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3 years ago

9

A roller coaster starts from rest at point a and continues to point b on a frictionless track. what best describes the changes i

n kinetic and potential energy as the car travels from point a to point b?

2 answers:

IceJOKER [234]3 years ago

5 0

Took a wild guess and got it correct. The answer is B Kinetic Energy increases and potential energy decreases.

saw5 [17]3 years ago

3 0

<span>At first when the car is at the datum point where is no elevation, the kinetic energy increases while the potential energy is zero. As it travel the path and goes upward the kinetic enrgy decreases while the potential energy increases. When it goes down again the kinetic enrgy increases again while the potential energy decreases </span>

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Explain a lever and a pivot in a full sentence

worty [1.4K]

Answer:

The lever is a movable bar that pivots on a fulcrum attached to a fixed point. The lever operates by applying forces at different distances from the fulcrum, or a pivot. As the lever rotates around the fulcrum, points farther from this pivot move faster than points closer to the pivot.

IF HELPED MARK AS BRAINLIEST

4 0

2 years ago

A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg

saveliy_v [14]

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

Following formulas will be used.

$x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2$

$F =ma$

$V(t) = V_{o} +at$

$x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2$

(a) Calculating velocity right before going up the ramp.

Wooden block is going on a straightaway and has net for on it.

$F_{n} =F-F_{s} = F-uF_{n} = 100N-0.4*9.8m/s^2*5kg$ = $81N$

and this force produces acceleration of

$a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .$

With this acceleration, wooden block would reach at the foot of ramp in.

$x(t) = 12m = 16.2m/s^2*t^2$

$t = 1.217s$

and final velocity will be

$v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.$

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

$V= V_{h}cos(20) = 18.5288m/s$

and deceleration along the ramp is

$a = \frac{F_{s} }{m}$

Where $F_{s}$ force of friction along the inclined plane.

$F_s = uF_n = u*m*a$

$a = 9.8m/s^2*cos(20) = 9.2089m/s^2$

is a component of g along normal of the inclined plane.

$F_{s} = 0.25*5kg*9.2089m/s^2$

$= 11.5112N$

$a = \frac{11.5112N}{5kg} = 2.3022m/s^2$

And with this deceleration time needed to get wooded block to stop is.

$v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0$

$t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s$

and in that time wooden block would travel

$x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m$

This is how up wooden box will go before coming to stop.

3 0

3 years ago

A fish scale, consisting of a spring with spring constant k=200N/m, is hung vertically from the ceiling. A 2.6 kg fish is attach

Olegator [25]

Answer:

Explanation:

The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.

Change in gravity potential energy = change in spring potential energy

mgh = 1/2kh^2

Assume gravity constant g is 10m/s^2

2.6*10*h = 1/2*200*h^2

100h^2 - 26h = 0

2h(50h - 13) = 0

h = 0 or h = 13/50 = 0.65m

h = 0 is before the spring is stretched

So the maximum distance is 0.65m.

3 0

2 years ago

Read 2 more answers

300 ml of a gas at 27°C is Cooled at -3°c at Constant pressure the final volume is plzz answer fast i will mark brainliest

Scorpion4ik [409]

Answer:V₁=300ml

T₁=27°C

V₂=?

T₂= -3°C

as we know

V₁T₁=V₂T₂

By putting values in formula

300ml×27°C=V₂×(-3°C)

300ml×27°C/-3°C=V₂

8100ml/-3=V₂

-2700ml=V₂

or V₂= -2700ml

4 0

3 years ago

If a car is moving on a road at 70 km/hr going due north, and then changes direction and starts traveling north-east staying at

alex41 [277]

Question: If a car is moving on a road at 70 km/hr going due north, and then changes direction and starts traveling north-east staying at 70 km/hr, what happens to its speed and velocity?

Answer: the velocity of the car changes, but the speed stays the same.

Explanation: basically velocity is speed with a direction and speed is the absolute value or magnitude of velocity.

question answered by

(jacemorris04)

4 0

3 years ago

Read 2 more answers