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zhenek [66]
2 years ago
5

Two cars are traveling along perpendicular roads, car A at 40 mi/hr, car B at 60 mi/hr. At noon, when car A reaches the intersec

tion, car B is 90 mi away, and moving toward it. At 1 p.m. the distance between the cars is changing, in miles per hour, at the rate of:
Physics
1 answer:
Serggg [28]2 years ago
3 0

Answer:

\frac{dD}{dt} = -4 miles/hour

negative sign indicates that the distance is decreasing with time

Explanation:

Let at any time t after noon that is 12 p.m.  

distance traveled by car A = 40t

distance traveled by car B = 90-60t

then distance between the two cars at time t

D^2= (40t)^2+(90-60t)^2............1

also, at time 1 p.m.

distance D^2= (40\times1)^2+(90-60\times1)^2

D=50 Km

differentiating equation 1 w.r.t. t we get

2D\frac{dD}{dt}= 2\times40t\times40+2(90-60t)(-60)

put t= 1 and D= 50 we get

2\times50\frac{dD}{dt}= 3200\times1-3600\times1

\frac{dD}{dt} = -4 miles/hour

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2 years ago
If a star with an absolute magnitude of -5 has an apparent magnitude of +5 ,then its distance is
klio [65]
You asked a question.  I'm about to answer it. 
Sadly, I can almost guarantee that you won't understand the solution. 
This realization grieves me, but there is little I can do to change it. 
My explanation will be the best of which I'm capable.


Here are the Physics facts I'll use in the solution:

-- "Apparent magnitude" means how bright the star appears to us.

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That's all the Physics.  The rest of the solution is just arithmetic.
____________________________________________________

-- The star in the question would appear M(-5) at a distance of
32.6 light years. 

-- It actually appears as a M(+5).  That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.

-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .

-- But brightness varies as the inverse square of distance,
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(32.6) · (100)^(1) light years

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I'll have to confess that I haven't done one of these calculations
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If somebody's health or safety depended on it, or the success of
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6 0
2 years ago
During takeoff, a plane goes from 0 to 50 m/s in 8s. What is its acceleration? Halos fast is it going after 5 s? How far had it
svlad2 [7]
A= 50/8 m/s^2 

<span>vf=at=50/8 * 5= 250/8 m/s at t=5sec </span>

<span>time to get to 50m/s </span>
<span>50=50/8*t or t=8 seconds </span>
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<span>distance= 400 m check that.</span>
4 0
2 years ago
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spayn [35]

Answer:

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3 years ago
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