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charle [14.2K]
2 years ago
10

The height of the tide measured at a seaside community varies according to the number of hours t after midnight. If the height h

, in feet, is currently given by the equation h=-1/2 t^2+6t-9, when will the tide first be at 6 ft?
Physics
1 answer:
antoniya [11.8K]2 years ago
6 0

Explanation:

Given that, the height of the tide measured at a seaside community varies according to the number of hours t after midnight. The height is given by the equation as :

h=-\dfrac{1}{2}t^2+6t-9

When the tide first be at 6 ft, put h = 6 ft in above equation as :

-\dfrac{1}{2}t^2+6t-9=6

-t^2+12t-18=0

On solving the above equation to find the value of t. It is equal to :

t = 3.551 seconds

or

t = 8.449 seconds

So, the tide of 6 ft is at  3.551 seconds and 8.449 seconds. Hence, this is the required solution.

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Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti
Alex

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = m_1=6.50 kg- 1.55 kg=4.95 kg

Velocity achieved by squid = v_1=2.60 m/s

Momentum gained by squid = P=m_1v_1

Mass of water = m_2=1.55 kg

Velocity by which water was released by squid = v_2

Momentum gained by water but in opposite direction = P'=m_2v_2

P = P'

m_1v_1=m_2v_2

v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E  =\frac{1}{2}m_1v_1^{2}

Kinetic energy of water = K.E' = \frac{1}{2}m_2v_2^{2}

Total kinetic energy created by this maneuver:

K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}

=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules

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3 years ago
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A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.
kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

F_k = \mu_k N = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

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