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Alex73 [517]
3 years ago
11

A 67-kg base runner begins his slide into second base when he is moving at a speed of 3.6 m/s. The coefficient of friction betwe

en his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base. (a) How much mechanical energy is lost due to friction acting on the runner? J
Physics
1 answer:
USPshnik [31]3 years ago
4 0

Answer:

434.16 Joules

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of person

From work-energy theorem

KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}\times 67\times (3.6^2-0)\\\Rightarrow KE=434.16\ Joules

The runner loses 434.16 Joules of mechanical energy.

W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow \mu mg=\frac{W}{s}\\\Rightarrow s=\frac{W}{\mu mg}\\\Rightarrow s=\frac{434.16}{0.7\times 67\times 9.81}\\\Rightarrow s=0.94364\ m

He slides 0.94364 m

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Consider two straight wires lying on the x-axis, separated by a gap of 4 nm. The potential energy in the gap is about 3 eV highe
Tcecarenko [31]

Answer: 1.3 ×10^-31

Explanation:

the required probability is P = e^(-2αL)

Firstly, evaluate (-2αL)

α= 1/hc √2mc^2 (U - E)

h= modified planck's constant

where,

(-2αL)

= -(2L)/(h/2π ) ×√2mc^2 (U - E)

= -(2L) / (hc^2/π )×√2mc^2 (U - E)

(hc^2/2pi) = 197*eV.nm (standard constant)

2*L = 8 nm

mc^2 = 0.511×10^6 eV

Where m = mass electron

C= speed of light

(-2αL) = [-8nm/(197 eV.nm)] × (1.022× 10^6 eV*×3 eV)^0.5

(-2αL) = -71.1

Probability = e^(-2αL) = e^-71.1 = 1.3 ×10^-31

Therefore, the probability that a ondu tion ele tron in one wire arriving at the gap will pass through the gap into the other wire 1.3 ×10^-31.

3 0
3 years ago
Absolute pressure in tank is P1 = 260 kPa and local ambient absolute pressure is P2 =100 kPa. If liquid density in pipe is 13600
Pavlova-9 [17]

Answer:

1.176m

Explanation:

Local ambient pressure(P1) = 100 kPa

Absolute pressure(P2)=260kPa

Net pressure=absolute pressure-local ambient absolute pressure

Net pressure=P1(absolute pressure)-P2(local ambient absolute pressure)

Net pressure=260-100=160kPa

Pressure= ρgh

160kPa=13600*10*h

h=\frac{160000}{136000}

h=1.176m

5 0
3 years ago
A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet
astraxan [27]

Answer:

Acceleration = 2.35 m/s^{2}

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = \frac{mg}{2} = 0.5 mg

Friction along incline = umg cos30° = mg 0.3\times \frac{\sqrt{3}}{2}

Friction along incline  = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = \frac{net force}{mass} = 0.24 g = 2.35 m/s^{2}

The height of incline = 8 m

Length of the inclined edge = 16 m

v^{2}=u^{2}+2as

v^{2}= 2\times 0.24 \times 9.8\times 16

v= 8.67 m/s

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3 years ago
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Answer:

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Explanation:

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3 years ago
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