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3 years ago

What variable can affect the amount of fruit produced by an apple tree

Physics

2 answers:

stich3 [128]3 years ago

8 0

A variable is a letter so just be like
13a or something like that

Gennadij [26K]3 years ago

8 0

In real life you don't label variables as letters, the other answer is wrong.
The real answer is something like a drought where apple trees have less nourishment to grow apples, or a heat wave or flash flood.
Hope this helps!

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The oscillating electric field in a plane electromagnetic wave is given by <img src="https://tex.z-dn.net/?f=%7B50%5Csin%20%28%5

Scorpion4ik [409]

Here, we are given with:

${:\implies \quad \longrightarrow \begin{cases}\sf E_{0}= 50\: V/m\\ \sf \nu = 2\times 10^{7}Hz\end{cases}}$

(a) So now, we can thus obtain

${:\implies \quad \sf \omega =2\pi \nu}$

${:\implies \quad \sf \omega =2\pi (2\times 10^{7})}$

${:\implies \quad \boxed{\bf{\omega = 4\pi \times 10^{7}\: s^{-1}}}}$

Now, finding $\lambda$

${:\implies \quad \sf \lambda =\dfrac{c}{\omega}=\dfrac{3\times 10^{8}}{4\pi \times 10^{7}}}$

${:\implies \quad \boxed{\bf{\lambda \approx 2.39\:m}}}$

Now, finding k

${:\implies \quad \sf k=\dfrac{2\pi}{\lambda}=\dfrac{200\pi}{239}}$

${:\implies \quad \boxed{\bf{k\approx 2.63\:m^{-1}}}}$

Thus, expression for the electric field is:

${:\implies \quad \boxed{\bf{E=50\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)}}}$

(b) Now, here

${:\implies \quad \sf B_{0}=\dfrac{E_{0}}{c}=\dfrac{50}{3\times 10^{8}}}$

${:\implies \quad \boxed{\bf{B_{0}\approx 16.67×10^{-7}\:T}}}$

Thus, expression for the magnetic field:

${:\implies \quad \boxed{\bf{B_{y}=16.67\times 10^{-7}\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)\:T}}}$

4 0

2 years ago

Corey runs a 100-meter race. 7 seconds after the race started Corey is 45 meters from the starting line and reaches his max spee

RUDIKE [14]

Answer:

Corey's max speed is $7 \frac{m}{s}$
the distance Corey's covers in z seconds is $7 \frac{m}{s} * z \ s$
$d (z) = 45 m + 7 \frac{m}{s} * z$
$d (x) = 45 m + 7 \frac{m}{s} * (x-7 s)$

Explanation:

<h3>Corey's max speed</h3>

For constant speed, we know:

$v=\frac{distance}{time}$

The distance between the 80 meters and the 45 meters is:

$distance = 80 m - 45 m = 35 m$

and the time it took to reach the 80 meter will be:

$time = 12 s - 7 s = 5 s$

So, Corey's max speed is

$v_{max}=\frac{35 m}{5 s} = 7 \frac{m}{s}$

<h3>How far runs Corey</h3>

As the velocity of Corey's is $v_{max}$ , the distance Corey's covers in z seconds is

$distance = v_{max} * z \ s$

$distance = 7 \frac{m}{s} * z \ s$

<h3>What is Corey's distance from the starting line</h3>

At time 7 + z seconds the distance will be the 45 meters he covers in the first part of the race plus the distance he traveled at constant speed. this is:

$d (z) = 45 m + v_{max} * z$

$d (z) = 45 m +7 \frac{m}{s} * z$

At time x ( x greater or equal to 7 seconds) the distance will be the 45 meters he covers in the first part of the race plus the distance he traveled at constant speed. this is:

$d (x) = 45 m + v_{max} * (x-7 s)$

$d (x) = 45 m + 7 \frac{m}{s} * (x-7 s)$

4 0

3 years ago

"a grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. the angular speed is then increased to 1

kotegsom [21]

The average angular speed of the grindstone is 10 rad/s

$\texttt{ }$

<h3>Further explanation</h3>

<em>Let's recall </em><em>Angular Speed</em><em> formula as follows:</em>

$\boxed{ \omega = \omega_o + \alpha t }$

$\boxed{ \theta = \omega_o t + \frac{1}{2} \alpha t^2 }$

$\boxed{ \omega^2 = \omega_o^2 + 2 \alpha \theta }$

$\boxed{ \theta = \frac{( \omega + \omega_o )}{2} t }$

<em>where :</em>

<em>ω = final angular speed ( rad/s )</em>

<em>ω₀ = initial angular speed ( rad/s )</em>

<em>α = angular acceleration ( rad/s² )</em>

<em>t = elapsed time ( s )</em>

<em>θ = angular displacement ( rad )</em>

$\texttt{ }$

<u>Given:</u>

radius of the grindstone = R = 4.0 m

initial angular speed = ω₀ = 8.0 rad/s

final angular speed = ω = 12 rad/s

elapsed time = t = 4.0 seconds

<u>Asked:</u>

average angular speed = ?

<u>Solution:</u>

<em>Firstly , we will calculate </em><em>angular displacement </em><em>as follows:</em>

$\theta = \frac{( \omega + \omega_o )}{2} t$

$\theta = \frac{ ( 12 + 8.0 ) }{2} \times 4.0$

$\theta = 10 \times 4.0$

$\boxed {\theta = 40 \texttt{ rad}}$

$\texttt{ }$

<em>Next , we could calculate the </em><em>average angular speed</em><em> as follows:</em>

$\texttt{average angular speed} = \theta \div t$

$\texttt{average angular speed} = 40 \div 4.0$

$\boxed{\texttt{average angular speed} = 10 \texttt{ rad/s}}$

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Moment of Inertia : brainly.com/question/13796477
The Ratio of the Moments of Inertia : brainly.com/question/2176655

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

4 0

3 years ago

Read 2 more answers

How long does it take for a truck accelerating at 1.5 m/s^2 to got from rest to 75 km/hr

Free_Kalibri [48]

Answer:

t = 13.9s

Explanation:

u = 0 m/s

v = 75 km/h

= 20.83 m/s

a = 1.5 m/s²

Using

v = u + at

20.83 = 0 + 1.5t

t = 13.9s

6 0

4 years ago

You are working on a laboratory device that includes a small sphere with a large electric charge Q. Because of this charged sphe

madam [21]

Answer:

the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.

Explanation:

We can answer this exercise using Gauss's law

Ф = ∫ e . dA = $q_{int}$ / ε₀

field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field

From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.

5 0

3 years ago