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3 years ago

How many pounds of meat does a cougar eat per day

1 answer:

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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Which is a gas at room temperature?

Sedaia [141]

Which is a gas at room temperature?
B) nitrogen

7 0

2 years ago

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Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is

marshall27 [118]

Answer:

$x=0.0049\ m= 4.9\ mm$

$d=0.01153\ m=11.53\ mm$

Explanation:

Given:

mass of the object, $m=0.75\ kg$

elastic constant of the connected spring, $k=150\ N.m^{-1}$

coefficient of static friction between the object and the surface, $\mu_s=0.1$

(a)

Let x be the maximum distance of stretch without moving the mass.

The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.

$f=k.x$

$\mu_s.N=k.x$

where:

N = m.g = the normal reaction force acting on the body under steady state.

$0.1\times (9.8\times 0.75)=150\times x$

$x=0.0049\ m= 4.9\ mm$

(b)

Now, according to the question:

Amplitude of oscillation, $A= 0.0098\ m$

coefficient of kinetic friction between the object and the surface, $\mu_k=0.085$

Let d be the total distance the object travels before stopping.

Now, the energy stored in the spring due to vibration of amplitude:

$U=\frac{1}{2} k.A^2$

This energy will be equal to the work done by the kinetic friction to stop it.

$U=F_k.d$

$\frac{1}{2} k.A^2=\mu_k.N.d$

$0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d$

$d=0.01153\ m=11.53\ mm$

is the total distance does it travel before stopping.

7 0

2 years ago

emmasim [6.3K]

Answer:

false

Explanation:

the president can over turn any ruling made by any governing body not just the supreme court

4 0

2 years ago

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A very long wire carries a uniform linear charge density of 5 nC/m. What is the electric field strength 13 m from the center of

kipiarov [429]

Answer:

E=6.91 N/C

Explanation:

Given that

Linear Charge density ,λ = 5 nC/m

Distance ,R= 13 m

We know that formula for long wire to find electric field

$E=\dfrac{\lambda }{2\pi \varepsilon _0R}$

E=Electric field

R=Distance

εo=8.85 x 10⁻¹² C²/N.m²

λ=Linear Charge density

Now by putting the values

$E=\dfrac{5\times 10^{-9}}{{2\times \pi \times 8.85\times 10^{-12}\times 13}}$

E=6.91 N/C

Therefore the electric filed at distance 13 m will be 6.91 N/C

5 0

3 years ago

If the number of unemployed workers is 19 million, the number in the working-age population is 500 million, and the unemployment

Darina [25.2K]

Answer:

The percentage of the population that is actively participating in the labor market, whether working or looking for a job, is 95% of the total working age population.

Explanation:

If the number of unemployed in a country is 19 million people, and that number represents an unemployment rate of 4%, to know the total number of economically active people in the country we must perform the following calculation:

(19,000,000 / 4) x 100 = X

475,000,000 = economically active population

However, the country has a working-age population of 500 million people, of which only 475 are actively participating in the labor market. To know the percentage that these 475 million people represent, we must perform the following cross multiplication:

500 = 100

475 = X

(475 x 100) / 500 = X

95 = X

The percentage of the population that is actively participating in the labor market, whether working or looking for a job, is 95% of the total working age population.

6 0

2 years ago