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BigorU [14]
3 years ago
10

Perfect square of 11650​

Physics
2 answers:
suter [353]3 years ago
7 0

Answer:

Since the area of the perfect square is 11650, and all of a squares sides ar equal, we just need to find the square root.

The square root of 11650 is 107.935166.

One side of the square is 107.935166

107.935166 x 107.935166 = 11650

(っ◔◡◔)っ ♥ Hope It Helps ♥

BartSMP [9]3 years ago
3 0

Answer:

107.94

Explanation:

Take the square root of 11650

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Answer:

If it is not an object in motion, all forces are balanced.

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If there is a gravitational force between all objects, why do we not feel or observe it?
Dmitry [639]
We do not feel the gravitational forces from objects other than the Earth because they are weak.
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Read 2 more answers
The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how
liraira [26]

Answer:

A

   x = 0.198456 \ m

B

    h  =  1.3061 \  m

C

 v =  5.06 \  m/s

D

  d = 4.0273 \  m

Explanation:

Considering the first question

From the question we are told that

   The spring constant is  k  =  32.50 N/m

    The potential energy is  PE  =  0.640 \ J

Generally the potential  energy stored in spring  is mathematically represented as   PE  =  \frac{1}{2}  *  k  *  x^2

=>    0.640=  \frac{1}{2}  * 32.50  *  x^2  

=>    x = \sqrt{0.03938}  

=>    x = 0.198456 \ m  

Considering the second question

 From the question we are told that

   The mass of the dart is  m =  0.050 kg

Generally from the law of energy conservation

         PE =  mgh

=>       0.640   =  0.050 *  9.8  *  h

=>      h  =  1.3061 \  m

Considering the third  question

   The height at which the dart was fired horizontally is  H  =   3.90\  m

Generally  from the law of energy conservation

         PE = KE

Here  KE is kinetic energy of the dart which is mathematical represented as

     KE  =  \frac{1}{2}  *  mv^2

=>      0.640 =  \frac{1}{2}  * 0.050 *  v^2

=>       v^2 = 25.6

=>       v =  5.06 \  m/s

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

       t  =  \frac{ 2 *  H }{g}

=>     t  =  \frac{ 2 * 3.90 }{9.8 }

=>     t  =  0.7959 \ s

Generally the  horizontal distance from the equilibrium position to the ground is  mathematically represented as

       d =  v  *   t

=>     d = 5.06  *   0.7959

=>     d = 4.0273 \  m

5 0
3 years ago
The 1.0-kg collar slides freely on the fixed circular rod. Calculate the velocity v of the collar as it hits the stop at B if it
soldi70 [24.7K]

Answer:

6.21 m/s

Explanation:

Using work energy equation then

U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})

where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity

Substituting 1 Kg for m, 0.4m for h, v_a as 0, 9.81 for g then

58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s

7 0
3 years ago
A diffraction grating with 750 slits per mm is illuminated by light which gives a first-order diffraction angle of 34.0°. What i
lesya [120]
<h2>Answer: 745.59 nm</h2>

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by the following equation:

dsin\theta_{n}=n\lambda (1)

Where:

d is the width of the slit

\lambda is the wavelength of the light  

n is an integer different from zero

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda (2)

We know:

\theta_{1}=34\°

In addition we are told the diffraction grating has 750 slits per mm, this means:

d=\frac{1mm}{750}

Solving (2) with the known values we will find \lambda:

\lambda=(\frac{1mm}{750})sin(34\°) (3)

\lambda=0.00074559mm (4)

Knowing 1mm=10^{6}nm:

\lambda=745.59nm  >>>This is the wavelength of the light, wich corresponds to red.

6 0
4 years ago
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