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2 years ago

DESPERATE!! WILL GIVE BARINLIST AND THANKS

Physics

2 answers:

Finger [1]2 years ago

3 0

Answer:

it grabs unto soil and keeps it clamped together it prevents this makes the soil harder to wash away

Murljashka [212]2 years ago

3 0

Answer:

When the leftovers are left in the autumn, they have two purposes:

- Preventing erosion; the roots remain in the ground, so they hold it tightly till the next planting of the crops.

- Fertilizing the soil with natural material; the roots will not rot quickly in the colder period of the year, but will start rotting much faster in the spring, thus they will enrich the soil.

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g If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of th

Reil [10]

Answer:

Speed; v = 17 m/s

Explanation:

We are given;

Radius; r = 110m

Angle; θ = 15°

Now, we know that in circular motion,

v² = rg•tanθ

Thus,

v = √(rg•tanθ)

Where,

v is velocity

r is radius

g is acceleration due to gravity

θ is the angle

Thus,

v = √(rg•tanθ) = √(110 x 9.8•tan15)

v = √(288.85)

v = 17 m/s

8 0

3 years ago

HELP PLEASE

4vir4ik [10]

Recall this kinematic equation:

a = $\frac{Vi+Vf}{Δt}$

This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).

a is the acceleration.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.

Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 25 m/s.

Δt = 10 s

Substitute the terms in the equation with the given values and solve for a:

a = $\frac{0+25}{10}$

7 0

3 years ago

Read 2 more answers

Jonah is trying to move his 22-kg desk by pushing on it with a force of 130 N, but his brother is leaning on it with a downward

Dafna11 [192]

Answer:

0.54

Explanation:

Draw a free body diagram. There are 5 forces on the desk:

Weight force mg pulling down

Applied force 24 N pushing down

Normal force Fn pushing up

Applied force 130 N pushing right

Friction force Fnμ pushing left

Sum of the forces in the y direction:

∑F = ma

Fn − mg − 24 = 0

Fn = mg + 24

Fn = (22)(9.8) + 24

Fn = 240

Sum of the forces in the x direction:

∑F = ma

130 − Fnμ = 0

Fnμ = 130

μ = 130 / Fn

μ = 130 / 240

μ = 0.54

6 0

3 years ago

Everyone's favorite flying sport disk can be approximated as the combination of a thin outer hoop and a uniform disk, both of di

Umnica [9.8K]

Answer:

The length of the boomerang is 0.364 m

Explanation:

The moment of inertia is:

$I=\frac{1}{2} m_{d} r^{2} +m_{h} r^{2}$

Where

md = 0.05 kg

mh = 0.12 kg

r = d/2 = 0.273/2 = 0.1365 m

$I=\frac{0.05*(0.1365)^{2} }{2} +(0.12)*(0.1365)^{2} =2.7x10^{-3} kgm^{2}$

The length of the boomerang is:

$L_{b} =\sqrt{\frac{12I}{m} } =\sqrt{\frac{12*2.7x10^{-3} }{0.245} } =0.364m$

5 0

3 years ago

A 5000-lb wrecking ball hangs from a 50-ft cable of density 10 lb/ft attached to a crane. Calculate the work done if the crane l

aniked [119]

Answer:

total work is = 52450 J

Explanation:

given data

mass = 5000-lb

density = 10 lb/ft

height = 50 ft

solution

as we will treat here cable and ball are separate

and

here work need to lift cable is

w = (10Δy )(9.8 y ) j

and

now summing all segment of cable

so passing limit Δy to 0

so total work need

= $\int\limits^{10}_0 {98y} \, dy$

= $[49 y^2]^{50}_0$

= 2450J

so lifting 5000 lb wrcking 50 m required additional 5000 + 2450

so total work is = 52450 J

3 0

3 years ago