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2 years ago

5

Consider a sign suspended on a boom that is supported by a cable, as shown. What is the proper equation to use for finding the n

et force in the y direction?
Fnety = (FT)(sin 32°) + Fg
Fnety = (FT)(sin 32°) – Fg
Fnety = (FT)(cos 32°) + Fg
Fnety = (FT)(cos 32°) – Fg

2 answers:

-Dominant- [34]2 years ago

8 0

Answer:

B: Fnety = (FT)(sin 32°) – Fg

Explanation:

oksian1 [2.3K]2 years ago

3 0

Fnety = (FT)(sin 32°) – Fg Or the answer B, I checked it.

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1. Suppose a tank filled with water has a liquid column with a height of 10 meters. If the area is 2 square meters (m²), what's

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Answers:

1. Firstly, we have to define that Pressure $P$ is Force applied $F$ per unit area $A$ . It is mathematically expressed as follows:

$P=\frac{F}{A}$ (1)

The unit of P is Pascal (Pa) which is equivalent to $\frac{kg}{ms^{2}}$ and also equivalent to $\frac{N}{m^{2} }$

There is also another expression of the Pressure in which it is dependent on the density $d$ of the liquid, the height $h$ of the container and the gravity force $g$ :

$P=d*h*g$ (2)

In this problem the liquid is water, and its known density is approximately:

$d=1000kg/m^{3}$

So, we have to substitute the values in equation (2) to obtain the pressure <u>(Being careful with the units)</u>:

$P=1000\frac{kg}{m^{3}}*10m*9.8\frac{m}{s^{2}}$

$P=98000Pa$

Then, we have to substitute this value in equation (1) and clear $F$ :

$F=P*A$

Finally:

$F=196000N$

2. For this problem, we will use equation (1) to find the Pressure. We already know the area $A$ and the force exerted by water in the container $F$ :

$P=\frac{F}{A}=\frac{900N}{3m^{2}}$

$P=300Pa$

3. In this case, equation (2) is the perfect way to find the hydrostatic pressure at any point at the bottom of the tank <u>(be careful with the units):</u>

$P=d*h*g$

$P=1000\frac{kg}{m^{3}}*7.5m*9.8\frac{m}{s^{2}}$

$P=73500Pa$

4. In this case, it's important to know that in fluids (in this case the water) the higher the fluid is, the lower the pressure. Then, if $P_{1}$ and $P_{2}$ are the respective pressures at the heights $h_{1}$ and $h_{2}$ , and knowing that the water density and the gravity force in this case are constants, we can use the following expression to solve this problem:

$P_{2}- P_{1} =d*g(h_{2}- h_{1})$ (3)

Where:

$P_{1}=1.5 kPa$ at $h_{1}=2m$

Note that $1kPa=1*1000 Pa$

And $P_{2}=?$ is unknown at a given height $h_{2}=6m$

Then, we have to substitute the values in equation (3) to find $P_{2}$ :

$P_{2}-1500Pa=1000\frac{kg}{m^{3}}*9.8\frac{m}{s^{2}} (6m-2m)$

Finally: $P_{2} =40700Pa$

5. In this case we have the area $A=0.75m^{2}$ and the mass of the piston $m=200kg$ , and we need to know the pressure $P$ .

We will use equation (1):

$P=\frac{F}{A}$

But, <u>do you remember that above we stated that pressure is the force applied over an area?</u>

Well, in this case we will use the following equation, in which the gravity force and the mass of a body are involved, to find $F$ :

$F=m*g=200kg*9.8\frac{m}{s^{2}}$

Then:

$F=1960N$

Now we can finally calculate $P$ :

$P=\frac{1960N}{0.75m^{2}}$

$P=2613.33Pa$

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Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit

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The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

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$Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s$

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

$Inlet volume = inlet velocity*area of circle=\pi r^{2}*inlet velocity$

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

$Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}$

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

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