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3 years ago

How do you calculate time from launch to explosion in a firework

Physics

2 answers:

xenn [34]3 years ago

8 0

Answer:

Explanation:

If a rocket (firework) is launched

with an initial velocity of 39.2 meters per second at a height of 1.6 meters above

the ground, the equation h = -4.9t2

+ 39.2t + 1.6 represents the rocket’s height h in

meters after t seconds. The rocket will explode at approximately the highest point.

Fire official require the fireworks to explode at a height greater than 76 feet in

order for debris to land in Lake Michigan and not on the festival sight. The event

planners assured the fire officials that the fireworks would reach a height greater

than 76 feet

Sholpan [36]3 years ago

4 0

Answer:

If a rocket (firework) is launched with an initial velocity of 39.2 meters per second at a height of 1.6 meters above the ground, the equation h = -4.9t2 + 39.2t + 1.6 represents the rocket's height h in meters after t seconds. The rocket will explode at approximately the highest point.

Explanation:

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A vessel with an unknown volume is filled with 10 kg of water at 90oC. Inspection of the vessel at equilibrium shows that 8 kg o

damaskus [11]

Hey there!

In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).

Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:

$x=\frac{m_{steam}}{m_{total}}$

Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:

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Now, at this temperature and pressure, the volume of a saturated vapor is 2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:

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2 years ago

A passenger bus is travelling 28.0 m/s to the right when the driver applies the brakes. The bus stops in 5.00 s. What is the acc

MAVERICK [17]

Change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right

Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s

Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right

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