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3 years ago

Fill-in-the-Blank

Physics

2 answers:

padilas [110]3 years ago

6 0

Answer:

Earthquakes

YES

DIA [1.3K]3 years ago

5 0

Answer:

The answer is earthquakes

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Sound travels fastest through A) space. B) cool air. C) warm air. D) a metal spoon.

Mademuasel [1]

A metal spoon. Because metal travels the fastest

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3 years ago

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A planar electromagnetic wave is propagating in the +x direction. At a certain point P and at a given instant, the electric fiel

jeka94

Answer:

$B=2.74\times 10^{-10}\ T$

Explanation:

It is given that,

A planar electromagnetic wave is propagating in the +x direction.The electric field at a certain point is, E = 0.082 V/m

We need to find the magnetic vector of the wave at the point P at that instant.

The relation between electric field and magnetic field is given by :

$c=\dfrac{E}{B}$

c is speed of light

B is magnetic field

$B=\dfrac{E}{c}\\\\B=\dfrac{0.082}{3\times 10^8}\\\\B=2.74\times 10^{-10}\ T$

So, the magnetic vector at point P at that instant is $2.74\times 10^{-10}\ T$ .

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3 years ago

The maximum displacement in an oscillatory motion is A = 0.49 m. Determine the position x at which the kinetic energy of the par

amid [387]

Answer:

x = 0.40 m

Explanation:

When the displacement is maximum, the particle is momentarily at rest, which means that at this point (assuming no friction present) all the mechanical energy is elastic potential, which can be written as follows:

$E_{tot} = U_{o} = \frac{1}{2} *k*A^{2} (1)$

Since in absence of friction, total mechanical energy must keep constant, this means that at any time, the sum of the kinetic and potential energy, must be equal to (1), as follows:

$E_{tot} = U_{o} = \frac{1}{2} *k*A^{2} = (KE)_{f} + U_{f} (2)$

If KEf = U/2f, replacing in (2), we get:

$E_{tot} = U_{o} = \frac{1}{2} *k*A^{2} = (U/2)_{f} + U_{f} = \frac{3}{2} *U_{f} (3)$

At any point, the elastic potential energy is given by the following expression:

$U_{f} = \frac{1}{2} *k*x^{2} (4)$

where k= spring constant (N/m) and x is the displacement from the

equilibrium position.

Replacing (4) in (3), simplifying and rearranging, we get:

$E_{tot} = U_{o} = \frac{1}{2} *A^{2} = \frac{3}{4} *x^{2} (5)$

Finally, solving for x, we get:

$x = \sqrt{\frac{2}{3} } * A = \sqrt{\frac{2}{3} } * 0.49m = 0.40 m (6)$

8 0

3 years ago

A brick is resting on the surface of a flat board. as one end of the board is slowly raised, what change, if any, is there in th

vovangra [49]

The normal force decreases, this is the frictional force. It will be counteracted by the force which accelerates the brick to slide downward opposite to the end where the board is raised. As the angle increases the force acting upon the brick opposite to the normal force will decrease.

5 0

4 years ago

A 2.00 kg, frictionless block s attached to an ideal spring with force constant 550 N/m. At t = 0 the spring is neither stretche

gladu [14]

Answer:

a) A = 0.603 m , b) a = 165.8 m / s² , c) F = 331.7 N

Explanation:

For this exercise we use the law of conservation of energy

Starting point before touching the spring

Em₀ = K = ½ m v²

End Point with fully compressed spring

$Em_{f}$ = $K_{e}$ = ½ k x²

Emo = $Em_{f}$

½ m v² = ½ k x²

x = √(m / k) v

x = √ (2.00 / 550) 10.0

x = 0.603 m

This is the maximum compression corresponding to the range of motion

A = 0.603 m

b) Let's write Newton's second law at the point of maximum compression

F = m a

k x = ma

a = k / m x

a = 550 / 2.00 0.603

a = 165.8 m / s²

With direction to the right (positive)

c) The value of the elastic force, let's calculate

F = k x

F = 550 0.603

F = 331.65 N

5 0

3 years ago