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2 years ago

A. What evidence have you discovered to explain how

Physics

2 answers:

frozen [14]2 years ago

7 0

Answer:

The change in ice alters the color of the lake surface — ice and snow cover are generally white, while the lake water itself is a deep blue.

Explanation:

hope it help

svp [43]2 years ago

3 0

Answer:

Well if it gets warmer the ice will weaken and if it gets hot enough will melt. Of course if the weather gets colder the ice will thicken.

Explanation:

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A simple generator is used to generate a peak output voltage of 25.0 V . The square armature consists of windings that are 7.0 c

4vir4ik [10]

Answer:

<h3>The 28 loops wound on the square armature</h3>

Explanation:

Peak output voltage $\epsilon _{peak} = 25$ V

Area of square armature $A = (7 \times 10^{-2} )^{2} = 49 \times 10^{-4}$

Magnetic field $B = 0.490$ T

Angular frequency $\omega = 2\pi f = 2 \pi \times 60 = 120\pi$

According to the law of electromagnetic induction,

$\epsilon _{peak} = NBA \omega$

Where $N =$ number of loops of wire.

$N = \frac{25}{49 \times 10^{-4} \times 0.49 \times 120\pi }$

$N = 27.6$ ≅ 28

Thus, 28 loops of wire should be wound on the square armature.

6 0

3 years ago

It is estimated that 26 large pizzas are about right to serve 66 students of a physics club meeting. How many pizzas would be re

Brums [2.3K]

Answer:

The answer to your question is: 15 pizzas

Explanation:

data

26 large pizzas ------ 66 students

? large pizzas -------- 38 students

Rule of three

x = 38 (26) / 66 = 14.96 ≈ 15 pizzas

5 0

3 years ago

The first asteroid to be discovered is Ceres. It is the largest and most massive asteroid in our solar system’s asteroid belt, h

stira [4]

Answer:

$4.81*10^{29}J$

Explanation:

Since the formula for kinetic energy of an object is:

$E_k = \frac{mv^2}{2}$

Where m is the mass of the object and v is the speed. We can substitute m = $3*10^{21}$ kg and v = 17900m/s:

$E_k = \frac{3*10^{21} * (17900)^2}{2} = 4.81*10^{29}J$

3 0

3 years ago

For this discussion, you will work in groups to answer the questions. In a video game, airplanes move from left to right along t

Mariulka [41]

Answer:

When fired from (1,3) the rocket will hit the target at (4,0)

When fired from (2, 2.5) the rocket will hit the target at (12,0)

When fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

When fired from (4,2.25) the rocket will hit the target at (40,0)

Explanation:

All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).

Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.

In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:

$y=2+\frac{1}{x}$

$y=2+x^{-1}$

$y'=-x^{-2}$

$y'=-\frac{1}{x^{2}}$

so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(1)^{2}}$

m=y'=-1

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-3=-1(x-1})$

$y-3=-1x+1$

$y=-x+1+3$

$y=-x+4$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-x+4=0$

and solve for x

x=4

so, when fired from (1,3) the rocket will hit the target at (4,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)

so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2)^{2}}$

$m=y'=-\frac{1}{4}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.5=-\frac{1}{4}(x-2})$

$y-2.5=-\frac{1}{4}x+\frac{1}{2}$

$y=-\frac{1}{4}x+\frac{1}{2}+\frac{5}{2}$

$y=-\frac{1}{4}x+3$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{4}x+3=0$

and solve for x

x=12

so, when fired from (2, 2.5) the rocket will hit the target at (12,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)

so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2.5)^{2}}$

$m=y'=-\frac{4}{25}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.4=-\frac{4}{25}(x-2.5})$

$y-2.4=-\frac{4}{25}x+\frac{2}{5}$

$y=-\frac{4}{25}x+\frac{2}{5}+2.4$

$y=-\frac{4}{25}x+\frac{14}{5}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{4}{25}x+\frac{14}{5}=0$

and solve for x

$x=\frac{35}{20}$

so, when fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)

so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(4)^{2}}$

$m=y'=-\frac{1}{16}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.25=-\frac{1}{16}(x-4})$