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2 years ago

14

A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 38 m/s when i

t reaches ground level. What was its speed when its height was half that of its starting point

1 answer:

inysia [295]2 years ago

5 0

Explanation:

look !

speed= 38m/s

start from rest= 0

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found the potential spring energy. not sure how to find the height of the block by knowing only the mass. found the weight of it

evablogger [386]

(h + .16) m g = 1/2 k x^2 total PE of block relative to where it stops

(h + .16) .82 * 9.8 = .5 * 120 * .16^2 PE released = PE of spring

8.04 h + 1.29 = 1.536

h = (1.536 - 1.29) / 8.04 = .031 m = 3.1 cm

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3 years ago

An electron in a TV picture tube is accelerated through a potential difference of 10 kV before it hits the screen. What is the k

laiz [17]

Answer:

10,000 eV

Explanation:

According to the law of conservation of energy, the kinetic energy gained by the electron is equal to its change in electric potential energy:

$K=\Delta U=q \Delta V$

where:

K is the kinetic energy of the electron

$q=1 e$ is the magnitude of the charge of the electron

$\Delta V$ is the potential difference through which the electron has been accelerated

For this electron in the TV, we have

$\Delta V=10 kV=10000 V$

Therefore, the kinetic energy of the electron in electronvolts is

$K=(1 e)(10000 V)=10000 eV$

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4 years ago

A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

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3 years ago

How can humans activities affect the frequency and impact of natural disasters

denpristay [2]

Forest fires can be an impact

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3 years ago

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What is the reaction force if a girl pulls on a cow?

Papessa [141]

Answer : B) The cow pulls back on the girl.

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3 years ago