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sweet-ann [11.9K]
2 years ago
8

Please show work if possible! Thank you!!

Physics
1 answer:
Leviafan [203]2 years ago
3 0

B. The normal force that is exerted on the coaster by the track at the lowest point is 4.7 x 10⁴ N.

<h3>Normal force exerted on the coaster at the lowest point</h3>

Fₙ = mg + mv²/r

where;

  • m is mass of the coaster
  • v is speed of the coaster
  • r is radius of the path

Fₙ = (2,000 x 9.8) + (2,000 x 18²)/24

Fₙ = 46,600 N

Fₙ =  4.7 x 10⁴ N

Thus, the normal force that is exerted on the coaster by the track at the lowest point is 4.7 x 10⁴ N.

Learn more about normal force here: brainly.com/question/14486416

#SPJ1

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John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2
Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function cos(\omega t +\phi) is defined between -1 and 1, So it is not possible obtain a value 2\pi greater.  

In addition, if you  move the function cosine a T Value, and T is the Period,  the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have f=1+cos(\omega t +\phi), the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the f=1+cos(\omega t +\phi), and its mean that the function f=cos(\omega t +\phi), in general, is not greater than cos(\omega t +\phi).

3 0
3 years ago
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ludmilkaskok [199]
The answer to this is D
6 0
3 years ago
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An elevator provides 21 000 w of power during a 12 s ride. how much work does the elevator do?
Mice21 [21]

Answer:

252000 J

Explanation:

W = Pt

W = (21000 w)(12 s)

W =252000 J

4 0
3 years ago
A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t
notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, F_f, backward

The frictional force can be written as

F_f = \mu mg

where

\mu=0.03 is the coefficient of kinetic friction

m=150 kg is the mass of the motorbike

g=9.8 m/s^2 is the acceleration of gravity

Therefore the net force is given by

\sum F = F - F_f = F - \mu mg

And substituting, we find

\sum F=250 - (0.03)(150)(9.8)=205.9 N

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have

\sum F = 205.9 N is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s

4)

For this part of the problem, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

a=1.37 m/s^2

Solving for t, we find

t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s

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brainly.com/question/11411375

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6 0
4 years ago
You have a stopped pipe of adjustable length close to a taut 62.0-cm, 7.25-g wire under a tension of 3910 N . You want to adjust
Ber [7]

Answer:

L = 6 cm

Explanation:

Second overtone of the wire is same as third harmonic

so its frequency is given as

f = \frac{3}{2L}\sqrt{\frac{T}{\mu}}

here we know that

\mu = \frac{M}{L}

\mu = \frac{7.25 \times 10^{-3}}{0.62}

\mu = 0.0117 kg/m

now we have

f = \frac{3}{2\times 0.62}\sqrt{\frac{3910}{0.0117}}

f = 1398.6 Hz

Now fundamental frequency of sound in a pipe is given as

f = \frac{v}{4L}

1398.6 = \frac{340}{4L}

L = 0.06 m

L = 6 cm

7 0
3 years ago
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