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zavuch27 [327]
4 years ago
13

A train left a station at noon, travelling at 40 mph. A second train left at the station at 3 PM traveling at 60 mph. At what ti

me will the second train catch up with the first
Physics
2 answers:
Arlecino [84]4 years ago
7 0

Answer:

9:00 pm

Explanation:

hope it helps

kolezko [41]4 years ago
4 0

Answer:

both train meet at 9.00 pm

Explanation:  

let t is the time after 1st train start  

we know distance = time * speed

D_1 = 40*t

D_2= 60*(t-3)                ( as second train depart after 3hr gap)  

we know

distance by one train = distance traveled by 2nd train i.e

D_1 = D_2

40*t=60*(t-3)  

40*t = 60*t - 60*3  

20*t = 180

t = 9  

hence both train meet at 9.00 pm

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The awnser is condensation
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Emily uses a rifle to shoot a bullet at a target. The bullet has a mass of 13 grams. The rifle has a mass of 3,500 grams. When s
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4 0
3 years ago
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3 years ago
A flat sheet of ice has a thickness of 2.20 cm. It is on top of a flat sheet of crystalline quartz that has a thickness of 1.50
kolezko [41]

Answer:

Distance_{vaccum}=5.19cm

Explanation:

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Total time = Time taken through ice + Time taken through quartz

Time taken through ice = Thickness of ice / (speed of light in ice)

T_{ice}=\frac{2.20\times 10^{-2} \times \mu _{ice}}{V_{vaccum}}

T_{quartz}=\frac{1.50\times 10^{-2} \times \mu _{quartz}}{V_{vaccum}}

Thus in the same time the it would had covered a distance of

Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}]

we have

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Applying values we have

Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542]

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6 0
4 years ago
A truck is moving around a circular curve at a uniform velocity of 13 m/s. If the centripetal force on the truck is 3,300 N and
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Answer: Option B.

Since here the truck is moving on a circular track, it will experience centripetal force.

F(centripetal) = m × acc 
or 
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where r is the radius of the track. 
m is the mass of truck
v is the speed  of the truck. 
Given: v = <span>13 m/s
m = </span><span>1,600 kg
</span>F = 3300 Newton

To find = radius of track=?
r = \frac{m v^{2} }{F}
r = \frac{1600*13*13}{3300}
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4 0
3 years ago
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