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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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a crane lifts four pallets of bricks each of which weigh 5000 N. the crane lifts each pallet a height of 30m. the crane takes 4

love history [14]

Answer:

625 W

Explanation:

Applying

P = W/t.................... Equation 1

Where p = power, W = Work, t = time

But,

W = Force (F) × distance (d)

W = Fd........................ Equation 2

Substitute equation 2 into equation 1

P = Fd/t.................... Equation 3

From the question,

Given: F = 5000 N, d = 30 m, t = 4 munites = (4×60) seconds = 240 seconds

Substitute these values into equation 3

P = (5000×30)/240

P = 625 Watt

3 0

2 years ago

A bowling ball is 21.6 cm in diameter. What is the angular speed of these ball whenit is moving at 3.0 m/s?

sergejj [24]

Answer:

Angular speed = 27.78 rad/s (Approx)

Explanation:

Given:

Diameter = 21.6 cm

Speed = 3 m/s

Find:

Angular speed

Computation:

Radius = 21.6 / 2 = 10.8 cm = 0.108 m

Angular speed = v / r

Angular speed = 3 / 0.108

Angular speed = 27.78 rad/s (Approx)

3 0

2 years ago

The diameter of the moon is a little less than the distance across the United States. Please select the best answer from the cho

dlinn [17]

Answer:

true

Explanation:

The diameter of the Moon is 3474 km. The distance across the United States, from Florida to Washington, is 4509.382 km.

6 0

3 years ago

Read 2 more answers

A block of weight w sits on a frictionless inclined plane, which makes an angle theta with respect to the horizontal, as shown.

zavuch27 [327]

Answer:

Explanation:

6 0

3 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago