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Rus_ich [418]
3 years ago
15

You have been asked to make a roller coaster more exciting. The owners want the speed at the bottom of the first hill doubled. H

ow much higher must the first hill be built?
Physics
1 answer:
qaws [65]3 years ago
4 0

Answer:

The height will be 4 times.

Explanation:

Given that,

The speed at the bottom of the hill doubled.

We need to calculate the height

Using conservation of energy

K.E_{t}+P.E_{t}=K.E_{b}+P.E_{b}

K.E_{b}=P.E_{t}

\dfrac{1}{2}mv^2=mgh

\dfrac{1}{2}m(4v^2)=mgh

Therefore,

P.E = mg(4h)

Here, m and g are constant

Hence, The height will be 4 times.

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PLEASE HELP ASAP WILL GIVE BRAINLIEST ​
stealth61 [152]

Explanation

(m) is measured in kilograms (kg)

<h2>(F) is measured in newtons (N)</h2>

<h3>acceleration (a) is measured in metres per second squared (m/s²)</h3>
4 0
3 years ago
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liubo4ka [24]

Answer:

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Explanation:

5 0
3 years ago
A 1200 kg car accelerates from 13 m/s to 17 m/s. find the change in momentum of the car.
mixas84 [53]
P=4800kgm/s
As
p=mΔv
where p is momentum, m is mass and v is velocity
Given values is
m =1200kg
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8 0
2 years ago
A rock is suspended by a light string. When the rock is in air, the tension in the string is 51.9 N . When the rock is totally i
Luden [163]

Answer:

\rho _{liquid}=1995.07kg/m^{3}

Explanation:

When the rock is immersed in unknown liquid the forces that act on it are shown as under

1) Tension T by the string

2) Weight W of the rock

3) Force of buoyancy due to displaced liquid B

For equilibrium we have T_{3}+B = W_{rock}

T_{3}+\rho _{Liquid}V_{rock}g=W_{rock}.....(\alpha)

When the rock is suspended in air for equilibrium we have

T_{1}=W_{rock}....(\beta)

When the rock is suspended in water for equilibrium we have

T_{2} + \rho _{water}V_{rock}g=W_{rock}.....(\gamma)

Using the given values of tension and solving α,β,γ simultaneously for \rho _{Liquid} we get

W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\

Solving for density of liquid we get

\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000

\rho _{liquid}=1995.07kg/m^{3}

5 0
3 years ago
The molecular mass of sodium oxide (Na2O) is
tresset_1 [31]
<span>The answer is 62 u. The molecular mass of a molecule (Mr) is the sum of atomic masses (Ar) of its elements. Ar(Na) = 23 u. Ar(O) = 16 u. Therefore, the molecular mass of sodium oxide is 62 u: Mr(Na2O) = 2 * Ar(Na) + Ar(O) = 2 * 23 u + 16 u = 46 + 16 = 62 u.</span>
7 0
4 years ago
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