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3 years ago

A transform boundary occurs where two tectonic plates _____.?

2 answers:

8 0

- - B. Move apart from one another.

This is because A transform boundary occurs where two tectonic plates split together and move the opposite sides from each other.

✡ Hope this helps! ✡

gladu [14]3 years ago

7 0

Where they slide over each other.

Transform boundaries are formed or occur when two plates slide past each other in a sideways motion. They do not tear or crunch into each other (but the rock in between them may be ground up) and therefore none of the spectacular features are seen such as occur in divergent and convergent boundaries.

In a transform boundary, neither plate is added to at the boundary nor destroyed. They are marked in some places by features like stream beds that have been split in half and the two halves moved in opposite directions.

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Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acti

Tems11 [23]

Answer:

The unknown force will be 18.116 lb.

Explanation:

Given that,

Three forces act on a flange as shown in figure.

The net force acting on the flange is a minimum.

$\dfrac{dF_{net}}{df}=0$

We need to calculate the unknown force

Using formula of net force

$\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}$

Put the value into the formula

$\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}$

$\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}$

The magnitude of net force,

$F_{net}=\sqrt{F_{x}^2+F_{y}^2}$

$F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}$

$F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}$

$F_{net}=\sqrt{F^2+4899.78+36.232F}$

On differentiating w.r.to F

$(\dfrac{dF_{net}}{dF})^2=2F+36.232$

$0=2F+36.232$

$F=-\dfrac{36.232}{2}$

$F=-18.116\ lb$

Negative sign shows the direction of force which is downward.

Hence, The unknown force will be 18.116 lb.

6 0

3 years ago

Identify the number of atoms on the reactant side of the chemical equation.

klio [65]

Hi, I believe this should belongs to the chemistry section, but nevertheless here is the solution :

Number of Fe atoms: (2) + (2) =4
Number of O atoms. : (3)
Number of H atoms: (3 X 2)=6
Total atoms: 4+3+6=13 atoms

5 0

3 years ago

Read 2 more answers

Calculate the amount of hcn that gives the lethal dose in a small laboratory room measuring 14 × 15 × 8.0ft. the density of air

vova2212 [387]

Since we are given the density and volume, then perhaps we can determine the amount in terms of the mass. All we have to do is find the volume in terms of cm³ so that it will cancel out with the cm³ in the density. The conversion is 1 ft = 30.48 cm. The solution is as follows:

V = (14 ft)(15 ft)(8 ft)(30.48 cm/1 ft)³ = 0.0593 cm³

The mass is equal to:
Mass = (0.00118g/cm³)(0.0593 cm³)
Mass = 7 grams of HCN

7 0

3 years ago

A campus bird spots a member of an opposing football team in an amusement park. The football player is on a ride where he goes a

kenny6666 [7]

Answer:

No..

Explanation:

As the bird releases the drop there is no internal force which will drive it into circular path but it will fall on tangent of the arc at the point of release because it has a tangential velocity same as bird. Path will be parabola in vertical plane.

As the person is on circular arc constantly moving it will never meet that drop.

6 0

3 years ago

the idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse

Finger [1]

Forward thrust has positive values and reverse thrust has negative values.

Thrust is a sudden push or pull in a certain direction.

Flight speed u = 150 km/h

1 km/h = $\frac{1}{3.6}$ km/s

therefore, 150 km/h = 41.67 km / s

The thrust force represents the horizontal or x-component of momentum equation:

$T = m_{exhaust} * U_{exhaust} - U_{flight}$

T = 50 * (150 - 41.67)

T = 5416.67 N

Therefore, the value of forward thrust is 5416.67 N.

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component,

thus thrust equation is:

$T = m_{exhaust} * U_{exhaust} - U_{flight}$

T = 50 * (0 - 41.67)

T = -2083.5 N

Therefore, the thrust force T is -2083.5 N in the reverse direction.

Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero because $U_{exhaust} = U_{flight} = 0\\$

T = 0

Therefore, there is no difference in two velocities in x direction.

The given question is incomplete, the complete question is,

"The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse thrust if the jet is properly deflected. Suppose that while the aircraft rolls down the runway at 150 km/h the idling engine consumes air at 50 kg/s and produces an exhaust velocity of 150 m/s.

a. What is the forward thrust of this engine?

b. What are the magnitude and direction (i.e., forward or reverse) if the exhaust is deflected 90 degree without affecting the mass flow?

c. What are the magnitude and direction of the thrust (forward or reverse) after the plane has come to a stop, with 90 degree exhaust deflection and an airflow of 40 kg/s?"

To know more about thrust,

brainly.com/question/14552836

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3 0

1 year ago