<span>Let's put it this way. Say you have a killer-whale and a penguin. Killer-whales are major predators to penguins. Now, say the killer-whale population increases. The penguins would be eaten more by the killer-whales, then causing a population decrease for the penguins. If the population decreases, they're won't be enough penguins, and they most likely will become extinct, as well as causing a population decrease for the killer-whales as well. Whereas, vis versa, they're were a killer-whale population decrease. The penguins would be less hunted, therefore, creating a population increase for the penguins.</span>
Answer:
13.33 or 13 1/3m/s (meters per second)
Explanation:
In physics, we use the basic units of meters and seconds. So first convert (km) into meters (m) and also hours and minutes into seconds (s). We end up with 120000m and 9000s. Then divide the 120000m by the 9000s and you end up with 13.33 or 13 1/3 m/s.
Answer:
<em>Height = 5.65 km</em>
Explanation:
is the circumference or we can say measures the boundary of hemisphere of friction-less ice that he is sitting on.
So, the height will be = 2 x 3.14 x (30)^2 = 5654.7 m = 5.65 km
Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:
I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
<span>I = 4 A</span>
Answer:
A)6.15 cm to the left of the lens
Explanation:
We can solve the problem by using the lens equation:
where
q is the distance of the image from the lens
f is the focal length
p is the distance of the object from the lens
In this problem, we have
(the focal length is negative for a diverging lens)
is the distance of the object from the lens
Solvign the equation for q, we find
And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is
A)6.15 cm to the left of the lens
