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alina1380 [7]
3 years ago
13

What is formula of baking soda.​

Physics
2 answers:
Anika [276]3 years ago
6 0

NaHCO3

That is the formula for baking soda

goldenfox [79]3 years ago
5 0

baking soda formula is NaHCO3

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Two waves are superposed. one wave has an amplitude of 5 cm, and other has an amplitude of 4 cm. What is the resultant amplitude
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If both waves have the same wavelength, then the amplitude of
their sum could be anything between 1 cm and 9 cm, depending
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If the waves have different wavelengths, then the resultant is a beat
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Draw the graphs for exothermic and endothermic reactions Label: axes, reactants, products, Energy of reaction, heat energy chang
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axes, reactants, products, Energy of reaction, heat energy

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A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above
babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

1)

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

h(t) = h_0 + ut - \frac{1}{2}gt^2 (1)

where

h_0 is the height at time t = 0

u is the initial vertical velocity

g=32 ft/s^2 is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

f(t)=-16t^2+44t+88 (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

h_0 = 88 ft

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

2)

The rock will hit the water when its height becomes zero, so when

f(t)=0

which means when

0=-16t^2+44t+88

First of all, we can simplify the equation by dividing each term by 4:

0=-4t^2+11t+22

This is a second-order equation, so we solve it using the usual formula and we find:

t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

3)

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

v=u-gt

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

v=0

which means

0=u-gt

Here we have

u=+44 ft/s (initial velocity)

g=32 ft/s^2 (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

t=\frac{u}{g}=\frac{44}{32}=1.38 s

4)

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

f(t)=-16t^2+44t+88

Substituting t = 1.38 s, we find:

f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m

So, the maximum height reached by the rock during its motion is

h_{max}=118.2 m

Which means 118.2 m above the water.

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Answer:

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