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Sladkaya [172]
3 years ago
8

A runner of mass 51.8 kg starts from rest and accelerates with a constant acceleration of 1.31 m/s^2 until she reaches a velocit

y of 5.47 m/s. She then continues running at this constant velocity. How long does the runner take to travel 165 m?
Physics
1 answer:
Stells [14]3 years ago
6 0

Answer:32.24 s

Explanation:

Given

mass of runner (m)=51.8 kg

Constant acceleration(a)=1.31 m/s^2

Final velocity (v)=5.47 m/s

Time taken taken to reach 5.47 m/s

v=u+at

5.47=0+1.31\times t

t=\frac{5.47}{1.31}=4.17 s

Distance traveled during this time is

s=ut+\frac{1}{2}at^2

s=\frac{1}{2}\times 1.31\times 4.17^2=11.42 m

So remaining distance left to travel with constant velocity=153.57 m

thus time =\frac{distance}{speed}

t_2=\frac{153.57}{5.47}=28.07 s

Total time=28.07+4.17=32.24 s

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5 0
4 years ago
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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 44.1m/s^2 . The acce
ollegr [7]
The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0 
y=44.1*100/2 = 2205m 
hence, the speed will be 
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7 0
3 years ago
A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a
UNO [17]

Answer:

P_{avg} = 6.283*10^{-9} \ W

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})

where;

E= \frac{d \phi}{dt}

E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

P_{avg} = 6.283*10^{-9} \ W

6 0
4 years ago
Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iro
belka [17]

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4 0
3 years ago
The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being inve
solmaris [256]

Answer:

the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

Explanation:

Given the data in the question;

we make use of the following expression;

hall Voltage VH = IB / ned

where I = 2.25 A

B = 0.685 T

d =  0.107 mm =  0.107 × 10⁻³ m

e = 1.602×10⁻¹⁹ C

VH = 2.59 mV = 2.59 × 10⁻³ volt

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so from the form; VH = IB / ned

VHned = IB

n = IB / VHed

so we substitute

n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )

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n = 3.4716 × 10²⁵ m⁻³

Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

5 0
3 years ago
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