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2 years ago

Before Collision:

Physics

1 answer:

marissa [1.9K]2 years ago

6 0

Answer:

4/3 m/s

Explanation:

Assuming momentum is conserved, the sum of products of mass and speed before the collision is the same as after:

(2000 kg)(4 m/s) +(4000 kg)(0 m/s) = (2000 +4000 kg)(Vt)

Vt = (8000 kg·m/s)/(6000 kg) = 4/3 m/s

The speed of the combined objects after the collision is 4/3 m/s.

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SOVA2 [1]

Answer:magnitude -5; angle 160°

Explanation:

Vector A is described as having magnitude 5 and angle -20°.

To get an equivalent vector, we either leave the magnitude at 5 and add 360° to the angle, or we reverse the magnitude to -5 and add 180° to the angle.

5 @ -20° = 5 @ 340°

5 @ -20° = -5 @ 160°

The third one is the answer.

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3 years ago

A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se

Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

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charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

$v = \frac{Bqr}{m}$

$v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}$

v = 661098.9 = 661099 m/s

(b)

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e / m = 1.76 x 10^14 C / kg

K = 0.5 x mv²

K = 0.5 x 9.1 x 10^-31 x 661099 x 661099

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K = 1.24 eV

So, the potential difference is

V = 1.24 V

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V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J

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K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

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$r=\frac{mv}{Bq}$

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7 0

3 years ago

#1 Not sure where to start. This is for AP Physics!

yaroslaw [1]

First,

$\rho=\dfrac mV$

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$2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V$

$\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3$

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So we have

$V=Ax$

where $A$ is the given area, so

$4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x$

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