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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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Answer:

the study of living organisms utilizing the scientific method.

Explanation:

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What affect, if any, does increasing the speed of the plunger have on the wavelength of the waves
mojhsa [17]

As the speed of the plunger increases, the wavelength of the waves decreases. The greater the frequency, the smaller the wavelength. The smaller the frequency, the greater the wavelength. When we increase the speed of the plunger, the frequency of the waves also increases, and just like with the size of the ball, it’s the speed of the plunger and the frequency of the waves are directly related.

6 0
2 years ago
A boy in a wheelchair (total mass 54.5 kg) has speed 1.40 m/s at the crest of a slope 2.10 m high and 12.4 m long. At the bottom
babymother [125]

Answer:

630.75 j

Explanation:

from the question we have the following

total mass (m) = 54.5 kg

initial speed (Vi) = 1.4 m/s

final speed (Vf) = 6.6 m/s

frictional force (FF) = 41 N

height of slope (h) = 2.1 m

length of slope (d) = 12.4 m

acceleration due to gravity (g) = 9.8 m/s^2

work done (wd) = ?

  • we can calculate the work done by the boy in pushing the chair using the law of law of conservation of energy

wd + mgh = (0.5 mVf^2) - (0.5 mVi^2) + (FF x  d)

wd = (0.5 mVf^2) - (0.5 mVi^2) + (FF x  d) - (mgh)

where wd = work done

m = mass

h = height

g = acceleration due to gravity

FF = frictional force

d = distance

Vf and Vi = final and initial velocity

wd =  (0.5 x 54.5 x 6.9^2) - (0.5 x 54.5 x 1.4^2) + (41 x 12.4) - (54.5 X 9.8 X 2.1)            

wd = 630.75 j

3 0
3 years ago
An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel s
ad-work [718]

Answer:

a

 \theta  =  0.0022 rad

b

 I  =  0.000304 I_o

Explanation:

From the question we are told that  

   The  wavelength of the light is \lambda  = 550 \ nm  =  550 *10^{-9} \ m

    The  distance of the slit separation is  d = 0.500 \ mm = 5.0 *10^{-4} \ m

 

Generally the condition for two slit interference  is  

     dsin \theta =  m \lambda

Where m is the order which is given from the question as  m = 2

=>    \theta  =  sin ^{-1} [\frac{m \lambda}{d} ]

 substituting values  

      \theta  =  0.0022 rad

Now on the second question  

   The distance of separation of the slit is  

       d =  0.300 \ mm  =  3.0 *10^{-4} \ m

The  intensity at the  the angular position in part "a" is mathematically evaluated as

      I  =  I_o  [\frac{sin \beta}{\beta} ]^2

Where  \beta is mathematically evaluated as

       \beta  =  \frac{\pi *  d  *  sin(\theta )}{\lambda }

  substituting values

     \beta  =  \frac{3.142  *  3*10^{-4}  *  sin(0.0022 )}{550 *10^{-9} }

    \beta  = 0.06581

So the intensity is  

    I  =  I_o  [\frac{sin (0.06581)}{0.06581} ]^2

   I  =  0.000304 I_o

3 0
3 years ago
Now we’ll use the component method to add two vectors. We will use this technique extensively when we begin to consider how forc
dimaraw [331]

Answer:

The magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

Explanation:

Since vector A has magnitude 50 cm and a direction of 30, its x - component is A' = 50cos30 = 43.3 cm and its y - component is A" = 50sin30 = 25.

Also, Since vector B has magnitude 35 cm and a direction of 110, its x - component is A' = 35cos110 = -11.97 cm and its y - component is A" = 35sin110 = 32.89 cm.

So, the vector sum R = A + B

The x-component of the vector sum is R' = A'+ B' = 43.3 cm + (-11.97 cm) = 43.3 cm - 11.97 cm = 31.33 cm

The y-component of the vector sum is R" = A"+ B" = 25 cm + 32.89 cm = 57.89 cm

So, the magnitude of R = √(R'² + R"²)

= √((31.33 cm)² + (57.89 cm)²)

= √(981.5689 cm² + 3,351.2521 cm²)

= √(4,332.821 cm²)

= 65.82 cm

≅ 65.8 cm

The direction of R is Ф = tan⁻¹(R"/R')

= tan⁻¹(57.89 cm/31.33 cm)

= tan⁻¹(1.84775)

= 61.58°

≅ 61.6°

So, the magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

4 0
2 years ago
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