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dedylja [7]
3 years ago
6

During a game of tug-of-war, one team exerts a force of 3,500 N to the left. The other team exerts a force of 3,800 N to the rig

ht. Which of the following statements is true about the game?
Select one:
a. The forces are balanced, and there will not be a winner.
b. The forces are unbalanced, and the second team will win.
c. The forces are balanced, and the first team will win.
d. The forces are unbalanced, and the first team will win.
Physics
2 answers:
NISA [10]3 years ago
8 0
B would be the correct answer.
Lunna [17]3 years ago
6 0
The answer is b hope this helps
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A 31.7 kg kid initially at rest slides down a frictionless water slide at 53.2 degrees, how fast is she moving in 3.45 s later?
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Answer:

34.55 m/s

Explanation:

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2 years ago
An organ pipe open at both ends has two successive harmonics with frequencies of 220 Hz and 240 Hz. What is the length of the pi
Harman [31]

Answer:

The  length is  l  =  8.6 \  m

Explanation:

From the question we are told that

   The frequencies of the two successive harmonics are f_1 =  220 \ Hz ,  f_2  =  240 \  Hz

   The speed of sound in the air is  v_s  =  343 \  m/s

Generally the frequency of a given harmonic is mathematically represented as

     f_n  =  \frac{n  v  }{2l}

Here  n defines  the position of the harmonics

Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as

    220  =  \frac{n v}{2l}

and  

     240  =  \frac{(n+1) v}{2l}

So

   \frac{(n + 1 ) v}{2l} - \frac{n v}{2l}  =  240-220

=>  \frac{v}{2l}  =  20

=>   l  =  8.6 \  m

7 0
3 years ago
A two-slit Fraunhofer interference-diffraction pattern is observed with light of wavelength 672 nm. The slits have widths of 0.0
ololo11 [35]

Answer:

Explanation:

In case of diffraction , angular width of central maxima =2 λ/d

λ is wave length of light and d is slit width

In case of interference , angular width of each fringe

= λ /D

D is distance between two slits

No of interference fringe in central diffraction fringe

=2 λ/d x D/λ = 2 x D /d = 2 x .24/.03 = 16.

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3 years ago
Why do tin roofs make creaking sounds on hot days?
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Explanation:

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3 years ago
Read 2 more answers
9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the
mojhsa [17]

Answer:

Moment about SHOULDER  ∑ τ = 3.17 N / m,

Moment respect to ELBOW   Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

         ∑ τ = I α

   

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

          ∑ τ = I α

           I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

            I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

            I_mass = m L²

we substitute

           ∑ τ = (⅓ m L² + M L²) α

let's calculate

          ∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

           ∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

         Στ = I α

         I = I_ forearm + I_mass

         I = ⅓ m (L-0.03)² + M (L-0.03)²

         

let's calculate

        i = ⅓ 2.3 0.47² + 0.5 0.47²

        I = 0.2798 Kg m²

        Στ = 0.2798 10

        Στ= 2.80 N m

3 0
3 years ago
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