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4 years ago

Where does friction oppose motion in a roller coaster car

1 answer:

7 0

In roller coasters, friction is a force that opposes motion and significantly slows the cars as they move on the track.<span> While it is easy to believe that friction is bad for the ride, it is one of the forces engineers consider in ensuring passengers have a safe ride</span>

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PLZ HELP AND FAST!!!!!!!!!!!!

9966 [12]

B is the anwer hoou[ that hrtoj

3 0

4 years ago

Read 2 more answers

Examine the following circuit diagrams. In which diagram will current flow through one resistor?

Hoochie [10]

Answer

the one resister will change because how fast the curcit is going because th rate will change f

Explanation:

4 0

4 years ago

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton

pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:

$F = K * \Delta{x}$

$2\:N = 4\:N/cm*\Delta{x}$

$4\:N/cm*\Delta{x} = 2\:N$

$\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}$

simplify by 2

$\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}$

$\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark$

Answer:

B.) 1/2 cm

_______________________

I Hope this helps, greetings ... Dexteright02! =)

7 0

3 years ago

Einstein's theory of general relativity is currently the best explanation of gravity.Why has this theory not been replaced with

alina1380 [7]

Answer:

Option C, It still explains the experimental evidence pertaining to gravity

Explanation:

Please find the attachment

8 0

3 years ago

EMERGENCY!! Doing my Physics homework now and I need help on this one please - all information provided, I'd be very VERY gratef

IrinaK [193]

(a) The vertical motion is accelerated by gravity. The horizontal component is constant (neglecting air resistance, if this is not a projectile motion, the horizontal component would also be accelerated)

(b) Vertical:

$v_y = 30\sin 40^\circ = 19.3\frac{m}{s}$

Horizontal:

$v_x = 30\frac{m}{s}\cos 40^\circ = 23.0 \frac{m}{s}$

$s_y = -\frac{1}{2}gt^2 +vt\,,\,\,\,s_y=0\\0 = -\frac{1}{2}gt^2 +vt\\\frac{1}{2}gt =v\\t = \frac{2v}{g}= \frac{2\cdot30\sin 40^\circ\frac{m}{s}}{9.8\frac{m}{s^2}}=3.9s$

The ball will be in the air for about 3.9 s.

(d) The range is the horizontal distance traveled. We know the ball is in the air for 3.9s and it moves with a horizontal velocity of 23 m/s. So:

$s_x = 23\frac{m}{s}\cdot 3.9 s = 89.7m$

The range is 89.7 meters.

6 0

3 years ago