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KonstantinChe [14]
3 years ago
12

Why is photosynthèses important to the biosphere

Physics
1 answer:
finlep [7]3 years ago
8 0

This exchange of food and energy makes the biosphere a self-supporting and self-regulating system.More often, however, the biosphere is described as having many ecosystems. Biosphere Reserves. People play an important part in maintaining the flow of energy in the biosphere

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Which object that attracts iron is?
Nutka1998 [239]
A magnet attracts iron
5 0
3 years ago
An automobile accelerates uniformly from a speed of 60 km/hr (16.67 m/s) to a speed of 80 km/hr (22.2 m/s) in 5 s. Determine the
monitta

Answer:

The acceleration is 1.106\ m/s^2 and the distance covered is 97.17 m.

Explanation:

Given that,

Initial speed of an automobile, u = 60 km/hr = 16.67 m/s

Final speed of an automobile, v = 80 km/hr = 22.2 m/s

Time, t = 5 s

We need to find the acceleration of the car and the distance traveled in this 5 sec interval.  Let a is the acceleration. Using the definition of acceleration as :

a=\dfrac{v-u}{t}\\\\a=\dfrac{22.2-16.67}{5}\\\\a=1.106\ m/s^2

Let d is the distance covered. Using the third equation of motion to find it as follows :

v^2-u^2=2as\\\\s=\dfrac{v^2-u^2}{2a}\\\\s=\dfrac{(22.2)^2-(16.67)^2}{2\times 1.106}\\\\s=97.17\ m

So, the acceleration is 1.106\ m/s^2 and the distance covered is 97.17 m.

7 0
3 years ago
What are 2 things Lithium, Beryllium, and Boron, have in common?
ivann1987 [24]

Answer:

larger atomic radii because they are on the far left, the s-block, of the periodic table. -The ionic radii increase as you go down groups on the periodic table. -Lithium and Beryllium are not examples of this trend because they are both in 2s.

8 0
3 years ago
What hanging mass will stretch a 3.0-m-long, 0.32 mm - diameter steel wire by 1.3 mm ? The Young's modulus of steel is 20×10^10
raketka [301]

Answer:

0.71 kg

Explanation:

L = length of the steel wire = 3.0 m

d = diameter of steel wire = 0.32 mm = 0.32 x 10⁻³ m

Area of cross-section of the steel wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.32 x 10⁻³)²

A = 8.04 x 10⁻⁸ m²

ΔL = change in length of the wire = 1.3 mm = 1.3 x 10⁻³ m

Y = Young's modulus of steel = 20 x 10¹⁰ Nm⁻²

m = mass hanging

F = weight of the mass hanging

Young's modulus of steel is given as

Y = \frac{FL}{A\Delta L}

20\times 10^{10} = \frac{F(3)}{(8.04\times 10^{-8})(1.3\times 10^{-3})}

F = 6.968 N

Weight of the hanging mass is given as

F = mg

6.968 = m (9.8)

m = 0.71 kg

7 0
4 years ago
Placing the north pole of a magnet near the south pole of another magnet results in:
MakcuM [25]

Answer:

An attractive force between the magnets. A is the correct answer if it was the north pole facing another north pole magnet the answer would have been a repulsive force between the magnets because the north pole of a magnet does not attract instead it separates from each other which is a repulsive force the best answer is A.

Explanation:

the southern pole of a magnet is a positive force while the northern pole of a magnet is negative .

Note: Magnet can attract each other only when they are facing the opposite directions a north pole will attract a south pole; the magnets pull on each other. But the two north poles will push each other away.

8 0
3 years ago
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