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avanturin [10]
3 years ago
10

During one year a population of blue jays has zero population growth. which must be true about the blue jay population at the en

d of the year compared to the beginning of the year?
Physics
2 answers:
creativ13 [48]3 years ago
6 0

Answer: Hello there!

the fact that the population has zero population growth over a year means that, if at the start of the year the population was X, at the end of the year the population, Y, can't be bigger than X.

Then the relation between the population at the end of the year, Y, and the start of the year, X, is:

Y ≥ X

aivan3 [116]3 years ago
3 0
It must be the same or less. Depending on if population growth means just how much it got bigger of if it can be negative (like -8% means it got smaller by 8% and being considered "growth").
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Natural gas is the answer.
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Allison wants to calculate the speed of a sound wave.
Fittoniya [83]
D.) Distance travelled / time ...........
5 0
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A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?
garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

U_i +K_i = U_f + K_f

where:

U_i is the initial potential energy, at the bottom

K_i is the initial kinetic energy, at the bottom

U_f is the final potential energy, at the top

K_f is the final kinetic energy, at the top

We can rewrite the equation as:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m is the mass of the hopper

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0 is the initial height

u is the launch speed of the hopper

h_f = 1.20 m is the maximum altitude reached by the hopper

v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s

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4 0
3 years ago
If the pitched ball was traveling 77 mph before stanton's bat hit it and 120 mph after his bat hit it, by what amount did the sp
Mice21 [21]

here we will use the concept of Newton's III law

as per Newton's III law the impulse given to the ball is same as the impulse lost by the bat

So here we will say

impulse gain by the ball = impulse lost by the bat

m_1(v_f - v_i) = m_2(\Delta v)

given that

m_1 = 5 ounce

m_2 = 32 ounce

For ball the change in speed will be

v_f - v_i = (120 - 77)mph

now from above equation

5\times (120 - 77) = 32 \times \Delta v

\Delta v = 6.72 mph

so speed of bat will decrease by 6.72 mph

3 0
3 years ago
A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula
Luda [366]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

V_{ab} = i R

V_{ab} = (0.1832) (65)

V_{ab} = 11.91 Volts

(c)

Power dissipated in the resister R is given as

P_{R} = i²R

P_{R} = (0.1832)²(65)

P_{R} = 2.18 Watt

Power dissipated in the internal resistance is given as

P_{r} = i²r

P_{r} = (0.1832)²(0.5)

P_{r} = 0.0168 Watt

5 0
3 years ago
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