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3 years ago

15

Suppose you observe that at night, the air just above the ground feels cooler than the air above it. then in the middle of a sun

ny day, you notice that the air close to the ground feels warmer. which process explains this?

1 answer:

yanalaym [24]3 years ago

6 0

The answer is convection

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Harrison wanted to find out what soil works best for growing roses. He grew them in potting soil, clay, sand, and soil he found

zepelin [54]

Answer:

This question is asking to identify the following variables:

Independent variable (IV): TYPE OF SOIL

Dependent variable (DV): HEIGHT AND NUMBER OF LEAVES

Control group: None in this experiment

Constant: SAME ROSE PLANT, SAME TIME INTERVAL (1 WEEK)

Explanation:

Independent variable in an experiment is the variable that is manipulated or changed by the experimenter in order to effect a measurable outcome. In this case, the independent variable is the TYPE OF SOIL used.

Dependent variable is the measurable variable that responds to changes made to the independent variable. In this experiment, the dependent variable is the HEIGHT AND NUMBER OF LEAVES of each rose.

Constants or control variable is the variable that is kept unchanged or constant for all groups throughout the experiment. In this experiment, the constants are SAME ROSE PLANT, SAME TIME INTERVAL (1 WEEK).

Control group are the groups that does not receive the experimental treatment. In this case, all the groups received the experimental treatment (different soil types). Hence, there is no control

4 0

3 years ago

The International Space Station (ISS) orbits Earth at an altitude of 4.08 × 105 m above the surface of the planet. At what veloc

alukav5142 [94]

This question involves the concepts of orbital velocity and orbital radius.

The orbital velocity of ISS must be "7660.25 m/s".

The orbital velocity of the ISS can be given by the following formula:

$v=\sqrt{\frac{GM}{R}}$

where,

v = orbital velocity = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m

R = 67.86 x 10⁵ m

Therefore,

$v=\sqrt{\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{67.86\ x\ 10^5\ m}}$

<u>v = 7660.25 m/s</u>

Learn more about orbital velocity here:

brainly.com/question/541239

3 0

2 years ago

A 1.3-kg model airplane flies in a circular path on the end of a 23-m line. The plane makes

storchak [24]

(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in

(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s

(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:

(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s

(c) The plane accelerates toward the center of the path with magnitude

<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²

(d) By Newton's second law, the tension in the line is

<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N

4 0

3 years ago

Consider the model above. It represents the electrical force. As r increases, the attractive force decreases. How would this mod

aivan3 [116]

Answer:

As we keep on increasing the radius the value of the gravitation force of attraction decreases and as we decrease the radius the gravitation force increases.

Explanation:

Like the coulombs law of electrostatics, the law of gravitation also depends inversely on the square of the value of r. Therefore, as we keep on increasing the value of r the value of the gravitation force decreases and as we decrease the value of the r the value of gravitation force increases.

Gravitation Force= $\frac{Gm_{1}m_{2} }{r^{2}}$

Coulombs's Law= $\frac{Kq_{1}q_{2} }{r^{2}}$

6 0

3 years ago

Read 2 more answers

A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac

Tpy6a [65]

Answer:

The force on one side of the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

$y=\dfrac{\Delta y}{\sin60}$

$y=\dfrac{2\Delta y}{\sqrt{3}}$

The area of strip is

$dA=\dfrac{9\times2\Delta y}{\sqrt{3}}$

We need to calculate the force on one side of the plate

Using formula of pressure

$P=\dfrac{dF}{dA}$

$dF=P\times dA$

On integrating

$\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}$

$F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}$

$F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})$

$F=3093529.3\ N$

Hence, The force on one side of the plate is 3093529.3 N.

7 0

3 years ago