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lana66690 [7]
2 years ago
8

PLEASE HELP!!! WOULD REALLY APPRECIATE!

Physics
1 answer:
Eva8 [605]2 years ago
3 0

Answer:

a. slope=rise/run

rise=0.02

run=-2

determined using the point (3,0.08) and (1,0.1) on the graph

slope=0.02/-2

= -0.01 or -1/100

b.area= area of trapizoid+ rectangle

((0.07+0.11)÷2)×4+1×0.07

0.36+0.07

=0.43$

c. the area represent the total cost after 5 hours

PLEASE MARK BRAINLIEST

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s_i is the initial position of the car which is certain height H

s_i is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

v^2 = 0 + 2g[H - 0]

=> v  =  \sqrt{ 2 g H}

b

H  =  9.86 \  m

Explanation:

Generally from third equation of motion we have that

v^2 =  u^2 + 2a[s_i - s_f]

Here v is the final speed of the car

u is the initial speed of the car which is zero

s_i is the initial position of the car which is certain height H

s_i is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

v^2 = 0 + 2g[H - 0]

=> v  =  \sqrt{ 2 g H}

When v  = 50 \  km/h = \frac{50 *1000}{3600} = 13.9 \  m/s we have that

13.9  =  \sqrt{ 2 g H}

=> H  =  \frac{13.9^2}{2 *  9.8}

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Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards​ Vernonville, and another car leaves Verno
iragen [17]

Answer:

Average speed of the car A = 70 miles per hour

Average speed of the car B = 60 miles per hour

Explanation:

Average speed of the car A is v_{A} =\frac{x_{A} }{t_{A} } (Equation A) and Average speed of the car B is v_{B} =\frac{x_{B} }{t_{B} } (Equation B), where x_{A} and x_{B} are the distances and t_{A} and t_{B} are the times at which are travelling the cars A and B respectively.

We have to convert the time to the correct units:

1 hour and 36 minutes = 96 minutes

96 minutes . \frac{1 hour}{60 minutes} = 1.6 h

From the diagram (Please see the attachment), we can see that at the time they meet, we have:

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v_{B} = \frac{208-x}{1.6h} (Equation D)

From Equation A and C, we have:

\frac{208-x}{1.6}+10 = \frac{x}{1.6}

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v_{A}  = \frac{112miles}{1.6h}

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Replacing x in Equation B:

v_{B}  = \frac{208miles-112miles}{1.6h}

v_{B}  = \frac{96miles}{1.6h}

v_{B}  = 60 miles per hour

3 0
3 years ago
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