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Citrus2011 [14]

3 years ago

15

In each of the following cases, determine where the car has no

1 answer:

valkas [14]3 years ago

8 0

Answer:

1.after

2.approaching

3.cruise

4.turning

Explanation:

if wrong sorry

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Pascal's Principle states that (a) If we apply pressure to a fluid in a sealed container, the pressure will be felt undiminished

nika2105 [10]

Answer:

a) If we apply pressure to a fluid in a sealed container, the pressure will be felt undiminished at every point in the fluid and on the walls of the container.

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area(A2) of the piston.

P=F/A

P1=P2

F1/ A1= F2/ A2

F2= F1* A2/ A1

The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

In an incompressible liquid, the volume and amount of mass does not vary when pressure is applied.

5 0

3 years ago

a train traveling at 30m/s comes to a complete stop in 20 sec.how far did the train go during the deceleration

elixir [45]

Answer:

600m

Explanation:

v=30m/s

t=20s

s=VT

20×30=600m

3 0

3 years ago

Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

3 0

3 years ago

A 65.0 kg football player is running at 10.0 m/s downfield when he catches a 0.420 kg football thrown in the same direction but

lukranit [14]

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

A car speeds over a hill past point A, as shown in the figure. What is the maximum speed the car can have at point A such that i

olchik [2.2K]

Answer:

see explanations below

Explanation:

At the point when the car leaves the track, the reaction on the road is zero, meaning that the centrifugal force equals the gravitation force, namely

mv^2/r = mg

Solve for v in SI units

v^2 = gr = 9.81 m/s^2 * 14.2 m = 139.302 m^2/s^2

v = sqrt(139.302) = 11.8 m/s

Answer: at 11.8 m/s (26.4 mph) car will leave the track.

3 0

3 years ago