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Levart [38]
2 years ago
15

if pete (mass =90.00kg) weighs himself and finds that he weighs 30.0 pounds, how far away from the earth is he?

Physics
1 answer:
Svet_ta [14]2 years ago
4 0

Here we know that mass of the person is 90 kg

His weight is given as 30 lbf

so here we can convert it into Newton as we know that

1 lbf = 4.45 N

Now from above conversion

30 lbf = 30 \times 4.45 = 133.45 N

now we can use this to find the gravity at this height

mg = 133.45

90\times g = 133.45

g = 1.49 m/s^2

now we know that with height gravity varies as

g' = g(\frac{R}{R+H})^2

1.49 = 9.8(\frac{6.37\times 10^6}{6.37\times 10^6 + h})^2

h = 1.0 \times 10^7 m

so above is the height from the surface of earth

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Answer:

14 m/s

Explanation:

u = 0, h = 10 m, g = 9.8 m/s^2

Use third equation of motion

v^2 = u^2 + 2 g h

Here, v be the velocity of ball as it just strikes with the ground

v^2 = 0 + 2 x 9.8 x 10

v^2 = 196

v = 14 m/s

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2 years ago
What do all waves carry
MAVERICK [17]

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the waves carry energy

Explanation:

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2 years ago
Approximate the human form as an unclothed vertical cylinder of 0.3 m diameter and 1.75 m length with a surface temperature of 3
Zolol [24]

Answer:

q=1118.865W

Explanation:

From the question we are told that

Diameter of the cylinder D=0.3m

Length of the cylinder L=1.75m

Surface temperature of cylinder T_s=30 \textdegree C=303k

Speed of air V=10m/s

Temperature of air T=20 \textdegree C

Generally the equation for Reynolds number is mathematically given by

Re_D=\frac{VD}{v}

where

v=15.27*10^-^6

Re_D=\frac{10*0.3}{15.27*10^-^6}

Re_D=1.96*10^5

Generally the equation for Nusselt number is mathematically given by

N_u=0.6(Re_D)^{0.6}(Pr)^{0.37}

where

Prandtl number

Pr=0.71

N_u=0.6(Re_D)^{0.6}(Pr)^{0.37}

N_u=0.6(1.96*10^5)^{0.6}(0.71)^{0.37}

N_u=791.53

Generally the equation for convective heat transfer is mathematically given by

h=Nu \frac{k}{D}

where

k=25.7*10^{-3}

h=791.53*\frac{25.7*10^{-3}}{0.3}

h=67.81W/m^2K

Generally the equation for surface area of a cylinder is mathematically given by

A=\pi DL

A=\pi *0.3*1.75

A=1.65m^2

Generally the equation for convective heat transfer is mathematically given by

q=h4(T_s-T_{inty)}

q=(67.81)*1.65*(30-20)\\

q=1118.865W

3 0
2 years ago
In each cycle, a heat engine performs 710 J of work and exhausts 1480 J of heat. What is the thermal effiency?
Tomtit [17]

Answer:

32%

Explanation:

For a heat engine, efficiency is work out divided by heat in:

η = Wₒ / Qᵢ

Since energy is balanced, heat in is the sum of work out and heat out:

Qᵢ = Wₒ + Qₒ

Therefore:

η = Wₒ / (Wₒ + Qₒ)

Given Wₒ = 710 J and Qₒ = 1480 J:

η = 710 / (710 + 1480)

η = 0.32

The thermal efficiency is 32%.

6 0
3 years ago
A certain reaction with an activation energy of 185 kJ/mol was run at 525 K and again at 545 K . What is the ratio of f at the h
Kazeer [188]

Answer: The ratio of f at the higher temperature to f at the lower temperature is 4.736

Explanation:

According to the Arrhenius equation,

f=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{f_2}{f_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

f_1 = rate constant at 525K

K_2 = rate constant at 545K

Ea = activation energy for the reaction = 185kJ/mol= 185000J/mol   (1kJ=1000J)

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 525 K

T_2 = final temperature = 545 K

Now put all the given values in this formula, we get

\log (\frac{f_2}{f_1})=\frac{185000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{525K}-\frac{1}{545K}]

\log (\frac{f_2}{f_1})=0.6754

(\frac{f_2}{f_1})=4.736

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 4.736

8 0
3 years ago
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