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zhenek [66]
3 years ago
13

What is a density dependent limiting factor?

Physics
1 answer:
Sliva [168]3 years ago
6 0
There are many types of density dependent limiting factors<span> such as:
 1) Availability of food
 2) Predation
 3) Disease
 4) Migration.

You should ask it in "Biology" instead of "Physics". Hope this helps!</span>
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Scientist A believes she has discovered a new procedure that is better than a procedure currently in use by physicians. She test
stepladder [879]

Answer:

B

Explanation:

3 0
3 years ago
Read 2 more answers
8. A rectangle is measured to be 6.4 +0.2 cm by 8.3 $0.2 cm.
mamaluj [8]

Answer:

a) The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty in its perimeter is 0.8 centimeters.

Explanation:

a) From Geometry we remember that the perimeter of the rectangle (p), measured in centimeters, is represented by the following formula:

p = 2\cdot (w+l) (1)

Where:

w - Width, measured in centimeters.

l - Length, measured in centimeters.

If we know that w = 6.4\,cm and l = 8.3\,cm, then the perimeter of the rectangle is:

p = 2\cdot (6.4\,cm+8.3\,cm)

p = 29.4\,cm

The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty of the perimeter (\Delta p), measured in centimeters, is estimated by differences. That is:

\Delta p = 2\cdot (\Delta w + \Delta l)  (2)

Where:

\Delta w - Uncertainty in width, measured in centimeters.

\Delta l - Uncertainty in length, measured in centimeters.

If we know that \Delta w = 0.2\,cm and \Delta l = 0.2\,cm, then the uncertainty in perimeter is:

\Delta p = 2\cdot (0.2\,cm+0.2\,cm)

\Delta p = 0.8\,cm

The uncertainty in its perimeter is 0.8 centimeters.

5 0
3 years ago
A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t
fredd [130]

Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

Explanation:

Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.

Let's look at the speed of the car

F = m a\\\\a =\frac{F}{m}

We use movies and find lips

\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t

The moment is defined by

\to p = m v

The moment change

\Delta p = m v - m v_0

Let's replace the speeds in this equation

\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0  + F t - m v_0\\\\\Delta p = F t

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

5 0
3 years ago
The height of the Washington Monument is measured to be 170.00 m on a day when the temperature is 35.0°C. What will its height b
N76 [4]

Answer:

169.98 m

Explanation:

\alpha = Thermal coefficient of marble = 2.5\times 10^{-6}/^{\circ}C

L_0 = Initial length = 170 m

\Delta T = Change in temperature = (-10-35)°C

Change in length is given by

\Delta L=\alpha L_0\Delta T\\\Rightarrow \Delta L=2.5\times 10^{-6}\times 170\times (-10-35)\\\Rightarrow \Delta L=-0.019125\ m

The height on the given day will be 170-0.019125=169.98\ m

3 0
4 years ago
If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo
lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

T( t ) = 20 + 75 e^{\dfrac{-t}{50}}

temperature at t = 0

T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}

T(0) = 95°C

temperature after half hour of cooling

T( t ) = 20 + 75 e^{\dfrac{-t}{50}}

t = 30 minutes

T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}

T( 30 ) = 20 + 75 \times 0.5488

T(30) = 61.16° C

average of first half hour will be equal to

T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt

T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30

T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30

T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]

T = \dfrac{1}{30}[600 - 2058.04 + 3750]

T = 76.39°C

4 0
3 years ago
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