A car traveling west at 70 miles per hour for 3 hours covers a distance of

Which type of boundary is modeled?

Katena32 [7]

A is convergent
B is

7 0

3 years ago

Which equation describes the fastest runner?

irina [24]

I think the answer will be A

7 0

3 years ago

Read 2 more answers

physics A river flows at a speed vr = 5.37 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shorelin

vova2212 [387]

Answer: Vb is the vector (-5.37m/s, 8.59 m/s), with a module 10.13m/s

then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9

Explanation:

We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.

We have Vr = (5.37m/s, 0m/s)

Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).

The velocity of the boat Vb will be:

Vb = (-c*sin(32). c*cos(32))

Then we should have that:

5.37 m/s - c*sin(32) = 0

c = (5.37/sin(32))m/s = 10.13 m/s

the velocity in the y-axis is:

10.13m/s*cos(32) = 8.59 m/s

So Vb = (-5.37m/s, 8.59 m/s)

the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.

7 0

3 years ago

If you know the amount of air pressure exerted on a tabletop, how can you calculate the force exerted on the tabletop? Multiply

babymother [125]

Answer:

Multiply the air pressure by the area of the tabletop.

Explanation:

The relationship between pressure, force and area is given by:

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

where in this case, p is the air pressure, F is the force exerted and A the area of the tabletop. By re-arranging the equation, we can solve for F, the force exerted:

$p=\frac{F}{A}\\p\cdot A=\frac{F}{A}\cdot A\\pA=F$

So, the correct answer is:

The force exerted on the tabletop can be found by multiplying the air pressure by the area of the tabletop.

7 0

3 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c

seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

$A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2$

So the volume flow rate along the pipe is

$\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s$

We can use the similar logic to find the cross-section area at the refinery

$A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2$

The radius of the pipe at the refinery is:

$A_r = \pi r^2$

$r^2 =A_r/\pi = 0.446/\pi = 0.141$

$r = \sqrt{0.141} = 0.377m$

So the diameter is twice the radius = 0.38*2 = 0.754m

6 0

3 years ago