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Trava [24]
3 years ago
10

A jogger maintained a constant velocity of 6 m/s while jogging. How far did the jogger travel if they ran at that velocity for a

time of 3 seconds?
Physics
1 answer:
myrzilka [38]3 years ago
4 0
18 i’m pretty sure, if it’s wrong don’t blame me xx good luck
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The PV system is operating in a location where the annual average daily incident solar energy (the insolation) on the array equa
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The average amount of solar energy incident on the PV per day is 10000 kWh/day.

<h3>Equation</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Let the PV array has an area equal to 50 square meters. Hence:

Average amount of solar energy incident on the PV per day = 200 kWh/m²/day * 50 m² = 10000 kWh/day.

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6 0
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What the unit of work?​
ra1l [238]

Answer:

yes

Explanation:

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3 years ago
How is a full moon different from a new moon?
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<span>A full moon is at its brightest, and here is no disk to be seen. New moons are barely visable.</span>
8 0
3 years ago
Read 2 more answers
A 16.5-kg crate starts at rest at the top of a 60.0° incline. The coefficients of friction are μs = 0.400 and μk = 0.300. The cr
irga5000 [103]

Answer:

t = 1.62 s

Explanation:

given,

mass of the block m₁ = 16.5 Kg

m₂ = 8 Kg

angle of inclination = 60°

μs = 0.400 and μk = 0.300

time to slide 2 m = ?

a) let a is the acceleration of the block m₁ downward.

Net force acting on m₂,

F₂ = T - m₂ g

m₂a = T - m₂ g

a = \dfrac{T}{m_2} - g.......(1)

net force acting on m₁

F₁ = m₁g sin(60°) - μ_k m₁g cos (60°) - T

m₁ a = m₁g sin(60°) - μ_k m₁g cos (60°) - T

a = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}.........(2)

from equations 1 and 2

\dfrac{T}{m_2} - g = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}

\dfrac{T}{m_2} +\dfrac{T}{m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)

 T\dfrac{m_1+m_2}{m_2\times m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)

 T = \dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{\dfrac{m_1+m_2}{m_2\times m_1}}

 T = {m_2\times m_1}\dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{{m_1+m_2}}  

T = {16.5\times 8}\dfrac{9.8 + 9.8 sin(60^0) - 0.3\times 9.8 cos (60^0)}{{16.5+8}}

T = 90.61 N

from equation (1)

a = \dfrac{90.61}{8} - 9.8.......(1)

a = 1.52 m/s²

let t is the time taken

Apply,

d = ut + 0.5 a t²

2 = 0 + 0.5 x 1.52 x t²

t = \sqrt{2.63}

t = 1.62 s

5 0
3 years ago
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