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3 years ago

A 0.150 kg baseball has 118 j of KE. how fast is the ball moving?(unit=m/s)

Physics

1 answer:

MrRissso [65]3 years ago

4 0

Answer:

Explanation 118 = (1/2) * 0.15 * v² 118 = 0.075 * v² v² = 1573.33 m/s ... since KE = m/2*V^2 , then : V = √2KE/m = √20*118/1.5 = 39.67 m//sec ( 142.8 km/h ; 88.75 mph).:

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A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250

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Answer:x=23.4 cm

Explanation:

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3 years ago

The eyes of some reptiles are sensitive to 850 nm light. If the minimum energy to trigger the receptor at this wavelength is 3.1

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Answer:

Minimum number of photons required is 1.35 x 10⁵

Explanation:

Given:

Wavelength of the light, λ = 850 nm = 850 x 10⁻⁹ m

Energy of one photon is given by the relation :

$E=\frac{hc}{\lambda}$ ....(1)

Here h is Planck's constant and c is speed of light.

Let N be the minimum number of photons needed for triggering receptor.

Minimum energy required for triggering receptor, E₁ = 3.15 x 10⁻¹⁴ J

According to the problem, energy of N number of photons is equal to the energy required for triggering, that is,

E₁ = N x E

Put equation (1) in the above equation.

$E_{1}=N\times\frac{hc}{\lambda}$

Substitute 3.15 x 10⁻¹⁴ J for E₁, 850 x 10⁻⁹ m for λ, 6.6 x 10⁻³⁴ J s for h and 3 x 10⁸ m/s for c in the above equation.

$3.15\times10^{-14} =N\times\frac{6.6\times10^{-34}\times3\times10^{8}}{850\times10^{-9}}$

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