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statuscvo [17]
3 years ago
10

I throw a stone off a cliff 32 feet above the water.  The height of the stone in terms of time (in seconds) is given by h(t) = -

16t2 + 128t + 32.  At what time will the stone splash into the water? Round your answer to nearest tenth of a second
Physics
1 answer:
devlian [24]3 years ago
5 0
<span>Your equation for the height of the stone at any time is   h(t) = -16t2<span> + 128t + 32 .

From your equation, we can tell that you're defining the upward direction as
positive.  We can also tell that you threw the stone upward, with an initial speed
as it left your hand of 128 feet per second, about 87 miles per hour ... a mighty toss indeed, and I think there's a man from the Chicago Cubs waiting outside
who'd like to talk to you.

Anyway, When the stone splashes into the water,  h(t) = 0 .
</span></span>
<span>-16t²<span> + 128t + 32 = 0</span></span>

Divide each side by -16 :

t² - 8t - 2 = 0

I don't see any easy way to factor the expression on the left,
so I have to use the quadratic formula to solve this equation.

t = 4 plus and minus √18 .

t = +8.24 seconds
t = -0.24 second

Mathematically, both numbers are valid solutions.But when you apply
the equation to a real world situation, only the positive 't' makes sense.
So  <u> t = 8.24 seconds</u>.
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Answer:

60a in a circuit with a 12v battery

Explanation:

60a in a circuit with a 12v battery

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3 years ago
As the scattering angle of the photon increases, what happens to the wavelength associated with the photon?
frez [133]

As the scattering angle of the photon increases, the wavelength associated with the photon increases.

<h3><u>Explanation:</u></h3>

The particle with quantum mechanical property is known as Compton wavelength. The wavelength of a photon increases during collision. When the scattering angle of the photon is 0 degree then the photon's wavelength increases by 0 and when the scattering angle  is 180 degree then the wavelength of  the photon will become double. This is known as Compton wavelength.

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3 years ago
A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The system is released with an init
AURORKA [14]

a) 19.4 cm

b) 3.2 m/s

Explanation:

a)

A horizontal spring-mass system has a motion called simple harmonic motion, in which the mass oscillates following a periodic function (sine or cosine) around an equilibrium position.

As the system oscillates back and forth, its total mechanical energy (sum of elastic potential energy and kinetic energy) will remain conserved (since we consider friction negligible). The elastic potential energy at any point is given by:

U=\frac{1}{2}kx^2

where

k is the spring constant

x is the displacement of the system

While the kinetic energy at any point is

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

So the total mechanical energy of the system is

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

For this system, when it is initially released,

m = 0.6 kg

k = 165 N/m

x = 7 cm = 0.07 m

v = 3 m/s

So the total energy is

E=\frac{1}{2}(0.6)(3)^2+\frac{1}{2}(165)(0.07)^2=3.1 J

Since friction is negligible, this total energy remains constant. Therefore, when the system reaches its maximum stretch during the motion, the kinetic energy will be zero and all the mechanical energy will be elastic potential energy; so we will have:

E=U=\frac{1}{2}kx_{max}^2

where x_{max} is the maximum stretch. Solving for x_{max},

x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(3.1)}{165}}=0.194 m

So, 19.4 cm.

b)

The maximum speed in a spring-mass oscillating system is reached when the kinetic energy is maximum, and therefore, since the total energy is conserved, when the elastic potential energy is zero:

U=0

which means when the displacement is zero:

x = 0

So, when the system is transiting through the equilibrium position.

Therefore, the total mechanical energy is equal to the maximum kinetic energy:

E=K=\frac{1}{2}mv_{max}^2

where

m is the mass

v_{max} is the maximum speed

Here we have:

E = 3.1 J

m = 0.6 kg

Therefore, solving for the maximum speed,

v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(3.1)}{0.6}}=3.2 m/s

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Read 2 more answers
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