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3 years ago

7

How many pounds of meat does a cougar eat per day

1 answer:

Luba_88 [7]3 years ago

4 0

A large male cougar living in the Cascade Mountains kills a deer or elk every 9 to 12 days, eating up to 20 pounds at a time and burying the rest for later.Except for females with young, cougars are lone hunters that wander between places frequented by their prey, covering as much as 15 miles in a single night.Cougars rely on short bursts of speed to ambush their prey. A cougar may stalk an animal for an hour or more

hope this helps in any way ! :)

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N2 + O2 → 2NO N-N triple bond: 941 kJ/mol O-O double bond: 495 kJ/mol N-O bond: 201 kJ/mol

kiruha [24]

Answer:

$\large \boxed{\text{761 kJ}}$

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

N₂ + O₂ ⟶ 2NO

N≡N + O=O ⟶ 2O-N=O

Bonds: 2N≡N 1O=O 2N-O + 2N=O

D/kJ·mol⁻¹: 941 495 201 607

$\begin{array}{rcl}\Delta H & = & \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\& = & 2 \times 941 +1 \times 495 - (2 \times 201 + 2\times 607)\\&=& 2377 - 1616\\&=&\textbf{761 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{761 kJ}}$}.$

3 0

2 years ago

Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.) With what initial v

lukranit [14]

Answer:

Vi = 94.64 m/s

Explanation:

I order to find out the initial velocity of the object, we can use third equation of motion:

2ah = Vf² - Vi²

where,

a = acceleration = -9.8 m/s²

h = maximum height covered by object = 460 m - 3 m = 457 m

Vf = Final Velocity = 0 m/s (since, object momentarily stops at highest point)

Vi = Initial Velocity = ?

Therefore,

2(-9.8 m/s²)(457 m) = (0 m/s)² - Vi²

Vi = √8957.2 m²/s²

<u>Vi = 94.64 m/s</u>

3 0

3 years ago

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

3 0

3 years ago

The social interaction theory explains that the purpose of aggression is to control another person’s behavior.

meriva

Sounds like true. ....

5 0

3 years ago

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Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.

kupik [55]

Answer:

2hrs and some mins

Explanation:

bc 2×36= 17 =)

3 0

3 years ago