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oee [108]
3 years ago
13

true or false? destructive interference occurs when a trough meets up with another trough given location along the medium

Physics
1 answer:
disa [49]3 years ago
4 0

Answer:Poopy-di scoop

Scoop-diddy-whoop

Whoop-di-scoop-di-poop

Poop-di-scoopty

Scoopty-whoop

Whoopity-scoop, whoop-poop

Poop-diddy, whoop-scoop

Poop, poop

Scoop-diddy-whoop

Whoop-diddy-scoop

Whoop-diddy-scoop, poop

Explanation:

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What is a lot like gases but the atoms are different because they are made up of free electrons and ions of elements
34kurt

plasmas  are a lot like gases



hope this helps.

7 0
2 years ago
A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the
marin [14]

Answer:

A)

0.395 m

B)

2.4 m/s

Explanation:

A)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

x = initial position of spring from equilibrium position = 0.21 m

v_{i} = initial speed of the cart = 2.0 ms⁻¹

A = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m

B)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

A = amplitude of the oscillation = 0.395 m

v_{o} = maximum speed at the equilibrium position

Using conservation of energy

Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}

5 0
3 years ago
Read 2 more answers
What factors affect potential energy
jok3333 [9.3K]
Mass ,gravity and height
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
Assuming the pick-up trucks, trailers and road conditions are exactly the same, which vehicle will take a longer distance to sto
mars1129 [50]

I’ve answered this before so I know the question is missing an important given and that given is: <span>1 has an empty trailer and the other has a fully loaded one. 

So, it would be the fully loaded trailer that would take a longer distance to stop because a lot of weight is being pulled, and when the brakes are started, the fully loaded trailer is more like pushing against the truck.</span>

6 0
2 years ago
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