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nekit [7.7K]
3 years ago
14

You ride a roller coaster with a loop-the-loop. Compare the normal force that the seat exerts on you to the force that Earth exe

rts on you when you are passing the bottom of the loop. Express your answer in terms of R (radius of the loop), vb (speed at the bottom of the loop), and constant g. Nbottom/mg
Physics
1 answer:
timofeeve [1]3 years ago
4 0

Answer:

N = mg + \frac{mv^2}{R}

Explanation:

At the bottom of the loop, the normal force is opposite to my weight.

I am making a circular motion. So,

F_{net} = \frac{mv^2}{R}

The relationship between the normal force, my weight, my speed and the radius of the loop is

N - mg = \frac{mv^2}{R}\\mg = N - \frac{mv^2}{R}\\ N = mg + \frac{mv^2}{R}

Here, my weight (mg) is constant. But the normal force is inversely proportional to my speed.

If my speed is zero, the normal force would be maximum and equal to my weight. If my speed is to much, then the normal force would be equally high too.

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A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is
ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0

Solving the above equation we get

t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89

So, the time the package was in the air is 1.97 seconds

3 0
2 years ago
The first mechanized industry was
Oxana [17]
C. Textiles

It was the first thing mechanized in the Industrial Revolution

3 0
2 years ago
The cars on an amusement-park ride travel at a constant velocity of 4.0 m/s on a circular track
lys-0071 [83]
V^2=u^2+2as
V=0
a =-u^2/2s
a=[4]^2/2[4]
a=-2m/s^2
7 0
2 years ago
NEED HELP!! ASAP JUST QUESTIONS 22, 23, 24, & 25
poizon [28]
22. reduction
25. Le Chatelier's principle
8 0
3 years ago
Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and
dexar [7]

Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
closest answer is 1.99
× 10^30

(it may vary
a little with rounding – the difference is less than 1%)


8 0
3 years ago
Read 2 more answers
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