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katrin [286]
3 years ago
9

A driver in a blue car traveling at 50 km h−1 sees a red car approaching in his rear view mirror. the red car is traveling at 60

km h −1 and is 30 m behind the blue car when first spotted.
a.how many seconds from the time the driver of the blue car first noticed it until the red car passes the blue car?
b.how much farther down the road will the blue car travel in this time?
Physics
1 answer:
diamong [38]3 years ago
8 0
The blue car in front travels at a slower speed compared to the red car behind. Eventually, the red car would have to overtake the blue car because it is much faster. First, let's compute the time it takes before the red car catches up to the blue car. The solution is as follows:

30 m = (60 km/h - 50 km/h)*(1000 m/1 km)*(1 h/3,600 s)*(t)
t = 10.8 seconds

After 10.8 seconds, the red car catches up to the blue car. With this amount of time, the blue car would still cover additional distance. That would be equal to:

Distance = Speed*time
Distance = (50 km/h)*(1 h/3600 s)*(10.8 s)
Distance = 0.15 km


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Answer:

Part A

The volume of water displaced is 4.1887902 × 10⁻³ m³

Part B

The buoyant force is approximately 40.93 N

Explanation:

From the question, we have;

The radius of the ball suspended (barely floating) in the water, r = 0.1 m

The density of the water, ρ = 997 kg/m³

Part A

The volume of the ball = The volume of a sphere = (4/3)·π·r³

∴ The volume of the ball = (4/3) × π × 0.1³ = 0.0041887902 m³ = 4.1887902 × 10⁻³ m³

Therefore;

The volume of water displaced, V = The volume of the ball = 4.1887902 × 10⁻³ m³

The volume of water displaced, V = 4.1887902 × 10⁻³ m³

Part B

The buoyant force = The weight of the water displaced = Mass of the water, m × The acceleration due to gravity, g

The buoyant force = m × g

Where;

g ≈ 9.8 m/s²

The mass of the water, m = ρ × V

∴ m = 997 kg/m³ × 4.1887902 × 10⁻³ m³ = 4.17622383 kg

The buoyant force = 4.17622383 kg × 9.8 m/s² ≈ 40.93 N.

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