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3 years ago

What factors affect where a plant or animal lives

Physics

1 answer:

alexdok [17]3 years ago

5 0

Depending on the animal the following can affect the animals or plants lives

-Sunshine
-Water
-Habitat

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Den pushes a desk 400 cm across the floor. He exerts a force of 10 N for 8 s to move the desk. What is his power output? (Power:

White raven [17]

Answer:

5 W

Explanation:

The formula of the power is:

● P = W/t

W is the work and t is the time needed to do it(in seconds)

Let's calculate first the work that the force exerced:

W = Vector F . Vector d

D is the distance ( here 400 cm wich is 4 m)

Make a representation to see how are the vectors F and V.(picture below)

The vector F and d are colinear since Den is pushing the desk on the ground.

● W = 4 × 10 = 40 J

J is Joule

■■■■■■■■■■■■■■■■■■■■■■■■■■

● P = W / t

● P = 40/ 8

● P = 5 W

7 0

3 years ago

Ok so I am taking a test I need help please

Aloiza [94]

Answer:

i know its definetly either clockwise or counter clockwise

Explanation:

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3 years ago

When magma cools quickly, what kind of texture or

Ksivusya [100]

Answer:

Explanation:

when magma cools Crystal's form because the solution is super saturated with respect to some minerals if the magma cools quickly the crystals do not have much time to form hence they are small and also the resulting rock is fine grained

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3 years ago

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crimeas [40]

B . it should be convert energy.

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3 years ago

Read 2 more answers

A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the

yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

x = ½ a₁ t²

x = ½ a₂ (t-4)²

Let's solve

a₁ t² = a₂ (t-4)²

a₁/a₂ t² = t² -2 4 t + 16

t² (1- 2.0 / 4.0) - 8 t +16

t² 0.5 - 8 t +16 = 0

t² -16 t + 32 = 0

Let's solve the second degree equation

t = [16 ±√( 16² - 4 32)] / 2

t = ½ (16 ± 11,3)

Solutions

t1 = 13.66 s

t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

6 0

3 years ago