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mezya [45]
2 years ago
13

1. Describe what is happening at point A when these 3 vectors act on point A. 2. Describe what happen to the resultant vector at

Point B when the 5N vector goes from 0 degree to 180 degrees. 3. A lawnmower is pushed with a constant force F. as 0 between vector F and the horizontal plane decreases describe what is happening to the Fx and Fy vector components of vector F. 4.Calculate the magnitude of the component of the force parallel to the ground.

Physics
1 answer:
zavuch27 [327]2 years ago
4 0

Answer: Don't know sorry

Explanation:

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Why was the microscope and its improvement an important part of the development of the cell theory?
netineya [11]

Answer:it’s is part of the cell theory because they where studying cells and to see it you need a microscope

Explanation:basically in the answer area

8 0
2 years ago
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From the last problem, what is the ratio of the ppm change in CO2 to the ppm change in CH4? Assume that the concentrations of CO
xxMikexx [17]

An additional 10ppm of methane will change the upward IR heat flux (in tropical

atmosphere) from 289.29 W/m2

to 286.211 W/m2

, which is a reduction of 3.08

W/m2

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An additional 10ppm of CO2 will change the upward IR heat flux from 289.29 W/m2

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3 0
3 years ago
The amount of radiant power produced by the sun is approximately 3.9 ✕ 1026 W. Assuming the sun to be a perfect blackbody sphere
fredd [130]

Answer:

T=5797.8  K

Explanation:

Given that

Power P = 3.9 x 10²⁶ W

Radius ,r= 6.96 x 10⁸ m

We know that ,From Plank's law

P = σ A T⁴

σ = 5.67 x 10 ⁻⁸

A= Area ,T= Temperature ( in Kelvin)

T=\left (\dfrac{P}{\sigma A} \right )^{\dfrac{1}{4}}

Now by putting the values

T=\left (\dfrac{P}{\sigma A}\right )^{\dfrac{1}{4}}

T=\left (\dfrac{3.9\times 10^{26}}{5.67\times 10^{-8}\times 4\times \pi \times (6.96\times 10^8)^2} \right )^{\dfrac{1}{4}}

T=5797.8  K

The temperature of surface will be 5797.8 K

4 0
2 years ago
A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio
valina [46]

Answer:

Explanation:

It is given that,

Mass of lump, m₁ = 0.05 kg

Initial speed of lump, u₁ = 12 m/s

Mass of the cart, m₂ = 0.15 kg

Initial speed of the cart, u₂ = 0

The lump of clay sticks to the cart as it is a case of inelastic collision. Let v is the speed of the cart and the clay after the collision. As the momentum is conserved in inelastic collision. So,

m_1u_1+m_2u_2=(m_1+m_2)v

v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}

v = 3 m/s

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3 years ago
Which layer of the atmoshere is the top layer of the thermoshere
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exosphere is the outer layer of the atmosphere

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