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Vlad [161]
3 years ago
12

(1) Predator-prey is the most important type of ecological relationship.

Physics
1 answer:
DIA [1.3K]3 years ago
4 0
1) True 2) True 3)True I hope I helped
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Town A lies 15 km north of town B. Town C lies 10 km west of town A. A small plane flies directly from town B to town C. What is
ZanzabumX [31]

Answer:

the correct answer is b

Explanation:

5 0
3 years ago
An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a
Morgarella [4.7K]

Answer:

0.37 m

Explanation:

The angular frequency, ω, of a loaded spring is related to the period, T,  by

\omega = \dfrac{2\pi}{T}

The maximum velocity of the oscillation occurs at the equilibrium point and is given by

v = \omega A

A is the amplitude or maximum displacement from the equilibrium.

v = \dfrac{2\pi A}{T}

From the the question, T = 0.58 and A = 25 cm = 0.25 m. Taking π as 3.142,

v = \dfrac{2\times3.142\times0.25\text{ m}}{0.58\text{ s}} = 2.71 \text{ m/s}

To determine the height we reached, we consider the beginning of the vertical motion as the equilibrium point with velocity, v. Since it is against gravity, acceleration of gravity is negative. At maximum height, the final velocity is 0 m/s. We use the equation

v_f^2 = v_i^2+2ah

v_f is the final velocity, v_i is the initial velocity (same as v above), a is acceleration of gravity and h is the height.

h = \dfrac{v_f^2 - v_i^2}{2a}

h = \dfrac{0^2 - 2.71^2}{2\times-9.81} = 0.37 \text{ m}

3 0
3 years ago
Use the values from PRACTICE IT to help you work this exercise. If the current in each wire is doubled, how far apart should the
melisa1 [442]
Which behavior would best describe someone who has good communication skills with customers ? a) Following up with some customers b) Talking to customers more than listening to them c) Repeating back what the customer says d) Interrupting customers frequently
8 0
3 years ago
I have a density of 1.61g/cm^3 and a mass of 28g. Find the missing value
Tamiku [17]

The missing value is the 'k' that should appear between '28' and 'g' .

Nobody has a mass of 28g.

My new grandson, who was born yesterday (January 29, 2019), had a mass of 4,039g at birth.  

Mother and son are happy and resting comfortably today, in Kfar Tapuach, Samaria, Israel.

8 0
3 years ago
How much heat (in kJ) is required to warm 13.0 g of ice, initially at -12.0 ∘C, to steam at 109.0 ∘C?
Sunny_sXe [5.5K]
<h2>Answer:</h2>

39.699 kJ

<h2>Explanation:</h2>

In this situation, there are a few transformations as follows;

(i) Heat required to warm the ice from -12°C to its melting point.

(ii) Heat required to melt the ice.

(iii) Heat required to boil the melted ice to boiling point (i.e to steam)

(iv) Heat required to vapourize the water

(v) Heat required to heat the steam from 100°C to 109.0°C

The sum of all the heat processes gives the heat required to warm the ice to steam;

<h3><em>Calculate each of these heat processes</em></h3>

<em>From (i);</em>

Let the heat required to warm the ice from -12.0°C to its melting point (0°C) be Q₁.

Q₁ = m x c x ΔT        -----------------------(i)

Where;

m = mass of ice = 13.0g

c = specific heat capacity of ice = 2.09 J/g°C

ΔT = final temperature - initial temperature = 0°C - (-12°C) = 12°C

Substitute these values into equation (i) as follows;

Q₁ = 13.0 x 2.09 x 12 = 326.04 J

<em>From (ii);</em>

Let the heat required to melt the ice be Q₂. This heat is called the heat of fusion and it is given by;

Q₂ = m x L        -----------------------(ii)

Where;

m = mass of ice = 13.0g

L = latent heat of fusion of ice = 333.6 J/g

Substitute these values into equation (ii) as follows;

Q₂ = 13.0 x 333.6

Q₂ = 4336.8 J

<em>From (iii);</em>

Let the heat required to boil the melted ice from 0°C to boiling point of 100°C be Q₃.

Q₃ =  m x c x ΔT        -----------------------(i)

Where;

m = mass of melted ice (water) which is still 13.0g

c = specific heat capacity of melted ice (water) = 4.2 J/g°C

ΔT = final temperature - initial temperature = 100°C - 0°C = 100°C

Substitute these values into equation (i) as follows;

Q₁ = 13.0 x 4.2 x 100 = 5460 J

<em>From (iv);</em>

Let the heat required to vaporize the water (melted ice) be Q₄. This heat is called the heat of vaporization and it is given by;

Q₄ = m x L        -----------------------(iv)

Where;

m = mass of ice = 13.0g

L = latent heat of vaporization of water = 2257 J/g

Substitute these values into equation (iv) as follows;

Q₄ = 13.0 x 2257

Q₄ = 29341 J

<em>From (v);</em>

Let the heat required to heat the steam from 100°C to 109°C be Q₅.

Q₅ =  m x c x ΔT        -----------------------(i)

Where;

m = mass of steam which is still 13.0g

c = specific heat capacity of steam = 2.01 J/g°C

ΔT = final temperature - initial temperature = 109.0°C - 100°C = 9°C

Substitute these values into equation (i) as follows;

Q₅ = 13.0 x 2.01 x 9 = 235.17J

<em>Finally:</em>

<em>Sum all the heat values together;</em>

Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Q = 326.04 + 4336.8 + 5460 + 29341 + 235.17

Q = 39699.01 J

Q = 39.699 kJ

Therefore, the amount of heat (in kJ) required is 39.699

7 0
2 years ago
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