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Sladkaya [172]
3 years ago
14

A roller coaster glides from rest from the top of an 80.0 meter hill. What is the speed of the roller coaster at the bottom of t

he hill?
Physics
1 answer:
shepuryov [24]3 years ago
4 0
So this is the case of energy conversion. From potential to kinetic.
As energy is conserve:
Epi+Eki=Epf+Ekf ; i for initial and f for final. At begining object at rest hence Eki=0 while at final, object on the lowest position, hence epf=0 therefore:
Epi=Ekf
M.g.h=1/2.m.v^2
Assume no mass change during the process, then
G.h=1/2.v^2
Hence v at bottom:
v=sqrt(2×g×h) assume g=10m/s^2
=sqrt(2×10×80)=sqrt(1600)=40m/s
Hence velocity in the bottom is 40 m/s

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3 years ago
PLEASE HELP
GREYUIT [131]

Answer:

h = 1.8 m

Explanation:

The initial velocity of the glove, u =- 6 m/s

We need to find the maximum height of the glove. Let it is equal to h. Using equation of kinematics. At the maximum height v = 0

v^2-u^2=2ah, h is the maximum height and a = -g

0^2-(6)^2=2\times (-10)\times h\\\\h=\dfrac{36}{20}\\\\h=1.8\ m

Hence, it will go up to a height of 1.8 m.

4 0
2 years ago
A submersible pump is put under the water at the bottom of a well and is used to push water up through a pipe. What minimum outp
Maslowich

Answer:

695800 N/m^2 or Pa

Explanation:

Height of the water from the ground H  =  71 m

Acceleration due to gravity g =9.8 m/s^2

density of water ρ= 1000 kg/m^3

The minimum output gauge pressure to make water reach height H

P= ρgH

= 1000×9.8×71= 695800 N/m^2 or Pa

5 0
3 years ago
A battery with an internal resistance ofrand an emf of 10.00 V is connected to a loadresistorR=r. As the battery ages, the inter
yanalaym [24]

Answer:

The current is reduced to half of its original value.

Explanation:

  • Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

        I_{1} = \frac{V}{R_{int} +r_{L} }

  • where Rint = r and RL = r
  • Replacing these values in I₁, we have:

       I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)

  • When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

       I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r}  (2)

  • We can find the relationship between I₂, and I₁, dividing both sides, as follows:

        \frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}

  • The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.  
7 0
3 years ago
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