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Zielflug [23.3K]
3 years ago
10

What processes formed the beaches of Florida?

Physics
1 answer:
alex41 [277]3 years ago
3 0
The processes Erosion and/or deposition
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skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If
scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

7 0
3 years ago
A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.
fgiga [73]

The vertical velocity of the projectile upon returning to its original is 17. 74 m/s

<h3>How to determine the vertical velocity</h3>

Using the formula:

Vertical velocity component , Vy = V * sin(α)

Where

V = initial velocity = 36. 6 m/s

α = angle of projectile = 29°

Substitute into the formula

Vy = 36. 6 * sin ( 29°)

Vy = 36. 6 * 0. 4848

Vy = 17. 74 m/s

Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s

Learn more about vertical velocity here:

brainly.com/question/24949996

#SPJ1

8 0
1 year ago
If the current through a 20-ω resistor is 8.0 a , how much energy is dissipated by the resistor in 1.0 h ?
fiasKO [112]
1 hour = 3600 seconds.
Energy dissipated = I²Rt = 8²×20×3600 = 4608000 J
4 0
2 years ago
A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. the box hits the end of the
Serjik [45]
The elastic potential energy of a spring is given by
U= \frac{1}{2}kx^2
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.

The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
W=-\Delta U =  \frac{1}{2}kx_i^2 -  \frac{1}{2}kx_f^2

In our problem, initially the spring is uncompressed, so x_i=0. Therefore, the work done by the spring when it is compressed until x_f is
W=- \frac{1}{2}kx_f^2
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
8 0
3 years ago
Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. It has a centripetal force. F acting on it. If bod
cluponka [151]

Answer:

The centripetal force on body 2 is 8 times of the centripetal force in body 1.

Explanation:

Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. The centripetal force acting on it is given by :

F=\dfrac{mv^2}{r}

Body 2 has a mass 2m and its moving in a circle of radius 4r at a speed 4v. The centripetal force on body 2 is :

F'=\dfrac{2m\times (4v)^2}{4r}\\\\F'=\dfrac{2m\times 16v^2}{4r}\\\\F'=8\dfrac{mv^2}{r}\\\\F'=8F

So, the centripetal force on body 2 is 8 times of the centripetal force in body 1.

8 0
2 years ago
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