Answer:
12.32 m/s
Explanation:
Using the formula of maximum height of a projectile,
H = u²sin²Ф/2g................... Equation 1
Where H = maximum height, u = initial velocity, Ф = angle of projection, g = acceleration due to gravity
make u the subject of the equation
u = √(2Hg/sin²Ф)............ Equation 2
Given: H = 2.3 m, Ф = 33°, g = 9.8 m/s²
Substitute into equation 2
u = √[(2×2.3×9.8)/sin²33°]
u =√ [45.08/(0.545)²]
u = 45.08/0.297
u = √(151.785)
u = 12.32 m/s
The final atmospheric pressure is 
Explanation:
Assuming that the temperature of the air does not change, we can use Boyle's law, which states that for a gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume. In formula,
where
p is the gas pressure
V is the volume
The equation can also be rewritten as
where in our problem we have:
is the initial pressure (the atmospheric pressure at sea level)
is the initial volume
is the final pressure
is the final volume
Solving the equation for p2, we find the final pressure:
Learn more about ideal gases:
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The car will take 300 m before it stops due to applying break.
<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
- As per Newton's equation of motion, V² - U² = 2aS
- V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
- Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
- So, 0² - 60² = 2×6× S
=> -3600 = -12S
=> S = 3600/12 = 300 m
Thus, we can conclude that the distance covered by the car is 300 m before it stopped.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?
Learn more about the Newton's equation of motion here:
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