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My name is Ann [436]

2 years ago

11

In surveying , supposing we can not pull the tape because it is passing through a shallow river. What will i do to obtain an acc

urate distance?

1 answer:

natulia [17]2 years ago

7 0

Answer:

Hold the tape in place and go down to the end.

Explanation:

You might be interested in

The absolute pressure in water at a depth of 9 m is read to be 185 kPa. Determine: a. The local atmospheric pressure b. The abso

ser-zykov [4K]

Answer:

a)Patm=135.95Kpa

b)Pabs=175.91Kpa

Explanation:

the absolute pressure is the sum of the water pressure plus the atmospheric pressure, which means that for point a we have the following equation

Pabs=Pw+Patm(1)

Where

Pabs=absolute pressure

Pw=Water pressure

Patm= atmospheric pressure

Water pressure is calculated with the following equation

Pw=γ.h(2)

where

γ=especific weight of water=9.81KN/M^3

H=depht

A)

Solving using ecuations 1 y 2

Patm=Pabs-Pw

Patm=185-9.81*5=135.95Kpa

B)

Solving using ecuations 1 y 2, and atmospheric pressure

Pabs=0.8x5x9.81+135.95=175.91Kpa

8 0

4 years ago

KVL holds for the supermesh, so we can write a KVL equation to generate the second equation we need to solve for the two unknown

kaheart [24]

Answer:

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

Explanation:

As the complete question is not given the complete question is found online and is attached herewith.

By applying KCL at node 1

$i_x+50mA=i_y\\i_x-i_y=0.05A$

Also

$V_{\Delta}=1K*i_y$

Now applying KVL on loop 1 as indicated in the attached figure

$1K*i_y+5K(i_y-i_z)+3K*i_x=0\\3i_x+6 i_y-5i_z=0$

Similarly for loop 2

$2V_{\Delta}+5K(i_z-i_y)=0\\2*1K*i_y+5K(i_z-i_y)=0\\2K*i_y+5K(i_z-i_y)=0\\3i_y-5i_z=0$

So the system of equations become

$i_x-i_y+0i_z=0.05\\3i_x+6i_y-5i_z=0\\0i_x-3i_y+5i_z=0$

Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as

$V_{\Delta}=1K*i_y\\V_{\Delta}=1K*-25 mA\\V_{\Delta}=-25 V$

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

8 0

4 years ago

What are the 4 types of electricity?

olchik [2.2K]

Answer:

Fossil Fuels 67% (Non-Renewable Source): Coal 41%, Natural Gas 21% & Oil 5.1%

Renewable Energy 16%

Mainly Hydroelectric 92%: Wind 6%, Geothermal 1%, Solar 1%

Nuclear Power 13%

Explanation:

3 0

3 years ago

determine the position d of the 6- kn load so that the average normal stress in each rod is the same.

Zinaida [17]

The load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2}$ area.

<h3>What is meant by torque?</h3>

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.

Let the beam is of length L

Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero

$N_1 * x=N_2 *(L-x)$

also, we know that stress at both ends are same

$\frac{N_1}{12}=\frac{N_2}{8}$

$2 * N_1=3 * N_2$

Now from two equations we have

$$\frac{3}{2} N_2 * x=N_2 *(L-x)$

solving the above equation we have

$$x=\frac{2}{5} L$

so the load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2}$ area.

The complete question is:

47. the beam is supported by two rods ab and cd that have cross-sectional areas of $$$12mm^2$ and $$$8mm^2$ , respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.

To learn more about torque refer to:

brainly.com/question/20691242

#SPJ4

7 0

2 years ago

Block A has mass of mA = 58kg and rests on a flat surface. The coefficient of static friction between the block and the surface

lesantik [10]

Answer:

The greatest mass, that the weight C can have such that block A does not move is approximately 23.259 kg

Explanation:

The given parameters are;

The mass of the block A= 58 kg

The coefficient of static friction between the block and the surface, $\mu _s$ = 0.300

The coefficient of static friction between the rope and the fixed peg, B $\mu _B$ = 0.310

Let T represent the tension in the rope

Therefore, when the rope is static, T = The normal reaction at the peg, B, $N_B$

The angle of inclination of the rope holding the block A = arctan(3/4) ≈ 36.87°

The length of the rope = √(0.4² + 0.3²) = 0.5

∴ sin(θ) = 3/5 = 0.6

cos(θ) = 4/5 = 0.8

The vertical component of the tension in the rope = T × sin(θ) = 0.6·T

The horizontal component of the tension in the rope = T × cos(θ) = 0.8·T

The friction force = μ×(W - 0.6·T) = 0.300×(58×9.8 - 0.6·T) = 170.52 - 0.18·T

The block will start to move when we have;

The horizontal component of the tension in the rope = The friction force

∴ 0.8·T = 170.52 - 0.18·T

0.8·T + 0.18·T = 170.52

0.98·T = 170.52

T = 170.52/0.98 = 174

Therefore, the tension in the rope = T = 174 N = The normal reaction at the peg, B $N_B$

The frictional force at the peg, $F_B$ = $\mu _B$ × $N_B$ = 0.310 × 174 N = 53.94 N

The weight of the mass, $m_c$ , $W_c$ = The frictional force at the peg, $F_B$ + The tension in the rope

∴ The weight of the mass, $m_c$ , $W_c$ = 53.94 N + 174 N = 227.94 N

Weight, W = Mass, m × The acceleration due to gravity, g, from which we have;

m = W/g

Where;

g = 9.8 m/s²

∴ $m_c$ = $W_c$ /g = 227.94 N/(9.8 m/s²) ≈ 23.259 kg.

The greatest mass, that the weight C can have such that block A does not move = $m_c$ ≈ 23.259 kg.

4 0

3 years ago