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Kamila [148]
3 years ago
14

What are mechanical waves ?​

Physics
1 answer:
Law Incorporation [45]3 years ago
7 0

Answer:

<h3>A mechanical wave is a wave that is an oscillation of matter and therefore, transfers energy through a medium...</h3>

<h3> <em>h</em><em>o</em><em>p</em><em>e</em><em> </em><em>i</em><em>t</em><em> </em><em>h</em><em>e</em><em>l</em><em>p</em><em>s</em><em>!</em><em>!</em></h3>

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A 19 g bullet is fired into the bob of a ballistic pendulum of mass 1.3 kg. When the bob is at its maximum height, the strings m
katovenus [111]

Answer:

217.43298 m/s

Explanation:

m_1 = Mass of bullet = 19 g

m_2 = Mass of bob = 1.3 kg

L = Length of pendulum = 2.3 m

\theta = Angle of deflection = 60°

u = Velocity of bullet

Combined velocity of bullet and bob is given by

v^2-u^2=2as\\\Rightarrow v=\sqrt{2aL(1-cos\theta)+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (1-cos60)+0^2}\\\Rightarrow v=3.13209\ m/s

As the momentum is conserved

m_1u=(m_1+m_2)v\\\Rightarrow u=\frac{(m_1+m_2)v}{m_1}\\\Rightarrow v=\frac{(0.019+1.3)\times 3.13209}{0.019}\\\Rightarrow v=217.43298\ m/s

The speed of the bullet is 217.43298 m/s

5 0
3 years ago
PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
aleksley [76]

Answer:

red is not at the bottom. its density is 0.9 not 9!

also I don't need the brainlest thanks

6 0
2 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
a ball of mass 16 kg on the end of a string is spun at a constant speed of 2 m/s in a horizontal circle with a radius of 1m. Wha
miss Akunina [59]

The work done by the centripetal force during om complete revolution is 401.92 J.

<h3>What is centripetal force?</h3>

Centripetal force is a force that acts on a body undergoing a circular motion and is directed towards the center of the circle in which the body is moving.

To Calculate the work done by the centripetal force during one complete revolution, we use the formula below.

Formula:

  • W = (mv²/r)2πr
  • W = 2πmv²................... Equation 1

Where:

  • W = Work done by the centripetal force
  • m = mass of the ball
  • v = velocity of the ball
  • π = pie

From the question,

Given:

  • m = 16 kg
  • v = 2 m/s
  • π = 3.14

Substitute these values into equation 1

  • W = 2×3.14×16×2²
  • W = 401.92 J

Hence, The work done by the centripetal force during om complete revolution is 401.92 J.
Learn more about centripetal force here: brainly.com/question/20905151

5 0
2 years ago
A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa
pav-90 [236]

Answer

given,

current (I) = 16 mA

circumference of the circular loop (S)= 1.90 m

Magnetic field (B)= 0.790 T

S = 2 π r

1.9 = 2 π r

r = 0.3024 m

a) magnetic moment of loop

    M= I A

    M=16 \times 10^{-3} \times \pi \times r^2

   M=16 \times 10^{-3} \times \pi \times 0.3024^2

   M=4.59 \times 10^{-3}\ A m^2

b)  torque exerted in the loop

 \tau = M\ B

 \tau = 4.59 \times 10^{-3}\times 0.79

 \tau = 3.63 \times 10^{-3} N.m

8 0
2 years ago
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