type of energy transformation is taking place when natural gas is used to heat water

bazaltina [42]

Thermal energy transformation is taking place.

3 0

3 years ago

A) Find the gravitational field strength of an asteroid with the mass of 3.2 * 10^3 kg and an average radius of 30 km when at a

MrMuchimi

a) $1.96\cdot 10^{-16} m/s^2$

The gravitational field strength near the surface of the asteroid is given by:

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the asteroid

R the radius of the asteroid

h is the distance from the surface

Substituting the data of the asteroid:

$M=3.2\cdot 10^3 kg$ is the mass

$R=30 km = 30000 m$ is the radius of the asteroid

$h=3 km = 3000 m$ is the distance from the surface

We find

$g=\frac{(6.67\cdot 10^{-11})(3.2\cdot 10^3)}{(30000+3000)^2}=1.96\cdot 10^{-16} m/s^2$

b) i) $5.53\cdot 10^9 s$

The acceleration of the astronaut popped out at 3 km from the surface is exactly that calculated at part a):

$g=1.96\cdot 10^{-16} m/s^2$

So, since its motion is at constant acceleration, we can find the time he takes to reach the surface using suvat equations:

$s=ut+\frac{1}{2}gt^2$

where

s = 3 km = 3000 m is his displacement to reach the surface

u = 0 is his initial velocity

t is the time

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(3000)}{1.96\cdot 10^{-16} m/s^2}}=5.53\cdot 10^9 s$

b) ii) $1.08\cdot 10^{-6} m/s$

Again, we can use another suvat equation:

$v=u+gt$

where

v is the final velocity

u is the initial velocity

g is the acceleration of gravity

t is the time

Since we have

u = 0

$t=5.53\cdot 10^9 s$

$g=1.96\cdot 10^{-16} m/s^2$

The velocity of the astronaut at the surface will be

$v=0+(1.96\cdot 10^{-16} m/s^2)(5.53\cdot 10^9)=1.08\cdot 10^{-6} m/s$

b) iii) 175 years

The duration of one year here is

$T=3.16\cdot 10^7 s$

And the time it takes for the astronaut to reach the surface of the asteroid is

$t=5.53\cdot 10^9 s$

Therefore, to find the number of years, we just need to divide the total time by the duration of one year:

$n=\frac{t}{T}=\frac{5.53\cdot 10^9 s}{3.16\cdot 10^7}=175$

So, the astronaut will take 175 years to reach the surface.

8 0

4 years ago

A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by V =[5.00m/s

beks73 [17]

Equation of velocity is given as

$V =[5.00m/s−(0.0180m/s^3)t^2]i^ + [2.00m/s+(0.550m/s^2)t]j^$

at t = 7.93 s

$v = 3.87 \hat i + 6.36 \hat j$

so the magnitude of the velocity is given as

$v = \sqrt{3.87^2 + 6.36^2}$

$v = 7.45 m/s$

Part b)

the direction of the velocity is given as

$\theta = tan^{-1}\frac{6.36}{3.87}$

$\theta = 58.7 degree$

part c)

for acceleration we know that

$a = \frac{dv}{dt}$

$a = -0.036 t\hat i + 0.550\hat j$

at t = 7.93 s

$a = -0.285\hat i + 0.550\hat j$

magnitude is given as

$a = \sqrt{0.285^2 + 0.550^2}$

$a = 0.62 m/s^2$

Part d)

for the direction of the motion

$\theta = tan^{-1}\frac{0.550}{-0.285}$

$\theta = 117.4 degree$

4 0

3 years ago