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Natali5045456 [20]
3 years ago
5

In 1939 or 1940, Emanuel Zacchini took his human-cannonball act to an extreme. After being shot from a cannon, he soared over th

ree Ferris wheels and into a net. Assume that he is launched with a speed of 26.5 m/s and at an angle of 53.0° and neglect air drag.
a) Treating him as a particle, calculate his clearance over the first wheel.
b) If he reach maximum height over the middle wheel, by how much did he clear it?
c) How far from the cannon should the net's center have been positioned?

Physics
1 answer:
Arlecino [84]3 years ago
8 0
 We first determine the vertex by using the formula,<span>-b/2a = vertex, in order to get the values for the t-coordinate. That is why we got 
</span>
v_y=26.5 sin(53)=21.163v_x=26.5 cos(53)=15.948 
then

let x=0since you are going to land on a 3m tally=-.5(9.8)t^2+ 21.163*t
y=0=-4.9t+21.163t=4.31
vx*4.31= total distance travelled=68.88m
Then for the first wheel, you have 15.948m=vxdetermine the time when he reaches 23 meters, that is

23/15.948=1.44218 sec

substitute t with1.44218 sec, then determine the height.
h(1.44218)=20.329

determine vertex by using a graphing calculatort=2.1594s h=22.85m
using the time value of the vertex, determine horizontal distance travelled
34.438m away from cannon
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steposvetlana [31]

Answer:

See explanation below

Explanation:

The question is incomplete. The missing part of this question is the following:

<em>"While the block is in contact with the spring and being brought to rest, what are (a)the work done by the spring force and (b) the increase in thermal energy of the blockfloor system? (c) What is the blocks speed just as it reaches the spring?"</em>

<em />

According to this we need to calculate three values: Work, Thermal Energy and Speed of the block when it reaches the spring.

Let's do this by parts.

<u>a) Work done by the spring:</u>

In this case, we need to apply the following expression:

W = -1/2 kx²  (1)

We know that k = 430 N/m, and x is the distance of compressed spring which is 5.8 cm (or 0.058 m). Replacing that into the expression:

W = -1/2 * 430 * (0.058)²

<h2>W = -0.7233 J</h2>

<u>b) Increase in thermal energy</u>

In this case we need to use the following expression:

ΔEt = Fk * x   (2)

And Fk is the force of the kinetic energy which is:

Fk = μk * N   (3)

Where μk is the coeffient of kinetic friction

N is the normal force which is the same as the weight, so:

N = mg (4)

Let's calculate first the Normal force (4), then Fk (3) and finally the chance in the thermal energy (2):

N = 4.8 * 9.8 = 47.04 N

Fk = 0.28 * 47.04 = 13.1712 N

Finally the Thermal energy:

ΔEt = 13.1712 * 0.058

<h2>ΔEt = 0.7639 J</h2>

<u>c) Block's speed reaching the spring</u>

As the block is just reaching the speed, the initial Work is 0. And the following expression will help us to get the speed:

V = √2Ki/m   (5)

And Ki, which is the initial kinetic energy can be calculated with:

Ki = ΔU + ΔEt   (6)

And ΔU is the same value of work calculated in part (a) but instead of being negative, it will be positive here. So replacing the data first in (6) and then in (5), we can calculate the speed:

Ki = 0.7233 + 0.7639 = 1.4872 J

Finally the speed:

V = √(2 * 1.4872) / 4.8

<h2>V = 0.7872 m/s</h2>

Hope this helps

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3 years ago
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SashulF [63]
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3 years ago
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A body of mass 400 kg is suspended at a lower end of a light vertical chain and is being pulled up vertically. Initially the bod
yKpoI14uk [10]

Answer:31.62 m/s

Explanation:

Given

mass of body m=400 kg

Pull on chain is F_1=6000g N=60 kN

Pull get smaller at the rate of F_2=360g N/m

Net Upward Force F=6000 g-360 g\times 10=24 kN

net acceleration a=\frac{F}{m}

a=\frac{24\times 1000}{m}

a=\frac{24000}{400}

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Hi please answer and show your work​
jek_recluse [69]

Answer:

\huge\boxed{\sf P.E. = 240\ MJ}

\huge\boxed{\sf K.E. = 19.6\ MJ}

Explanation:

<u>Given:</u>

Mass = m = 200,000 kg

Vertical Distance = h = 120 m

Speed = v = 14 m/s

Acceleration due to gravity = g = 10 m/s²

<u>Required:</u>

1) Gravitational Potential Energy = P.E = ?

2) Kinetic Energy = K.E. = ?

<u>Formula:</u>

1) P.E. = mgh

2) K.E. = \displaystyle \frac{1}{2} mv^2

<u>Solution:</u>

1) P.E. = (200,000)(10)(120)

P.E. = 240,000,000 Joules

P.E. = 240 Mega Joules

P.E. = 240 MJ

2) K.E. = 1/2 (200000)(14)^2

K.E. = (100000)(196)

K.E. = 19,600,000 Joules

K.E. = 19.6 MJ

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
4 0
3 years ago
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