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shtirl [24]
3 years ago
3

When farm animals eat the grass on land until there is almost no ground cover, it is called ______.

Physics
2 answers:
Minchanka [31]3 years ago
0 0

Answer:

When farm animals eat the grass on land until there is almost no ground cover, it is called overgrazing

Afina-wow [57]3 years ago
0 0

Answer:

Its called over grazing

Explanation:

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Does the force of kinetic friction depend on the weight of the block? explain.
drek231 [11]
Yes for an object moving on a horizontal plane, R = mg (where mg = weight). therefore, for an object moving on a horizontal plane: F = μmg



7 0
3 years ago
Solve for the length of the inclined plane if the angle equals 19.45 degrees.
mel-nik [20]

The length of the inclined plane is approximately 12 ft

The situation forms a right angle triangle.

<h3>Right triangle</h3>

Right triangle have one of its angle as 90 degrees.

Therefore,

The length of the inclined plane is the hypotenuse of the triangle. The length of the inclined plane can be found using trigonometric ratios.

height = 4 ft

angle(∅) = 19.45°

sin 19.45 = 4 / h

h = 4 / 0.33298412235

h = 12.0125847796

h = 12 ft

Therefore, the length of the inclined plane is approximately 12 ft

learn more on inclined plane:brainly.com/question/14163589?referrer=searchResults

5 0
2 years ago
Where would a car traveling on a roller coaster have the most kinetic energy ? and why?
goldenfox [79]

Answer:

As the car travels up the coaster it is gaining potential energy.

Explanation:

Because It has the greatest in amount of potential energy at the top of the coaster. when the car travels down the roller coaster it obtains speed and kinetic energy.

4 0
3 years ago
Read 2 more answers
A 700 kg car makes a turn going at 30 m/s with radius of
Gnoma [55]

Answer:5250 N

Explanation: ig:iihoop.vince

5 0
3 years ago
F = 2.0*10^20 N,
natka813 [3]

F=\dfrac{Gm_1m_2}{r^2}

With the given values of F,G,m_1,r, we have

2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}

Try dealing with the powers of 10 first: On the right, we have

\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}

Meanwhile, the other values on the right reduce to

\dfrac{6.67\times5.98}{3.8^2}\approx2.76

Then taking units into account, we end up with the equation

2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2

Now we solve for m_2:

m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}

m_2=7.25\times10^{22}\,\mathrm{kg}

or, if taking significant digits into account,

m_2=7.3\times10^{22}\,\mathrm{kg}

5 0
3 years ago
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