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Olin [163]
3 years ago
12

In a circus act, a 70 kg clown is shot from a cannon with an initial velocity of 17 m/s at some unknown angle above the horizont

al. A short time later the clown lands in a net that is 4.1 m vertically above the clown's initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?
Physics
1 answer:
EleoNora [17]3 years ago
7 0

Answer:

K_{2}=7302.4J

Explanation:

Given the initial velocity of the clown, his mass and final height we can calculate the final kinetic energy using the <em><u>conservation of total mechanical energy</u></em>

K_{1}+U_{1}=K_{2}+U_{2}

K_{2}=K_{1}+U_{1}-U_{2}

K_{2}=\frac{1}{2}mv_{1}^{2}+mgh_{1}-mgh_{2}

Since h_{1}=0

K_{2}=\frac{1}{2}mv_{1}^{2}-mgh_{2}

K_{2}=\frac{1}{2}(70kg)(17m/s)^2-(70kg)(9.8m/s^2)(4.1m)=7302.4J

K_{2}=7302.4J

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Angelina_Jolie [31]
That would be a savanna climate
7 0
3 years ago
SP1b.
nata0808 [166]

Answer:

2 m/s^2, west

Explanation:

Vf=final velcoity

Vi=initial velocity

t=timw

a =  \frac{vf - vi}{t}

=

\frac{15 - 25}{5}

= - 2 m/s^2

The - changes direction and makes it opposite

2 m/s, west

3 0
3 years ago
You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 11-m-hi
True [87]

Answer:

m = maximum mass of the coaster = 410 kg

d = maximum spring compression = 2.3 m

h = maximum height of the track = 11 m

H = maximum difference in height of the track = 19 m

g = acceleration by gravity = 9.8 m/s²

k = spring constant (without safety margin) = ?

K = spring constant (with safety margin) = ?

V = maximum speed of the coaster = ?

The gravitational potential energy of the coaster on the top of the 11 m high hill (relative to its initial starting point) is:

PEg = m g h

PEg = (410 kg) (9.8 m/s²) (11 m)

PEg = 44198 J

To reach that height, the elastic potential energy stored in the spring must be the same, so:

PEg = PEe = k d² / 2

(44198 J) = k (2.3 m)² / 2

k = 16710 N/m

Adding 14% to that value, you get:

K = 1.14 (16710 N/m)

K = 19045 N/m - answer spring constant

When fully compressed, the elastic potential energy stored in the spring is:

PEe = K d² / 2

PEe = (19045 N/m) (2.3m)² / 2

PEe = 51326 J

The difference in height between the starting point and the lowest point of the track is:

Δh = H - h

Δh = (19 m) - (11 m)

Δh = 8 m

So the initial gravitational potential energy of 330 kg coaster, relative to the lowest point, is

PEg = m g Δh

PEg = (340 kg) (9.8 m/s) (8 m)

PEg = 26656 J

The total energy of the coaster at its starting point (again, relative to the lowest point) is:

TE = PEe + PEg

TE = (51326J) + (26656 J)

TE = 77982J

At the lowest point of the track, all that energy is converted to kinetic energy, so the speed at that point will be:

TE = KE = m V² / 2

(77982 J) = (340kg) V² / 2

V = 21.46 m/s - answer maximum speed

4 0
3 years ago
Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys
max2010maxim [7]

Answer:

The spring was compressed the following amount:

\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount \Delta x), and the total final energy is the addition of the kinetic energies of both masses:

E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2

E_i=E_f\\

\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }

8 0
3 years ago
7. Answer each of the following questions if the student applied a 550N force to a 100kg box on the same surface. What would be
blagie [28]
F=ma
550=100a
a=550/100
a=5.5 m/s²
5 0
3 years ago
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