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MissTica
3 years ago
11

Please answer these diagrammatic questions ASAP and please no spam answers​

Physics
1 answer:
SVEN [57.7K]3 years ago
3 0

Answer:

i. The pressure of due to the water, <em>P</em>, is given according to the following equation;

P = ρ·g·h

Where;

ρ = The density of the water (a constant) = 997 kg/m³

g = The acceleration due to gravity = 9.81 m/s²

h = The height of the water (minimum h = h₁, maximum h = h₂)

The pressure is directly proportional to the water height, and we have;

The pressure, <em>P</em>, will be maximum when the water height, <em>h</em>, is maximum or h = h₂, which is the level DC

ii. The thrust = The force acting on the body = Pressure × Area

The maximum areas exposed to the water are on side AB and DC

However, the pressure at level DC, which is the location of the maximum pressure, is larger than the pressure at level AB, therefore, the maximum thrust will be at the level DC

Explanation:

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The voltage in a circuit is given by the equation V = IR . In this equation , Vis the voltage, the current, and Ris the resistan
Mumz [18]

Answer:

The choice D.

R =  \frac{v}{I}

I hope I helped you^_^

8 0
2 years ago
Two similar fans are operating in a room. Fan 1 makes a squeaking noise while running. Fan 2 is silent.
REY [17]

Answer:

a

Explanation:

because it has more energy

8 0
2 years ago
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A 2300 kg truck has put its front bumper against the rear bumper of a 2500 kg suv to give it a push. with the engine at full pow
valentina_108 [34]
F=ma
m=total mass = 2300kg+2500kg=4800
F=18000N
a=?
a=F/m
a=18000/4800
a=3.8m/s^2
Final answer
7 0
3 years ago
Read 2 more answers
A motorboat traveling with a current can go 160 km in 4 hours. against the current it takes 5 hours to go the same distance. Fin
MatroZZZ [7]
<h2>Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.</h2>

Explanation:

Let speed of motor boat be m and speed of current be c.

A motorboat traveling with a current can go 160 km in 4 hours.

   Distance = 160 km

   Time = 4 hours

    Speed = m + c

   We have

            Distance = Speed x Time

            160 = (m+c) x 4

            m + c = 40     --------------------- eqn 1

Against the current it takes 5 hours to go the same distance.

   Distance = 160 km

   Time = 5 hours

    Speed = m - c

   We have

            Distance = Speed x Time

            160 = (m-c) x 5

            m - c = 32     --------------------- eqn 2

eqn 1 + eqn 2

           2m = 40 + 32

             m = 36 km/hr

Substituting in eqn 1

               36 + c = 40

                      c = 4 km/hr

Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.

3 0
3 years ago
1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant
Vladimir79 [104]

Answer:

The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction (v_j) = 550 mph

Velocity of jet stream from west to east direction (v_s)=80\ mph

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph

Similarly, the velocity of the stream is, \vec{v_s}=80\vec{i}

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle \theta with the x axis in the third quadrant.

The direction is given as:

\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

5 0
2 years ago
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