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3 years ago

How can we predict how high a roller coaster can go knowing its velocity?

1 answer:

6 0

“to predict the speed that a coaster will reach before it is ever placed on the track. ... When coaster goes up by height h then its gravitational potential energy increases by the amount m.g.h where m = mass of coaster, h = height of coaster and g = gravitational acceleration due to Earth”

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Two metal plates 4 cm long are held horizontally 3 cm apart in a vacuum, one being vertically above other. The upper plate is at

kati45 [8]

Answer:

the answer to your question is 4 cm long

Explanation:

5 0

3 years ago

A baseball leaves a bat with a horizontal velocity of 20 m/s. In a time of 0.25 s, How far will it have traveled horizontally?

Maurinko [17]

Distance traveled by the ball is given by

$distance = speed \times time$

here we know that

speed = 20 m/s

times = 0.25 s

now we have

$distance = 20 \times 0.25$

$distance = 5 m$

so ball will travel 5 m distance in the given interval of time

6 0

3 years ago

Can some help me plz

kakasveta [241]

Answer:

D. I'm guessing

The gold-foil experiment showed that the atom consists of a small, massive, positively charged nucleus with the negatively charged electrons being at a great distance from the centre.

5 0

3 years ago

Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are

Sergio039 [100]

Answer:

$21678.47223\ lbf-in^2$

$383.1109\ lbf-in^2$

Explanation:

d = Diameter of column = 0.5 inch

$A_c$ = Area of concrete = $119.4\ in^2$

The strain in the system is conserved

$\dfrac{F_sL}{A_sE_s}=\dfrac{F_cL}{A_cE_c}\\\Rightarrow F_c=\dfrac{F_sA_cE_c}{A_sE_s}\\\Rightarrow F_c=\dfrac{F_s \times 119.4\times 4.1\times 10^6}{8\times \dfrac{\pi \dfrac{1}{2^2}}{4}\times 29\times 10^6}\\\Rightarrow F_c=10.74658F_s$

Now

$F_c+F_s=50000\\\Rightarrow 10.74658F_s+F_s=50000\\\Rightarrow F_s=\dfrac{50000}{11.74658}\\\Rightarrow F_s=4256.55807\ lbf$

$F_c=10.74658F_s\\\Rightarrow F_c=10.74658\times 4256.55807\\\Rightarrow F_c=45743.44182\ lbf$

Stress is given by

$\sigma_s=\dfrac{4256.55807}{\pi \dfrac{1}{2^2}}{4}\\\Rightarrow \sigma_s=21678.47223\ lbf-in^2$

The stress in the steel is $21678.47223\ lbf-in^2$

$\sigma_c=\dfrac{45743.44182}{119.4}\\\Rightarrow \sigma_s=383.1109\ lbf-in^2$

The stress in the steel is $383.1109\ lbf-in^2$

4 0

3 years ago

A student analyzes data of the motion of a planet as it orbits a star that is in deep space. The orbit of the planet is consider

Valentin [98]

Explanation:

The orbit of the planet is considered to be stable and does not change over time. This means that the force of gravity and inertia are perfectly balanced. Leading to forward motion of Planet and moons under force of gravity of some other large body(Most probably stars).This is in turns leads to formation of orbit.

8 0

3 years ago

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