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Gekata [30.6K]
3 years ago
8

How long does it take an automobile traveling in the left lane of a highway at 60.0 km/h to overtake (become even with) another

car that is traveling in the right lane at 20.0 km/h when the cars' front bumpers are initially 55 m apart?
Physics
1 answer:
pashok25 [27]3 years ago
8 0

Answer:

4.95 seconds

Explanation:

It is given that the speed of the cars are constant and not accelerating

The relative speed between the two cars will be the difference in their speeds

S_r=60-20\\\Rightarrow S_r=40\ km/h

Converting to m/s

1\ km/h=\frac{1}{3.6}\ m/s

\\\Rightarrow 40\ km/h=40\times \frac{1}{3.6}\ m/s\\ =\frac{100}{9}\ m/s\\ =11.11\ m/s

Time = Distance / Speed

Time=\frac{55}{\frac{100}{9}}=4.95\ seconds

Time it would take the faster car to cross the slower car is 4.95 seconds

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58 cm/s

Explanation:

0.5×129=0.5×(-45)+1.5×V

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Lifting a box off the floor is an example of what type of force?
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Gravitational I think would be the answer, Hope this helps!
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A car has the velocity of 2.35 after 4.67 it's velocity is 9.89 what is the average acceleration
kiruha [24]
Average acceleration  =  (change in speed) / (time for the change) .
  
Change in speed = (ending speed) - (beginning speed)

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3 years ago
Differenciate between fundamentals quantities and derived quantities
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Answer:

Fundamental quantities are the base quantities of a unit system, and they are defined independent of the other...

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3 years ago
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A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
3 years ago
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