A container of water is lifted vertically 3.0m

At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c

fiasKO [112]

Answer:

497.00977 N

3742514.97005

Explanation:

$\rho$ = Density of water = 1000 kg/m³

C = Drag coefficient = 0.09

v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

$F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N$

The drag force on the dolphin's nose is 497.00977 N

at 20°C

$\mu$ = Dynamic viscosity = $1.002\times 10^{-3}\ Pas$

Reynold's Number

$Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005$

The Reynolds number is 3742514.97005

8 0

3 years ago

What is angular velocity

butalik [34]

Answer:

angular velocity(ω) is the rate change of angular displacement.

ω=θ/t and it SI unit is rad/s

Explanation:

this is very similar with the definition of linear velocity (rate of change of displacement). it specifies the angular speed of an object and the axis about which the object is rotating.

7 0

3 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

3 years ago

Three out of four crashes happen at a speed of __________ miles per hour or less.

Nitella [24]

<span>The answer is 45 miles per hour or even less. Most crashes occur at a speed of 45 miles per hour or even less, and most of these accidents occur close to our homes. These crashes may also be caused by different factors, such as being drunk or sudden occurrences that are not controllable, which is why it is best to precede with caution when driving at crowded areas.</span>

5 0

3 years ago

A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from

o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

$\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}$