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3 years ago

What two gases probably dominated precambrian earth's atmosphere?

1 answer:

8 0

I think these gases are water vapor and nitrogen. As the temperature rises, these water vapor molecules, would condense and form the oceans we have. Also, it was said that in the early atmosphere, nitrogen is very abundant and even today the composition of air is 79% by volume.

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Jake drags a sled of mass 18 kg across the snow. The sled is attached to a rope that he pulls with a force of 70 N at an angle o

Advocard [28]

If the 30 N on the rope were pulled straight up, it would offset the force of gravity ( m g = 10 kg * 9.8 N/kg = 98 N) , leaving a net force up from the ground on the sled of 98-30 = 68 N. Since the rope is pulled at the angle of 25o, only part of the force is in the upward direction, (30N)(sin(25) = (30)(.423) = 12.7. So the net force becomes 98 N down offset by 12.7 up or 98-12.7 = 85.3 N. Ah, there it is: C.

8 0

3 years ago

Read 2 more answers

Gravity on the surface of the moon is only 1/6 as strong as gravity on the Earth. What is the weight of a 19 kg object on the Ea

Dahasolnce [82]

The weight of anything in any place is

(mass of the thing) x (acceleration of gravity in that place).

-- On Earth, the acceleration of gravity is about 9.807 m/s²

Weight of 19 kg of mass is (19 kg) x (9.807 m/s²) = <em>186.3 newtons</em>

-- On the Moon, the acceleration of gravity is about 1.623 m/s²

Weight of the same 19 kg of mass is (19 kg) x (1.623 m/s²) = <em>30.8 newtons</em>

7 0

3 years ago

The magnitude of a force is:

lys-0071 [83]

Answer:

Explanation:

force is how hard it is pulled or pushed

3 0

2 years ago

If an object absorbs all colors but red, we see

julia-pushkina [17]

[I researched for you, since I am not in that particular level to know that knowledge yet. I assure this is accurate info :)]

The answer is A, red.
"Remember, the color you see is light REFLECTING off the surface of that object. If all colors are absorbed in to the surface EXCEPT red, red must be reflected, and you'll see red." - Yahoo User @Chap

8 0

3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago