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Artyom0805 [142]
3 years ago
5

What is surface tension​

Physics
1 answer:
uysha [10]3 years ago
5 0

Answer:

Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. Surface tension allows insects, usually denser than water, to float and slide on a water surface.

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How does nuclear energy work? Explain
fgiga [73]
There are many processes to get nuclear energy. Nuclear energy is basically energy from an atom. For example fission is where the nucleus of an atom ( typically radioactive atoms ) gets split then energy is released ( typically heat). And in radioactive decay radiation is released from an radioactive atom. Hope this helps
6 0
3 years ago
Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1.05 m.At what minimum angle relative to the c
kiruha [24]

Answer:

15.19°, 31.61°, 51.84°

Explanation:

We need to fin the angle for m=1,2,3

We know that the expression for wavelenght is,

\lambda = \frac{c}{f}

Substituting,

\lambda = \frac{344}{1250}

\lambda = 0.2752m

Once we have the wavelenght we can find the angle by the equation of the single slit difraction,

sin\theta = \frac{m \lambda}{W}

Where,

W is the width

m is the integer

\lambda the wavelenght

Re-arrange the expression,

\theta = sin^{-1} \frac{m\lambda}{W}

For m=1,

\theta = sin^{-1} \frac{1 (0.2752)}{1.05}= 15.19\°

For m=2,

\theta = sin^{-1} \frac{2 (0.2752)}{1.05}= 31.61\°

For m=3,

\theta = sin^{-1} \frac{3 (0.2752)}{1.05}= 51.84\°

<em>The angle of diffraction is directly proportional to the size of the wavelength.</em>

5 0
3 years ago
One problem when using rigid metal conduit in a residence is that the installation of the conduit may a weaken the structure b r
worty [1.4K]
The correct answer is A. Installation of rigid metal conduit requires grounding and the grounding equipment used may weaken the structure.
8 0
3 years ago
Read 2 more answers
Ella makes this table to organize her notes on whether atoms gain or lose energy during the changes of state.
solong [7]

Answer:

The answer is D.

7 0
3 years ago
Read 2 more answers
uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station
Alex787 [66]

Answer:

The magnitude of the tangential velocity is v= 0.868 m/s

The magnitude of the resultant acceleration at that point is  a = 4.057 m/s^2

Explanation:

From the question we are told that

      The mass of the uniform disk is m_d = 40.0kg

       The radius of the uniform disk is R_d = 0.200m

       The force applied on the disk is F_d = 30.0N

Generally the angular speed i mathematically represented as

             w = \sqrt{2 \alpha  \theta}

Where \theta is the angular displacement given from the question as

           \theta  = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}

                 =1.257\  rad

   \alpha is the angular acceleration which is mathematically represented as

                    \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

    The moment of inertial is mathematically represented as

                     I = \frac{1}{2} m_dR^2_d

Substituting values

                    I = 0.5 * 40 * 0.200^2

                        = 0.8kg \cdot m^2

Considering the equation for angular acceleration

               \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

Substituting values

               \alph\alpha = \frac{(30.0)(0.200)}{0.8}

                   = 7.5 rad/s^2

Considering the equation for angular velocity

    w = \sqrt{2 \alpha  \theta}

Substituting values

     w =\sqrt{2 * (7.5) * 1.257}

         = 4.34 \ rad/s

The tangential velocity of a given point on the rim is mathematically represented as

                 v = R_d w

Substituting values

                    = (0.200)(4.34)

                     v= 0.868 m/s

The radial acceleration at hat point  is mathematically represented as

            \alpha_r = \frac{v^2}{R}

                  = \frac{0.868^2}{0.200^2}

                 = 3.7699 \ m/s^2

The tangential acceleration at that point is mathematically represented as

               \alpha _t = R \alpha

Substituting values

           \alpha _t = (0.200) (7.5)

                 = 1.5 m/s^2

The magnitude of resultant acceleration at that point is

                 a = \sqrt{\alpha_r ^2+ \alpha_t^2 }

Substituting values

                a = \sqrt{(3.7699)^2 + (1.5)^2}

                   a = 4.057 m/s^2

         

7 0
3 years ago
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