How can people help stop erosion on beach

A 5 kg block is released from rest at the top of a quarter- circle type curved frictionless surface. The radius of the curvature

malfutka [58]

Answer:

a. 186.2 J b. 8.63 m/s c. 190 m d. 43.2 s e. 186.2 J

Explanation:

a. From conservation of energy, the potential energy loss of block = kinetic energy gain of the block.

So, U + K = U' + K' where U = initial potential energy of block = mgh, K = initial kinetic energy of block = 0, U' = final potential energy of block at bottom of curve = 0 and K' = final kinetic energy of block at bottom of curve.

So, mgh + 0 = 0 + K'

K' = mgh where m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s², h = initial height above the ground of block = radius of curve = 3.8 m

So, K' = 5 kg × 9.8 m/s² × 3.8 m = 186.2 J

b. Since the kinetic energy of the block K = 1/2mv² where m = mass of block = 5 kg, v = velocity of block at bottom of curve

So, v = √(2K/m)

= √(2 × 186.2 J/5 kg)

= √(372.4 J/5 kg)

= √(74.48 J/kg)

= 8.63 m/s

c. To find the stopping distance, from work-kinetic energy principles,

work done by friction = kinetic energy change of block.

So ΔK = -fd where ΔK = K" - K' where K" = final kinetic energy = 0 J (since the block stops)and K' = initial kinetic energy = 186.2 J, f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance

ΔK = -fd

K" - K' = - μmgd

d = -(K" - K')/μmg

Substituting the values of the variables, we have

d = -(0 J - 186.2 J)/(0.02 × 5 kg × 9.8 m/s²)

d = -(- 186.2 J)/(0.98 kg m/s²)

d = 190 m

d. Using v² = u² + 2ad where u =initial speed of block = 8.63 m/s, v = final speed of block = 0 m/s (since it stops), a = acceleration of block and d = stopping distance = 190 m

So, a = (v² - u²)/2d

substituting the values of the variables, we have

a = (0² - (8.63 m/s)²)/(2 × 190 m)

a = -74.4769 m²/s²/380 m

a = -0.2 m/s²

Using v = u + at, we find the time t that elapsed while the block is moving on the horizontal track.

t = (v - u)/a

t =(0 m/s - 8.63 m/s)/-0.2 m/s²

t = - 8.63 m/s/-0.2 m/s²

t = 43.2 s

e. The work done by friction W = fd where

= μmgd where f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance = 190 m

W = 0.02 × 5 kg × 9.8 m/s² × 190 m

W = 186.2 J

5 0

3 years ago

Which statement about tornadoes is correct? a. They are a very common part of thunderstorms. b. Seventy-five percent of the worl

Anit [1.1K]

As strange as it may seem, around 75% world’s tornadoes

occur in the United States. It's actually because of the unique

geography of North America, and how the winds and storms

get funneled up through the US. (choice-b)

It's true that when there's a tornado, it usually comes along with a thunderstorm. But tornadoes are way way LESS common than thunderstorms.

Their main danger is from the high winds in the center of the funnel.

The rain is minor compared to the wind damage.

Tornadoes are not really the strongest storms on Earth.

Welll ... I guess that could depend on what "strong storm" means.

Sure, a tornado is incredibly vicious and powerful. But it's only a few

miles wide, and it only lasts for a few hours and then it's all over.

The tropical cyclones ... hurricanes, monsoons, typhoons ... don't have

winds as fast as tornadoes, but they can be a thousand miles wide and

last for weeks.

7 0

4 years ago

Read 2 more answers

How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a po

miss Akunina [59]

Incomplete question as the charge density is missing so I assume charge density of 3.90×10^−12 C/m².The complete one is here.

An electron is released from rest at a distance of 0 m from a large insulating sheet of charge that has uniform surface charge density 3.90×10^−12 C/m² . How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00×10−2 m from the sheet?

Answer:

Work=1.06×10⁻²¹J

Explanation:

Given Data

Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²

Charge density σ=3.90×10⁻¹² C/m²

The electron moves a distance d=3.00×10⁻²m

Electron charge e=-1.6×10⁻¹⁹C

To find

Work done

Solution

The electric field due is sheet is given as

E=σ/2ε₀

$E=\frac{3.90*10^{-12}C/m^{2} }{2(8.85*10^{-12}C^{2} /N.m^{2} )}\\ E=0.22V/m$

Now we need to find force on electron

$F=eE\\F=(1.6*10^{-19}C )(0.22V/m)\\F=3.525*10^{-20}N$

Now for Work done on the electron

$W=F*d\\W=(3.525*10^{-20} N)(3.00*10^{-2}m)\\W=1.06*10^{-21}J$

4 0

4 years ago

2

Xelga [282]

Answer:

About 7.67 m/s.

Explanation:

Mechanical energy is always conserved. Hence:

$\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}$

Where <em>U</em> is potential energy and <em>K</em> is kinetic energy.

Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:

$\displaystyle U_i = K_f$

Substitute and solve for final velocity:
$\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\ 2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ & =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}$

In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.

8 0

2 years ago

A steel beam that is 6.50 m long weighs 336 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

FinnZ [79.3K]

Answer:

Explanation:

When beam is balanced and not rotating with Suki standing on it , let reaction force on the supports be R₁ and R₂. Then

R₁ +R₂ = 336 + 590

= 929

Now the moment beam begins to tip , reaction on distant support R₁ = 0

only R₂ will exists on the support near to Suki.

Taking torque about this support of weight of beam acting from the middle point and weight of suki of 590N ,who is x distance from the support towards the other end.

336 x 1.5 = 590 x

x = .85 m

ie , from second support , Suki can not go beyond a distance of .85 m towards the second end.

4 0

3 years ago