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OleMash [197]
3 years ago
13

Which two conditions must exist in order for radiation to occur?

Physics
1 answer:
melomori [17]3 years ago
7 0

Answer:Which condition must exist in order for conduction to occur between two substances? The molecules of the substances must be touching. Both substances must be the same temperature. At least one substance must be a liquid.

Explanation:

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The chart shows the percentage of different elements in the human body.
GenaCL600 [577]
I believe that answer is nitrogen.
8 0
3 years ago
Two identical objects, A and B, are sitting on a table. If the net force on object A is 5 N and the net force on object B is 10
sveta [45]

If the net force on object A is 5 N and the net force on object B is 10 N, then object B will accelerate more quickly than object A provided the mass of both objects are same.

Answer: Option C

<u>Explanation: </u>

According to Newton’s second law of motion, any external force applied on an object is directly proportional to the mass and acceleration of the object. In order to state this law in terms of acceleration, it is stated that acceleration exhibited by any object is directly proportional to the net force applied on the object and inversely proportional to the mass of the object as shown below:

                      \text {Acceleration of the object } \propto \frac{\text {Net force on the object}}{\text {Mass of the object}}

So if two objects A and B are identical which means they have same mass, then the acceleration attained by the object will be directly proportionate to the net forces exerted on the objects only.

Thus if the force applied is more for one object, then the object will be exhibiting more acceleration compared to the other one. So as object B is experiencing a net force of 10 N which is greater than the net force experiences by object A, then the object B will be accelerating more quickly compared to the object A's acceleration.

7 0
3 years ago
Describe the shape of the light waves that would be created from a single, uncovered light bulb
lilavasa [31]
Curvy lines with light glooming
6 0
3 years ago
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.
S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }

Where:

\omega - Angular frequency, measured in radians per second.

m - Mass of the physical pendulum, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

d - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

I_{O} - Moment of inertia with respect to pivot point, measured in kg\cdot m^{2}.

In addition, frequency and angular frequency are both related by the following formula:

\omega =2\pi\cdot f

Where:

f - Frequency, measured in hertz.

If f = 0.658\,hz, then angular frequency of the physical pendulum is:

\omega = 2\pi \cdot (0.658\,hz)

\omega = 4.134\,\frac{rad}{s}

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}

I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}

Given that m = 1.15\,kg, g = 9.807\,\frac{m}{s^{2}}, d = 0.425\,m and \omega = 4.134\,\frac{rad}{s}, the moment of inertia associated with the physical pendulum is:

I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}

I_{o} = 0.280\,kg\cdot m^{2}

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

8 0
3 years ago
A 2.0 kg bucket is attached to a horizontal ideal spring and rests on frictionless ice. You have a 1.0 kg mass
bogdanovich [222]

Answer:

x = A cos (w \sqrt{2y_{o}/g})

a) maximun  Ф= \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) minimun     Ф = \frac{\pi }{2} - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

Explanation:

For this exercise let's use kinematics to find the time it takes for the mass to reach the floor

         y = y₀ + v₀ t - ½ g t²

   

as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)

      0 = y₀ - ½ g t²

      t = \sqrt{2y_{o}/g}

The bucket-spring system has a simple harmonic motion, which is described by

     x = A cos wt

in this expression we assumed that the phase constant (Ф) is zero

let's replace the time

     x = A cos (w \sqrt{2y_{o}/g})

this is the distance where the system must be for the mass to fall into it.

a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes

          w = \sqrt{k/m}

In the initial state

         w = \sqrt{k/2}

When the mass changes

         w ’= \sqrt{k/3}

the displacement in each case is

         x = A cos (wt)

for the new case

        x ’= A cos (w’t + Ф)

the phase constant is included to take into account possible changes due to the collision of the mass.

we see that this maximum expressions when the cosine is maximum

        cos (w´t + Ф) = 1

         w’t + Ф = 0

        Ф = -w ’t

        Ф = - \sqrt{k/3} \sqrt{2y_{o}/g}

       \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) the function is minimun if

        cos (w’t + fi) = 0

        w’t + Ф = π / 2

        Ф = π / 2 - w ’t

        Ф = \frac{\pi }{2} - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

8 0
3 years ago
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