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2 years ago

On a cold winter day, the flow of heat is from the outside in. A.True B.False

Physics

2 answers:

fredd [130]2 years ago

8 0

Your answer s obviously false

lbvjy [14]2 years ago

7 0

Answer:

False

Explanation:

ooga booga, cold winter day cold outside, heat not outside ooga booga.

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The concentration of Biochemical Oxygen Demand (BOD) in a river just downstream of a wastewater treatment plant’s effluent pipe

shtirl [24]

Answer:

The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L

Explanation:

Let the initial concentration of the BOD = C₀

Concentration of BOD at any time or point = C

dC/dt = - KC

∫ dC/C = -k ∫ dt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = -kt + b (b = constant of integration)

At t = 0, C = C₀

In 1 = 0 + b

b = 0

In (C/C₀) = - kt

(C/C₀) = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C₀ = 75 mg/L

k = 0.05 /day

C = 75 e⁻⁰•⁰⁵ᵗ

So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day

We calculate how many days it takes the river to reach 50 km downstream

Velocity = (displacement/time)

15 = 50/t

t = 50/15 = 3.3333 days

So, we need the C that corresponds to t = 3.3333 days

C = 75 e⁻⁰•⁰⁵ᵗ

0.05 t = 0.05 × 3.333 = 0.167

C = 75 e⁻⁰•¹⁶⁷

C = 63.5 mg/L

5 0

2 years ago

A muon is traveling at 0.988

AlexFokin [52]

As per Einstein's relation of relativity

$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$

here we know that

$m_0 = 207 m_e = 207 \times 9.11 \times 10^{-31} kg$

$m_0 = 1.88 \times 10^{-28} kg$

now here we know that

$v = 0.988 c$

now from above equation mass of the muon is given as

$m = \frac{1.88 \times 10^{-28}}{\sqrt{1 - 0.988^2}}$

$m = 1.22 \times 10^{-27} kg$

now for the momentum of muon we can use

$P = mv$

$P = 1.22 \times 10^{-27} \times 0.988(3 \times 10^8)$

$P = 3.62 \times 10^{-19} kg m/s$

so above is the momentum of muon

6 0

3 years ago

A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05

7nadin3 [17]

Answer:

$W_{drag} = 4.223\,J$

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

$U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}$

The work done on the ball due to drag is:

$W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})$

$W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})$

$W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]$

$W_{drag} = 4.223\,J$

7 0

2 years ago

Read 2 more answers

Lonnie pitches a baseball of mass 0.500 kg. The ball arrives at home plate with a speed of 35.0 m/s and is batted straight back

Andreyy89

Answer:

Explanation:

The impulse equation is

Δp = FΔt, where Δp = final momentum - initial momentum, F is the Force exerted on an object, and Δt is the change in time. In this equation,the entire right side defines the impulse. In other words, FΔt is the impulse; thus the change in momentum an object experiences is due to its change in impulse and is directly proportional to it.

Therefore, once we find the change in momentum, that is the impulse the object experiences. Δp = final momentum - initial momentum, where

p = mv and p is momentum.

$p_f=(.500)(50.0)$ so