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ozzi
3 years ago
14

A fisherman casts his lure at an angle of 33 degrees above the horizontal. The lure reaches a maximum height of 2.3 m. Assuming

no frictional forces, what was the initial velocity the fisherman gave the lure when he cast it?
Physics
1 answer:
omeli [17]3 years ago
3 0

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

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Golf Cart Batteries are an example of an electrochemical cell that is powered by the reaction between Lead and Sulfuric Acid (Lead-Acid battery). This type of reaction occurs on larger scales than an alkaline battery, and thus can generate a variety of powers depending on how many instruments are present within the battery. The reaction looks like this:

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3 0
3 years ago
A sound wave has a frequency of 265 hz and a wavelength measured at 1.3
pochemuha

Answer:

V = f λ     speed of wave in terms of frequency and wavelength

t = S / V      time for wave to travel a distance S

t = 91.4 m / 344.5 m/s = .265 sec        time to travel 91.4 m

8 0
2 years ago
A primary succession does not include any invasive species <br><br> True or false
yan [13]
To be referenced, it would be true
7 0
3 years ago
When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th
Veseljchak [2.6K]
Given:
F = ax
where
x = distance by which the rubber band is stretched
a =  constant

The work done in stretching the rubber band from x = 0 to x = L is
W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2}  [x^{2} ]_{0}^{L} =  \frac{aL^{2}}{2}

Answer:  \frac{aL^{2}}{2}

4 0
3 years ago
A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24
alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

I=0.8*m*r^{2}

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}

m=40kg

moment of inertia

M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}

Kinetic energy of the rotation motion

K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J

Kinetic energy translational

K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J

Total kinetic energy  

K=3317.79J+4147.2J\\K=7464.99J

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m

6 0
3 years ago
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