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nikklg [1K]
3 years ago
3

Assume that the x components of the blocks' momenta at a certain moment are p1x and p2x. find the x component of the velocity of

the center of mass (vcm)x at that moment. express your answer in terms of m1, m2, p1x, and p2x.
Physics
1 answer:
lys-0071 [83]3 years ago
4 0

Answer:

Explanation:

The amount of movement is the product of the mass of a body because of its speed, the speed is that of the center of mass; The equation is:

    p = mv

Where v is the velocity of the center of mass of the body, what we have to do is clear the velocity

      V = p / m

  It is our case for each body and as the amount of movement is a vector we must write it for each axis

   V1x = p1x / m1

   V2x = p2x / m2

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At what pressure will the mean free path in room-temperature (20?c) nitrogen be 1.5 m ?
kondaur [170]
In physical chemistry, the mean free path is the average distance between the atoms during collision. Its formula is

 Mean \ free \ path \ = \frac{RT}{ \sqrt{2}  \pi  d^{2} N_{A}P  }

where d is the diameter of the Nitrogen atom (d = 310 x 10^-7 m), Na is Avogadro's number (Na=6.022 x 10^23), R is the gas constant (8.314 J/mol-K), T is the absolute temperature (T= 20 + 273 = 293 K). Substituting the values,

1.5 = \frac{(8.314)(293)}{ \sqrt{2} \pi (310 \ x\ 10^{-7} )^{2} (6.022 \ x\ 10^{23} )P }
P = 6.32 x 10^-13 Pascals
6 0
4 years ago
In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward
BaLLatris [955]

Answer:

= 4.3 × 10 ⁻¹⁴ m

Explanation:

The alpha particle will be deflected when its kinetic energy is equal to the potential energy

Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C

Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C

Kinetic energy of  the alpha particle = 5.28 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV)

=  8.459 × 10⁻¹³

k electrostatic force constant = 9 × 10⁹ N.m²/c²

Kinetic energy = potential energy =   k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus

r = (  9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 8.459 × 10⁻¹³

= 4.3 × 10 ⁻¹⁴ m

3 0
4 years ago
Starting at t = 0 a net external force in the +x-direction is applied to an object that has mass 5.00 kg. A graph of the force a
Reptile [31]

Answer:

  15√2 N

Explanation:

The acceleration is given by ...

  a = F/m = 5t/5 = t . . . . meters/second^2

The velocity is the integral of acceleration:

  v = ∫a·dt = (1/2)t^2

This will be 9 m/s when ...

  9 = (1/2)t^2

  t = √18 . . . . seconds

And the force at that time is ...

  F = 5(√18) = 15√2 . . . . newtons

6 0
3 years ago
Accelerating charges radiate electromagnetic waves. calculate the wavelength of radiation produced by a proton in a cyclotron wi
levacccp [35]

B = magnetic field in the cyclotron = 0.400 T

q = magnitude of charge on a proton = 1.6 x 10⁻¹⁹ C

m = mass of the proton = 1.67 x 10⁻²⁷ kg

f = frequency of revolution of proton in the cyclotron = ?

v = speed of electromagnetic waves = 3 x 10⁸ m/s

λ = wavelength of electromagnetic wave = ?

Frequency of revolution of proton in the cyclotron is given as

f = qB/(2πm)

inserting the values

f = (1.6 x 10⁻¹⁹)(0.400)/(2 (3.14) (1.67 x 10⁻²⁷))

f = 6.1 x 10⁶ Hz

wavelength of electromagnetic wave is given as

λ = v/f

λ = (3 x 10⁸)/(6.1 x 10⁶)

λ = 49.2 m

7 0
3 years ago
Read 2 more answers
What is the magnetic field strength required to make a proton with a speed of 5.0 x 10^(5) m/s follow a circular path of radius
Nuetrik [128]

Answer:

Magnetic field strength required for this is 0.25 T

Explanation:

As we know that the proton moves in circular path in uniform magnetic field

so the radius of the path of the circle is given as

R = \frac{mv}{qB}

here we know that

q = 1.6 \times 10^{-19}C

m = 1.6 \times 10^{-27} kg

R = 0.0200 m

v = 5 \times 10^5 m/s

now we have

B = \frac{1.6 \times 10^{-27} (5 \times 10^5)}{(1.6 \times 10^{-19})(0.02)}

so we have

B = 0.25 T

8 0
4 years ago
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