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ladessa [460]
3 years ago
13

Parts of earth that are hit directly by the suns rays have

Physics
1 answer:
AVprozaik [17]3 years ago
6 0
Parts of earth that are hit directly are usually near the equator and have a warm and lush tropical environment for example Central America and south America for the most part have a tropical environment  because of humidity rainfall and direct sunlight
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What organisms apart from algae would have needed to be present to remove carbon dioxide from the early atmosphere??
QveST [7]
Plants in general I think. 
photosynthetic organisms.
7 0
4 years ago
State the Boyle's Law. If the graph for Boyle's law is straight line through the origin then state the quantity taken along: a)
astraxan [27]

Answer:

(a) X- axis represent (1/P) inverse of pressure

(b) Y-axis represent (V) volume

Explanation:

Boyle's Law States that at constant temperature, the volume of a fixed mass of gas is inversly proportional to its pressure. It can be expressed mathematically as

V α 1/P

VPα1

removing the proportionality sign and introducing a constant

VP = k

V₁P₁ = V₂P₂

Where V₁ and V₂ are initial and final volume respectively, P₁ and P₂ are initial and final pressure respectively.

If the graph for Boyle's law is straight line through the origin,

(a) X-axis represent (1/P) inverse of pressure

(b) Y- axis represent (V) volume.

7 0
3 years ago
In the graph below, why does the graph stop increasing after 30 seconds?
stepladder [879]

Answer:

The answer is "Option C".

Explanation:

It's evident from the figure below that after thirty minutes, not no more hydrogen can be created because all of the reactants have converted into products.

hydrogen gas created in cm cubes per period x = 20 seconds, y = 45 centimeters squared, and so on.

A reaction's terminus (the graph's flat line) indicates that no further products are being created during the reaction.

6 0
3 years ago
A constant net force is applied to an object. Suddenly, the object's mass doubles, but the net force remains the same. For this
____ [38]

<u>D: Half</u>

We know that,

F = m.a, where F is the force, m is the object's mass and a is the acceleration.

In the first case, we observed that a1 = F/m.

In the second case, we observed that the mass has been doubled, so a2 = F/2m .

By the ratio of the two cases, we get

a1/a2 = F/m / F/2m

or, a1/a2 = 2

or, a1 = 2.a2

or, a1/2 = a2

Therefore, the acceleration gets <u>half</u> of it's original measurement.

8 0
3 years ago
Work out the following a) Find β = v/c for a person walking at 3 mile/hr and a truck moving at 65 mile/hr. (b) Find γ-1, where γ
balu736 [363]

Explanation:

It is given that,

Speed of the person, v = 3 mile/hr = 1.34 m/s

Speed of the truck, v' = 65 mile/hr = 29.05 m/s

(a) Since, \beta =\dfrac{v}{c}

For the person, \beta =\dfrac{1.34}{3\times 10^8}=4.47\times 10^{-9}    

For the truck, \beta =\dfrac{29.05}{3\times 10^8}=9.68\times 10^{-8}

(b) The relativistic factor is given by :

\gamma=\dfrac{1}{\sqrt{1-\beta^2}}

For very small velocity, \beta

\gamma=(1-\beta ^2)^{-1/2}\approx 1+\dfrac{1}{2}\beta ^2

\gamma-1=\dfrac{\beta^2}{2}

For the person :

\gamma-1=\dfrac{(4.47\times 10^{-9})^2}{2}=9.99\times 10^{-18}

For the person :

\gamma-1=\dfrac{(9.68\times 10^{-8})^2}{2}=4.68\times 10^{-15}

Hence, this is the required solution.

3 0
3 years ago
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