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kenny6666 [7]
3 years ago
8

A. A PH202 student lives next to a construction site and sees a crane with a wrecking ball demolish the building next door. The

wrecking ball swings along the wall between her house and the neighbor’s house. In an effort to determine the length of the cable on the wrecking ball the student builds a pendulum using a random rock and a string. Her pendulum turns out to be 0.500m long. While she plays with her pendulum she realizes that the wrecking ball swings back and forth in the same amount of time that it takes the rock to complete 5 full oscillations. What is the length of the cable on the wrecking ball?
Physics
1 answer:
netineya [11]3 years ago
8 0

Answer:

The length of cable is 12.5 m

Explanation:

Since, the wrecking ball completes 1 oscillation, in the same time, as it takes for the rock to complete 5 oscillations.

Therefore,

Time Period of Wrecking Ball = 5 (Time Period of Rock)

Since,

Time Period of  Pendulum = 2π√(L/g)

Therefore,

2π√(L₁/g) = 5[2π√(L₂/g)]

√L₁ = 5√L₂

Squaring on both sides:

L₁ = 25 L₂

where,

L₁ = Length of Cable = ?

L₂ = Length of string = 0.5 m

Therefore,

L₁ = 25 (0.5 m)

<u>L₁ = 12.5 m</u>

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padilas [110]

The characteristics of the projectile launch allows to find the results for the questions about the movement of the ball are:

  • In the attached we see the vertical speed decreases with height and the curve is of the parabolic type.
  • In the movement is several dimensions, each one is independent of the others, the movement in the x axis does not affect the movement in the y axis.

Kinematics studies the motion of bodies looking for relationships between position, velocity and acceleration. In the case of vertical and projectiles launch  the acceleration on the vertical axis is the acceleration of gravity directed downward.

In the attachment we can see the position of the ball for two distances in the case of projectile launching.

We can see that the speed of the ball decreases with height according to the relation

          y = go t - ½ g t²

Where y is the height, g is the initial vertical velocity, g is the acceleration of gravity and t is time.

In all movements in various dimensions we assume that each movement in an . axis is independent

In the case of projectile launching, on the vertical axis there is an acceleration of gravity and on the horizontal axis there is no acceleration, the only parameter that this gives the two movements is the time, which is a scalar.

In conclusion, using the characteristics of the projectile launch, we can find the results for the questions about the movement of the ball are:  

  • In the movement is several dimensions, each one is independent of the others, the movement in the x axis does not affect the movement in the y axis.
  • In the attached we see the vertical speed decreases with height and the curve is of the parabolic type.

Learn more about projectile launch here: brainly.com/question/24888457

4 0
2 years ago
According to Newton's Second Law of Motion, an object will accelerate if you apply what kind of force? Question 1 options: Frict
Bumek [7]

An unbalanced force is required to accelerate an object according to Newton's Second Law of Motion.

<h3>What does Newton's Second Law of Motion state?</h3>

It states that the force applied to the object is equal to the product of mass and acceleration.

F = ma

  • An object will accelerate when the net force applied on the object is more than zero or unbalanced.
  • The acceleration is the change in the direction or speed of the object. To achieve acceleration the force must be greater in a direction.

  • When force is greater in one the object move in that direction which is known as acceleration.

Therefore, an unbalanced force is required to accelerate an object according to Newton's Second Law of Motion.

Learn more about Newton's Second Law of Motion.:

brainly.com/question/25810165

5 0
2 years ago
A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw
meriva

Answer:

As  28m/s = 28m/s

Explanation:

r = the radius of the curve

m =  the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

a = \frac{v^{2}}{r}

therefore

\mu mg = m \frac{v^{2}}{r} \\\\ \mu =  \frac{v^{2}}{rg}

v = \sqrt{\mu rg}

\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s

As  28m/s = 28m/s

8 0
3 years ago
An 8.0 cm object is 40.0 cm from a concave mirror that has a focal length of 10.0 cm. Its image is 16.0 cm in front of the mirro
svet-max [94.6K]
 We can rearrange the mirror equation before plugging our values in. 
1/p = 1/f - 1/q. 
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p  <-- cross multiplication
13.33cm = p

Now that we have the value of p, we can plug it into the magnification equation.

M=-16/13.33=1.2
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So the height of the image produced by the mirror is 9.6cm.
6 0
2 years ago
. Each valve A, B, and C, when open, releases water into a tank at its own constant rate. With all three valves open, the tank f
olasank [31]

Answer:

Time taken by A and B is 1.2 hr.

Explanation:

Given that

Time taken by tank when all(A+B+C) are open = 1 hr

Time taken by tank when A+C are open = 1.5 hr

Time taken by tank when B+C are open = 2 hr

If we treat as filling of tank is a work then

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Lets take work is 1 unit

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1 = 1.5(1/a+1/c)           ----------2

1 = 2(1/b+1/c)             --------3

From equation 1 and 3

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a=2

Form equation 2

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c=6

From equation 3

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So time taken by

A is alone to fill tank is 2 hr

B is alone to fill tank is 3 hr

C is alone to fill tank is 6 hr

So time\ taken\ by\ A\ and\ B =\dfrac{2\times 3}{2+3} hr

Time taken by A and B is 1.2 hr.

7 0
2 years ago
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