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IRINA_888 [86]
3 years ago
5

vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses

strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by 2.76 m from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as that of the net into which he is shot. He takes 2.60 s to travel the horizontal distance of 20.8 m between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.
Physics
1 answer:
dsp733 years ago
3 0

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}

(This is correct because the horizontal motion has acceleration zero). Then:

v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}

Then, plugging in the given values, we obtain:

k=\frac{(61.2kg)(15.0m/s)^2}{(2.76m)^2}\\\\k=1808N/m

Finally, the effective spring constant of the firing mechanism is 1808N/m.

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If all of the forces acting on an object are balanced, then:
kondaur [170]

A. the direction the object is moving in will not change.

B. the acceleration of the object will be 0 m/s2

Explanation:

We can answer this problem by using Newton's second law of motion:

\sum F = ma

where

\sum F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, all the forces acting on the object are balanced, therefore the net force is zero:

\sum F=0

which means that also the acceleration is zero:

a=0

Acceleration is equal to the rate of change of velocity: therefore, zero acceleration means that the velocity of the object does not change. We can now analyze the given statements:

A. the direction the object is moving in will not change.  --> TRUE, because the velocity is not changing.

B. the acceleration of the object will be 0 m/s2  --> TRUE, as we stated above

c. the object will not be in motion.  --> FALSE: we just know that its velocity is constant, but it can be different from zero

D. the velocity of the object will be 0 m/s. --> FALSE, for the same reason stated in C

Learn more about Newton's second law of motion:

brainly.com/question/3820012

#LearnwithBrainly

3 0
3 years ago
A force does 30000 J of work along a distance of 9.5m. Find the applied force.
Elina [12.6K]

Answer: F=3158N

Explanation:

Work is the product of force applied and the distance the object moves along the force applied. Work is measured in joules and the equation is as follows

W = F x d

So that F = W/d

F = 30000 J / 9.5m

F = ~3158 N

8 0
3 years ago
What is the acceleration of an object that goes from 45m/s to 10 m/s in 5 seconds?
kipiarov [429]

Answer:

\boxed {\boxed {\sf a= -7 \ m/s^2}}

Explanation:

Acceleration is the change in velocity over time.

a= \frac {v_f-v_i}{t}

The object accelerates <em>from</em> 45 meters per second <em>to </em>10 meters per second in 5 seconds. Therefore,

v_f=10 \ m/s \\v_i= 45 \ m/s \\t= 5 \ s

Substitute the values into the formula.

a= \frac{ 10 \ m/s - 45 \ m/s}{5 \ s}

Solve the numerator.

a= \frac { -35 \ m/s}{5 \ s}

Divide.

a= -7 \ m/s/s

a= -7 \ m/s^2

The acceleration of the object is -7 meters per square second. The acceleration is negative because the object's velocity decreases and the object slows down.

5 0
2 years ago
Explain why the life cycle of a star can be compared to the life style of a human
Vladimir79 [104]
A star is born when clouds of dust and elements are gathered together in a certain space due to gravity, more and more mass and therefore pressure builds. When the pressure becomes enough to overcome the electronic repulsive force between two hydrogen nuclei, they are forced together and massive amounts of energy are given off forming helium atoms. This energy is then used to fuse other nuclei together. This could be compared to the way human life starts, where instead of 2 nuclei joining together to start a life cycle, two gametes, or sex cells are joined together. Also at the start of both a star and persons life, we are weak and we gain strength until we reach the height of our existence, then humans slowly become less efficient at doing what they do until eventually they cannot sustain themselves any further.
6 0
2 years ago
A body moves in a straight line. At time, it's acceleration is given by a = 12t + 1. When t =0, the velocity of the body v is 2
madreJ [45]

Answer:

v = 6t² + t + 2, s = 2t³ + ½ t² + 2t

59 m/s, 64.5 m

Explanation:

a = 12t + 1

v = ∫ a dt

v = 6t² + t + C

At t = 0, v = 2.

2 = 6(0)² + (0) + C

2 = C

Therefore, v = 6t² + t + 2.

s = ∫ v dt

s = 2t³ + ½ t² + 2t + C

At t = 0, s = 0.

0 = 2(0)³ + ½ (0)² + 2(0) + C

0 = C

Therefore, s = 2t³ + ½ t² + 2t.

At t = 3:

v = 6(3)² + (3) + 2 = 59

s = 2(3)³ + ½ (3)² + 2(3) = 64.5

5 0
2 years ago
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