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Studentka2010 [4]
3 years ago
6

A 10kg roller coaster does a loop-de-loop and it’s your job to make sure it is safe. The people will fall out if the cart’s spee

d is less than 40m/s when it goes upside down. How high up does the car have to be at point A to make it? Fill in everything else as well.
Location



PE



KE

TE



A



h =





v = 0m/s







B



h = 0m





v =







C



h = 15m

PE=m*g*h



v = 40m/s

KE=0.5*m*v2=

TE=PE+KE



D



h = 0m





v =







E



h = 10m





v =







F



h = 20m





v =
Physics
1 answer:
Bingel [31]3 years ago
6 0

Answer:

3

Explanation:

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Answer: the water level would rise since the pebble displaces minimal water compared to the boat.

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How long will it be before 82.0<br> % of an isotope with a half life of 1250<br> y decays?
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Can you give more information
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A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between
SashulF [63]

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

\Delta W = \Delta KE

\Delta W = \frac{1}{2} mv^2

Here,

m = mass

v = Velocity

Our values are given as,

m = 79.7kg

v = 4.77m/s

Replacing,

\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2

\Delta W = 907J

Therefore the mechanical energy lost due to friction acting on the runner is 907J

6 0
3 years ago
When a charge of 8 C flows past any point along a circuit in 2 seconds, the current is ________ A?
RoseWind [281]
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3 years ago
A pitcher throws a baseball with a mass of 143 g horizontally at a speed of 38.8 m/s (87 mi/h). The hitter's bat is in contact w
SIZIF [17.4K]

Answer:

F = −10093.41 N

Explanation:

Given that,

Mass of a baseball, m = 143 g = 0.143 kg

Initial speed of the baseball, u = +38.8 m/s

The hitter's bat is in contact with the ball for 1.20 ms and then travels straight back to the pitcher's mound at a speed of 45.9 m/s, v = -45.9 m/s

We need to find the average force exerted on the ball by the bat. So, Force is given by :

F=ma

a is acceleration

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.143\times (-45.9-(38.8))}{1.2\times 10^{-3}}\\\\F=-10093.41\ N

So, the average force exerted on the ball by the bat has a magnitude of 10093.41 N.

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3 years ago
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