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4 years ago

In which season of the year is a forest most likely to look very stark?

Physics

2 answers:

kolbaska11 [484]4 years ago

8 0

B. Winter Would be ur answer

Hope I helped:P

Zina [86]4 years ago

3 0

<h3><u>Answer;</u></h3>

B. winter

<h3><u>Explanation;</u></h3>

<em><u>Winter happens to be the coldest season of the year in polar and temperate zones. This season occurs before spring and after autumn each year. </u></em>
<em><u>Winter temperatures are often below freezing. Since the winter is cooler, this encourages trees to shed their leaves. Trees and plants stop growing during this season.</u></em>
<em><u>Therefore, it is during winter season when forests would look very stark.</u></em>

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What is crossing-over in geneitcs and why is it important

slava [35]

Crossing over, or recombination, is the exchange of chromosome segments between nonsister chromatids in meiosis. Crossing over creates new combinations of genes in the gametes that are not found in either parent, contributing to genetic diversity.

8 0

3 years ago

A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

sladkih [1.3K]

Answer:

The kinetic energy is $KE = 7.59 *10^{10} \ J$

Explanation:

From the question we are told that

The radius of the orbit is $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

The gravitational force is $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

$KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

$v = \sqrt{\frac{G M}{r^2} }$

=> $v^2 = \frac{GM }{r}$

substituting this into the equation

$KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

$F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and m is the mass of the satellite

Now looking at the formula for KE we see that we can represent it as

$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

=> $KE = \frac{ 1}{2} *F_g * r$

substituting values

$KE = \frac{ 1}{2} *6600 * 2.3*10^{7}$

$KE = 7.59 *10^{10} \ J$

7 0

3 years ago

1. Johnny wants to know where the water line is in a dark well. He drops a penny into the well and counts until he hears the pen

777dan777 [17]

Answer:

3 feet down

Explanation:

i think

8 0

3 years ago

If 6.24×10¹⁸ electron pass through a wire in 1s how many pass through it during a time interval of 2hr, 47min and 10s

levacccp [35]

Answer = 6.24x10^18 x ((2 x 3600) + (47 x 60) + 10)

3 0

3 years ago

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f

vampirchik [111]

Answer:

$\large \boxed{\text{30 rev/s}}$

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

I₁ = 3.5 kg·m²s⁻¹

ω₁ = 6.0 rev·s⁻¹

I₂ = 0.70 kg·m²s⁻¹

Calculation:

$\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}$

8 0

3 years ago