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3 years ago

How does a longshore current affect a beach?

2 answers:

8 0

Longshore currents are affected by the velocity and angle of a wave. When a wave breaks at a more acute (steep) angle on a beach, encounters a steeper beach slope, or is very high, longshore currents increase in velocity. ... This process, known as “longshore drift,” can cause significant beach erosion.

raketka [301]3 years ago

3 0

Answer:

"Longshore currents are affected by the velocity and angle of a wave. When a wave breaks at a more acute (steep) angle on a beach, encounters a steeper beach slope, or is very high, longshore currents increase in velocity. ... This process, known as “longshore drift,” can cause significant beach erosion."

Credits to ocean service

Explanation:

Hope this helps

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A charge of -3.40 nC is placed at the origin of an xy-coordinate system, and a charge of 2.45 nC is placed on the y axis at y =

gavmur [86]

Answer:

The magnitude of this force is $1.866\times10^{-4}\ N$

The direction of this force is 55.69°.

Explanation:

Given that,

Charge at origin $q_{1}= -3.40\ nC$

Charge at y axis $q_{2}= 2.45\ nC$

Distance on y axis = 4.25 cm

Third charge $q_{3}= 5.00\ nC$

Distance on x axis = 2.90 cm

(a). We need to calculate the force F₁₃

Using formula of force

$F_{13}=\dfrac{kq_{1}q_{3}}{r^2}$

Put the value into the formula

$F_{13}=\dfrac{9\times10^{9}\times(-3.40\times10^{-9})\times5.00\times10^{-9}}{(2.90\times10^{-2})^2}$

$F_{13}=-0.00018192\ N$

$F_{13}=-1.82\times10^{-4}\ N$

We need to calculate the force F₁₂

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{r^2}$

Put the value into the formula

$F_{12}=\dfrac{9\times10^{9}\times(-3.40\times10^{-9})\times2.45\times10^{-9}}{(4.25\times10^{-2})^2}$

$F_{12}=-0.00004150\ N$

$F_{12}=-4.15\times10^{-5}\ N$

The magnitude of this force is

The total force exerted on this charge by the other two charges.

$F=\sqrt{F_{13}^2+F_{12}^2+2F_{13}F_{12}\cos \theta}$

Here, $\theta=90$

$F=\sqrt{F_{13}^2+F_{12}^2}$

Put the value into the formula

$F=\sqrt{(-1.82\times10^{-4})^2+(-4.15\times10^{-5})^2}$

$F=0.0001866\ N$

$F=1.866\times10^{-4}\ N$

(c). We need to calculate the direction of this force

Using formula of direction

$\tan\theta=\dfrac{y}{x}$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{4.25}{2.90})$

$\theta=55.69^{\circ}$

Hence, The magnitude of this force is $1.866\times10^{-4}\ N$

The direction of this force is 55.69°.

3 0

4 years ago

A plane flies 1,995 miles in a southwesterly direction from Baltimore to Phoenix in 5.00 hours . What is the velocity of the pla

disa [49]

displacement of the plane is given as

$d = 1995 miles$

time taken by the plane

$t = 5.00 hours$

now the velocity is given as

$v = \frac{displacement}{time}$

$v = \frac{1995}{5}$

$v = 399 mi/h$

so the velocity of airplane is 399 mi/h towards South West

4 0

3 years ago

Which of the following statements is true? A. Both warming up and cooling down or important. B. It is more important warm up the

Alenkasestr [34]

Answer:

Both warming up and cooling down or not important

8 0

3 years ago

From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.10 m/s and angle

GarryVolchara [31]

Answer:

Part a)

$y = 88.5 m$

Part b)

$v_x = 7.7 m/s$

$v_y = 2.5 m/s$

Part c)

equation of position in x direction is given as

$x = v_x t$

equation of position in y direction is given as

$y = v_y t + \frac{1}{2}gt^2$

Part d)

$x = 30.8 m$

Part e)

H = 88.5 m

Part f)

t = 1.2 s

Explanation:

As we know that ball is projected with speed

v = 8.10 m/s at an angle 18 degree below the horizontal

so we will have

$v_x = 8.10 cos18$

$v_x = 7.7 m/s$

$v_y = 8.10 sin18$

$v_y = 2.5 m/s$

Part a)

Since it took t =4 s to reach the ground

so its initial y coordinate is given as

$y = v_y t + \frac{1}{2}a_y t^2$

$y = 2.5(4) + \frac{1}{2}(9.81)(4^2)$

$y = 88.5 m$

Part b)

components of the velocity is given as

$v_x = 8.10 cos18$

$v_x = 7.7 m/s$

$v_y = 8.10 sin18$

$v_y = 2.5 m/s$

Part c)

equation of position in x direction is given as

$x = v_x t$

equation of position in y direction is given as

$y = v_y t + \frac{1}{2}gt^2$

Part d)

distance where it will strike the floor is given as

$x = v_x t$

$x = 7.7 \times 4$

$x = 30.8 m$

Part e)

Height from which it is thrown is same as initial y coordinate of the ball

so it is given as

H = 88.5 m

Part f)

time taken by ball to reach 10 m below is given as

$y = v_y t + \frac{1}{2}gt^2$

$10 = 2.5t + \frac{1}{2}(9.81) t^2$

t = 1.2 s

7 0

3 years ago

Suppose a 96.10 kg hiker has ascended to a height of 1841 m above sea level in the process of climbing Mt. Washington. By what p

lbvjy [14]

Answer:

0.029%

Explanation:

Weight on the surface is given by,

$F_W = mg = (96.1Kg)(9.8m/s^2)=941.78N$

The acceleration due to gravity at given height is,

$g_h=\frac{gR^2}{R+h^2}$

The weight in another height is equal to

$F_h=mg_h$

$F_h= m*\frac{gR^2}{R+h^2}$

$F_h = \frac{F_WR^2}{(R+h)^2}$

$F_h=\frac{(941.78)(6.371*10^6)}{(6.371*10^6+1841)}$

$F_h = 941.50N$

The change of the Weight is,

$\frac{F-F_h}{F}*100=\frac{941.78-941.50}{941.78}*100=0.029\%$

6 0

3 years ago