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Trava [24]
4 years ago
15

The scientist most often credited with the idea that matter can have wave-like properties is:

Physics
2 answers:
Flura [38]4 years ago
6 0
I believe the correct answer from the choices listed above is option B. The scientist most often credited with the idea that matter can have wave-like properties is de Broglie. He <span>related E=mc^2, E=hf and p=mv in order to acquire the de Broglie wavelength, whose formula λ=h/p where h is Planck's constant and p is momentum of the particle.</span>
love history [14]4 years ago
6 0

the correct answer is de Broglie. I am sure of it.

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The wavelength of a wave is the distance between
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The anwser is a even bigger wave is coming
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4 years ago
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What is the average acceleration of a jet ski that goes from rest to 10 m/s west in 4 s? Take the direction west to be positive,
skad [1K]

|acceleration| = (change in speed) / (time for the change)

                       = (10 m/s - 0)          /  (4 s)

                       =        (10 / 4) (m/s²)

                       =            2.5 m/s² .

The direction of the acceleration is west.
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3 years ago
Ahmad pulls a sled across level snow by 75 N . If the sled moves 15 m , how much work does Ahmad do on the sled?
Nataly_w [17]

Answer:

1125 J

Explanation:

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= 75 x 15

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3 years ago
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7 0
4 years ago
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

5 0
3 years ago
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