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Snowcat [4.5K]
2 years ago
5

he drag characteristics of a torpedo are to be studied in a water tunnel using a 1 : 7 scale model. The tunnel operates with fre

shwater at 20 ˚C, whereas the prototype torpedo is to be used in seawater at 15.6 ˚C. To correctly simulate the behavior of the prototype moving with a velocity of 53 m/s, what velocity is required in the water tunnel?
Physics
1 answer:
dusya [7]2 years ago
4 0

Answer:20.03 m/s

Explanation:

Given

L_r=1:7

velocity of Prototype v_p=53 m/s

Taking Froude number same for both flow as it is a dimensionless number for different flow regimes in open Flow

(\frac{v_m}{\sqrt{L_mg}})=(\frac{v_p}{\sqrt{L_pg}})

v_m=v_p\times \sqrt{\frac{L_m}{L_p}}

v_m=53\times \frac{1}{\sqrt{7}}

v_m=20.03 m/s

           

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yKpoI14uk [10]

Answer:

m = 3.91 kg

Explanation:

Given that,

Mass of the object, m = 3.74 kg

Stretching in the spring, x = 0.0161 m

The frequency of vibration, f = 3.84 Hz

When the object is suspended, the gravitational force is balanced by the spring force as :

mg=kx

k=\dfrac{mg}{x}

k=\dfrac{3.74\times 9.8}{0.0161}

k = 2276.52 N/m

The frequency of vibration is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

m=\dfrac{k}{4\pi^2f^2}

m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}

m = 3.91 kg

So, the mass of the object is 3.91 kg. Hence, this is the required solution.

8 0
3 years ago
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
3 years ago
Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis.
amm1812

Answer:

38 cm from q1(right)

Explanation:

Given, q1 = 3q2 , r = 60cm = 0.6 m

Let that point be situated at a distance of 'x' m from q1.

Electric field must be same from both sides to be in equilibrium(where EF is 0).

=> k q1/x² = k q2/(0.6 - x)²

=> q1(0.6 - x)² = q2(x)²

=> 3q2(0.6 - x)² = q2(x)²

=> 3(0.6 - x)² = x²

=> √3(0.6 - x) = ± x

=> 0.6√3 = x(1 + √3)

=> 1.03/2.73 = x

≈ 0.38 m = 38 cm = x

8 0
3 years ago
A ship sets out to sail to a point 120 km due north. an unexpected storm blows the ship to a point 100 km due east of its starti
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If you draw the problem, it would look like that shown in the attached picture. The total length the ship will now travel can be solved using the Pythagorean theorem. The solution is as follows:

d = √(120)²+(100)²
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So, the ship will have to travel 156.2 km to the northwest direction.

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nika2105 [10]
Moment about the pivot must be equal for the seesaw to balance. Initially, the first cat and the bowl are at 2 m from the pivot.

The moment due to cat = 5.3*2 = 10.6 kg.m
The moment due to bowl = 2.5*2 = 5 kg.m
The unbalanced moment = 10.6 - 5 = 5.6 kg.m

Therefore, the 3.7 kg cat should stand at a distance x from the pivot in left to balance the 5.6 kg.m.
That is,
3.7*x = 5.6 => x = 5.6/3.7 = 1.5134 m to the left (on the side of the bowl)
7 0
3 years ago
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