The graph below plots the temperature and luminosity of stars on the main sequence. (2 points)

Which "spheres" are interacting when water evaporates from plants

Novay_Z [31]

The plant grows in the solid part of earth, the lithosphere. When water evaporates from the plant, it enters the hydrosphere, the portion if earth on kand and in the air that contains water. The atmosphere is part of the hydrosphere.

8 0

2 years ago

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A ball is dropped from rest at the top of a 6.10 m

natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

$\frac{v_1}{v}$ = e ( coefficient of restitution ) = $\frac{1}{\sqrt{10} }$

and

$\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }$

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

$e = \sqrt{\frac{h_1}{6.1} }$

$e = \sqrt{\frac{h_2}{h_1} }$

So on

$e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }$

$(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}$ = .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0

2 years ago

A train starting from rest picks up a speed of 20 m/s in 20 s while travelling on a straight path. It continues to move at the s

Andrej [43]

Answer:

$1300 m$

Explanation:

As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from $20 s$ to $70 s$ is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.

The total distance covered by the train during the entire journey is the area of the speed-time graph.

Area $=\frac{1}{2}(20\times 20)+ 50\times 20+\frac{1}{2}(20\times 10)$

$=200+1000+100$

$=1300$

As velocity is in $m/s$ and time is in $s$ so the unit of area is $m$

Hence, the total distance is $1300m$ .

8 0

3 years ago

How does the electric force between two charged participles change if one particle’s charge is reduced by a factor of 3?

Eva8 [605]

Answer:

ick

Explanation:

Me down daddy

3 0

3 years ago

A light, rigid rod is 55.8 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at res

VMariaS [17]

To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have

$KE = PE$

$\frac{1}{2} mv^2 = mgh$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8)(2(55.8*10^{-2}))}$

$v = 4.67m/s$

Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s

4 0

3 years ago