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seraphim [82]
2 years ago
5

How can people help stop erosion on beach

Physics
1 answer:
seraphim [82]2 years ago
7 0

Answer:

<h3>since erosion is unavoidable the problem becomes discovering ways to prevent it. present beach erosion prevention methods include sand bags,vegetation,seawalls,sand dunes,and sand fences.</h3>
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If there are 60 min in 1hr, and in 1min there are 60 sec; Solve the following problem using dimensional analysis.: 3,800 hrs. to
GREYUIT [131]
13680000 should be the number of seconds
6 0
3 years ago
What is your least favorite candy? Explain
aev [14]

Answer:

My least favorite is whoppers.....Trust me i love chocolate, but not when it taste like chalk......

7 0
3 years ago
Read 2 more answers
A fly travels along the x-axis. His starting point is x = 16 m and his ending point 15x = - 25 m. His flight lasts 4.0 seconds H
gogolik [260]

Explanation:

Starting position at x  = 16m

Ending position at x  = -25m

Time of flight  = 4s

Unknown:

Distance flown  = ?

Displacement  = ?

Speed  = ?

Velocity  = ?

Solution:

To find the distance flown, we should understand that the body is moving on the x - plane;  

  So distance  = 16 + 25  = 41m

Displacement is 41m to the left or -x axis

Speed is the distance divided by the time taken;

      Speed  = \frac{distance}{time}   = \frac{41}{4}   = 10.25m/s

Velocity is 10.25m/s along -x axis

5 0
3 years ago
Steel train rails are laid in 12.0-m-long segments placed end to end. The rails are laid on a winter day when their temperature
Yanka [14]

Answer:

a) Space = 6.05 x 10^{-3} m = 0.605 cm

b) Stress= -100.8 x 10^{6} Pa

Explanation:

<em>1) Data Given</em>

L = 12 m , T_i = -9 C \degree, T_f = 33 C \degree

<em>2) Calculate the space using Linear thermal expansion formula</em>

We need to use Linear thermal expansion formula since the space created would be a change on 1 dimension, the increase of the temperature will increase the length of the steel.  The formula is given by:

\Delta L = L_i \alpha_{steel} \Delta T

We have everything except the \alpha_{steel} , so we look for this on a book and we find that \alpha_{steel} = 1.2 x 10^{-5} C^{-1}, so we can replace.

\Delta L = 12 m (1.2 x 10^{-5} C^{-1}) (33 C \degree -(-9 C \degree)) = 12 m (1.2 x 10^{-5} C^{-1}) 42 C \degree =6.048 x 10^{-3} m = 0.6048cm

<em>3) Calculate the stress of the steel </em>

The Stress is the ratio of applied force F to a cross section area - defined as

\sigma = \frac{F_n}{A}

Since we don't have the force and the Area, we need to look for another way to find the stress.

For this we can use the concept called Young's Modulus, defined as : "the mechanical property that measures the stiffness of a solid material", and the formula for this is given by:

Y =\frac{F L}{A \Delta L} (1)

Solving \frac{F}{A} from the previous formula we have this:

\frac{F}{A}  = (Y  Δ L)/L  (2)

From the <em>Linear thermal expansion formula</em> we can solve like this

\frac{\Delta L}{L} =  α  ΔT  (3)

And replacing equation (3) into equation (2) we have:

\frac{F}{A}  = Y α ΔT (4)

We have that the Young's Modulus for the steel is 20x10^{10} Pa, so replacing into equation (4)

\frac{F}{A} = 20x10^{10} Pa (1.2x10^-5 C^-1) (42C) = 100.8 *10^{6} Pa  

That represent the absolute value for the Stress, the sign on this case would be negative since there is a compression.

3 0
4 years ago
2. A certain object revolves at a rate of 30 rpm. Please determine the frequency and period of this
schepotkina [342]

Answer:

The time period of the motion is, T = 0.03 s

The frequency of the rotation is, f = 30 Hz

Explanation:

Given data,

The rotational speed of an object, ω = 30 rpm

                                                          ω = 188.5 rad/s

The time period of motion is,

                            T = 2π / ω

Substituting the given values in the above equation

                               = 2π / 188.5

                            T = 0.03 s

The time period of the motion is, T = 0.03 s

The frequency of rotation,

                                f = 1 /T

                                   = 1 / 0.03

                                   = 30 Hz

Hence, the frequency of the rotation is, f = 30 Hz

7 0
4 years ago
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