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3 years ago

How are animals of the coniferous forest well adapted to long, cold winters?

Physics

1 answer:

skelet666 [1.2K]3 years ago

3 0

They have thick body coverings

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1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.

Lisa [10]

Answer:

(D) x = 93.8 m

Explanation:

v^2 = v0^2 + 2ax

(20 m/s)^2 = (5 m/s)^2 + 2(2 m/s^2)x

Solving for x,

x = 93.8 m

4 0

3 years ago

How could Trevor increase the amplitude of the motion

iren [92.7K]

Answer:

He could pull the mass down farther.

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3 years ago

What else is produced when sodium carbonate decomposes

zaharov [31]

Answer:

The chemical symbol for sodium carbonate is . The decomposition of sodium carbonate is: The decomposition of sodium bicarbonate, will result in the formation of sodium oxide, and carbon dioxide, . Hence, carbon dioxide, will produce with sodium oxide, on decomposition of

Hope this will help you

4 0

3 years ago

Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.

Alexxx [7]

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called Electrostatic Force.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

$F=\frac{k.q.Q}{r^{2}}$

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

$F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}$

$F_{12}=536.02.10^{-3}$ N

Force between 2 and 3:

$F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}$

$F_{23}=2711.63.10^{-3}$ N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

$F_{T}=2711.63.10^{-3}-536.02.10^{-3}$

$F_{T}=2175.61.10^{-3}$ N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, $F_{13}=F_{23}$ . Suppose distance from 1 to 3 is x, then from 2 to 3 is $x-0.301$

Calculating:

$\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}$

$\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}$

$\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}$

$x^{2}=0.67x^{2}-0.40x+0.061$

$0.33x^{2}+0.40x-0.061=0$

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

4 0

2 years ago

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words: