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Nataly [62]
2 years ago
8

Explain why a roller coaster’s second hill cannot be taller than the first hill.

Physics
2 answers:
koban [17]2 years ago
8 0

The reason behind the second hill in the roller coaster not taller than the first hill is the frictional force acting on the roller coaster.

<u>Explanation: </u>

The height of the first hill in the roller coaster is taller with the decrease in the subsequent hills as the frictional forces reduces the kinetic energy.  

Coming down the first tallest hill, the speed and kinetic energy both increases as the motion is in line with gravity, but due to conservation of energy, the potential energy decreases. And as the potential energy decreases, the height of hill has to be decreased in order to complete the motion.

max2010maxim [7]2 years ago
4 0
There will not be enough momentum from the first hill to cross another hill if he same or larger size because of the way potential energy and kinetic energy works it will not be able go as high as it could go on he fist hill.
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Please answer this question brainliest ko promise
Paul [167]

Answer:

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7 0
2 years ago
How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners
Natasha_Volkova [10]

Answer:

Potential\ Energy=Work \ Done=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V=\frac{kq}{r}

So,

Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}

Where:

k is Coulomb Constant=8.99*10^9 Nm^2/C^2

q is the charge on electron=-1.6*10^-19 C

r is the distance=3.0*10^{-10}m

For 3 Electrons Potential Energy or work Done is:

Potential\ Energy=3*\frac{kq^2}{r}

Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

7 0
3 years ago
What is the acceleration of a 10 kg mass pushed by a 5 N force
hichkok12 [17]

Answer:

The formula is a = F m so in this case a = 5 10 = 0.5 m s 2

Explanation:

3 0
2 years ago
Please help i'm going to throw up from stress
Eddi Din [679]

Answer:

Explanation:

First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.

120.g = .1200 kg of ethyl alcohol. Now for the formula:

t_f=\frac{(m_{H2O}*spheat_{H2O}*temp_{H2O})+(m_{EtOH}*spheat_{EtOH}*temp_{EtOH})}{(m_{H2O}*spheat_{H2O})+(m_{EtOH}*spheat_{EtOH})} where spheat is specific heat.

Filling that horrifying-looking formula in with some values:

16.0=\frac{(x*4182*20.0)+(.1200*2440*10.0)}{(x*4182)+(.1200*2440)} and

16.0=\frac{83640x+2928}{4182x+292.8} and

16(4182x + 292.8) = 83640x + 2928 and

66912x + 4684.8 = 83640x + 2928 and

1756.8 = 16728x so

x = .105 kg and the amount of water added is 105 g

4 0
3 years ago
A sled is pulled with a horizontal force of 18 n along a level trail, and the acceleration is found to be 0.39 m/s2. an extra ma
Natali5045456 [20]
From Newton's second law of motion, it is identified that the net force applied to the object with mass m, will make it move with an acceleration of a. This can be mathematically translated as,
                        F = m x a
To solve for the mass of the sled, we derive the equation above such that,
                        m = F / a
Substituting,
                       m = (18 N) / (0.39 m/s²)
                          m = 46.15 kg

Then, we add to the calculated mass the mass of the extra material.
                      total mass = 46.15kg + 4.5 kg
                       total mass = 50.65 kg
We solve for the normal force of the surface to the object by calculating its weight.
                     F₂ = (50.65 kg)(9.8 m/s²)
                     F₂ = 496.41 N

The force that would allow barely a movement for the object is equal to the product of the normal force and the coefficient of kinetic friction.
                     F = (F₂)(c)
                      c = F/F₂

Substituting,
                      c = 18 N/496.41 N
                       c = 0.0362

<em>ANSWER: c = 0.0362</em>

5 0
2 years ago
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