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1 year ago

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Skater N, 58.9 kg, is moving north at a speed of 7.8 m/s when she collides with Skater E, 72.6 kg, moving east at 3.5 m/s. The t

wo skaters are stuck together. In what direction do they move after the collision?
a. 61 degrees c. 32 degrees

b. 45 degrees d. 27 degrees

1 answer:

Alik [6]1 year ago

4 0

The direction of the skaters after the collision is 61 degrees.

<h3>Momentum of the skaters</h3>

The momentum of the skaters is calculated as follows;

P(N) = 58.9 x 7.8 = 459.42 kgm/s

P(E) = 72.6 x 3.5 = 254.1 kgm/s

<h3>Direction of the skaters after collision</h3>

The direction of the skaters after the collision is calculated as follows;

$\theta = tan^{-1} (\frac{459.42}{254.1} )\\\\\theta = 61^ 0$

Thus, the direction of the skaters after the collision is 61 degrees.

Learn more about direction of collision here: brainly.com/question/25891473

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What landform is created when magma squeezes between rock layers to form a hardened slab of rock?

Zigmanuir [339]

<span>D. sill

A sill forms when hot magma squeezes between layers of rock to form a hardened sheet of rock. A sill is basically a tabular sheet intrusion between old layers of sedimentary rock, volcanic beds, or in the direction of foliation inherent in metamorphic rock. The word sill means a concordant intrusive sheet.</span>

5 0

2 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

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Answer:

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Explanation:

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