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2 years ago

Two waves collide and the temporary combined waves that results is smaller that the original wave.

1 answer:

5 0

A.

Destructive interference is when two waves cancel each other out or when the crest of one wave passes through the trough of another wave.

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The relationship between mass and inertia is described by newton's second law of motion. true or false

Mariana [72]

False. Inertia and mass is not described in Newton’s second law of motion but in Newton’s first law of motion. Newton’s first law of motion or sometimes referred to as the law of inertia. In Newton’s first law indicates that an object at rest will remain at rest unless acted by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

4 0

3 years ago

If you move 50 meters in 10 seconds, what is your speed

iogann1982 [59]

Explanation:

My answer didn't save :(

8 0

2 years ago

17. (16 pts) You set up a two-slit experiment with a laser that produces light with a

Natali [406]

when the two waves interfere with eachother to make a dark spot the periodic difference of the two waves is π . the wave length for 2π is 600nm

. ie. for π difference it is 300nm

4 0

3 years ago

A spring has a force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is

kenny6666 [7]

Answer:

$f2/f1 = \sqrt{2}$

Explanation:

From frequency of oscillation

$f = 1/2pi *\sqrt{k/m}$

Initially with the suspended string, the above equation is correct for the relation, hence

$f1 = 1/2pi *\sqrt{k/m}$

where k is force constant and m is the mass

When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

$f2 = 1/2pi *\sqrt{2k/m}$

Employing f2/ f1, we have

$f2/f1 = \sqrt{2}$

3 0

2 years ago

A charged particle of mass 0.0050 kg is subjected to a 5.0 T magnetic field which acts at a right angle to its motion. If the pa

Irina18 [472]

Answer:

0.01 C

Explanation:

Applying,

F = qvBsinФ................ Equation 1

Where F = Force on the charged particle, q = charge on the particle, v = velocity, B = magnetic field, Ф = angle

Since the charged particle noves in a circle,

F = mv²/r................. Equation 2

Where m = mass of the particle, v = velocity of the particle, r = radius of the circle

Substitute equation 2 into equation 1

mv²/r = qvBsinФ

make q the subject of the equation

q = mv/(rBsinФ)............. Equation 3

Given: m = 0.005 kg, v = 2 m/s, r = 0.2 m, B = 5 T, Ф = 90° (Act at right angle)

Substitute these values into equation 3

q = (0.005×2)/(0.2×5×sin90°)

q = 0.01/(1)

q = 0.01 C

5 0

2 years ago