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Marina86 [1]
3 years ago
9

Please help me on this

Physics
1 answer:
Fynjy0 [20]3 years ago
3 0

combustion of fossil fuels would be the correct answer when dealing with the alteration of the carbon cycle.

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How much heat is required to heat 2 kg of water from 25°C to 40°C?
Dominik [7]

Answer:

126000 J

Explanation:

Applying,

Q = cm(t₂-t₁).................. Equation 1

Where Q = Amount of heat, c = specifc heat capacity of water, m = mass of water, t₁ = Initial temperature, t₂ = Final temperature.

From the question,

Given: m = 2 kg, t₁ = 25°C, t₂ = 40°C

Constant: c = 4200 J/kg.°C

Substitute these value into equation 1

Q = 2×4200(40-25)

Q = 2×4200×15

Q = 126000 J

5 0
3 years ago
During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot ca
Dmitriy789 [7]

Answer:

1472.98 m

Explanation:

Data provided:

Speed of circular looping, v = 340 m/s

Acceleration, a = 8g

here,

g is the acceleration due to the gravity = 9.81 m/s²

Now,

the centripetal acceleration is given as,

a=\frac{v^2}{r}

r is the radius of the loop

on substituting the respective values, we get

8\times9.81=\frac{340^2}{r}

or

r = 1472.98 m

5 0
3 years ago
What is a direct result of having greenhouse gases in the atmosphere?
Shalnov [3]

Answer:

Well this is tough. I'm not sure but you are smart and can push through it. YOU DONT need the someone telling you the answer when it is inside you.

hope this helps p

3 0
2 years ago
Read 2 more answers
I need the right answer ASAP NO LINKS!!!
Jet001 [13]

Answer:

models are only used by scientists

5 0
2 years ago
Suppose astronomers find an earthlike planet that is twice the size of Earth (that is, its radius is twice the radius of Earth).
JulijaS [17]

Answer:

4 times the mass of Earth

Explanation:

M_1 = Mass of Earth

M_2 = Mass of the other planet

r = Radius of Earth

2r = Radius of the other planet

m = Mass of object

The force of gravity on an object on Earth is

F=\frac{GM_1m}{r^2}

The force of gravity on an object on the other planet is

F=\frac{GM_2m}{(2r)^2}

As the forces are equal

\frac{GM_1m}{r^2}=\frac{GM_2m}{(2r)^2}\\\Rightarrow M_1=\frac{M_2}{4}\\\Rightarrow M_2=4M_1

So, the other planet would have 4 times the mass of Earth

6 0
3 years ago
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