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Dmitrij [34]
3 years ago
11

Junior slides across home plate during a baseball game. If he has a mass of 115 kg, and the coefficient of kinetic friction betw

een him and the ground is 0.35, what is the force of friction acting on him?
Physics
2 answers:
uysha [10]3 years ago
6 0

Answer:

like the other person said, it is 395

Explanation:

tatyana61 [14]3 years ago
4 0

Weight Force of Junior = m g = 115kg x 9.81 m/s^2 = 1128.15N then compute for the friction force


Friction Force= WF x (coefficient of kinetic friction) = 1128.15N x 0.35 =  394.8525N or 395N

 

But you can compute in a straightway:

Solution:

= 115 x 9.81 x 0.35

= 394.85

= 395 N

 

It will still give the same results.

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An ordinary flashlight battery has a potential difference of 1.2 V between its positive and negative terminals. How much work mu
Maru [420]

The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.

Given:

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4 0
2 years ago
Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher
siniylev [52]

Answer:

v = 4.44 \times 10^5 m/s

Explanation:

By Einstein's Equation of photoelectric effect we know that

h\nu = W + \frac{1}{2}mv^2

here we know that

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now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

W = \frac{hc}{351 nm}

here we know that

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now we have

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now by energy equation above when photon of 303 nm incident on the surface

\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2

4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2

(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2

8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2

v = 4.44 \times 10^5 m/s

6 0
3 years ago
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Answer:

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