Which word best describes the symmetry of a starfish

A device that measures potential is a(n) circuit.

Darina [25.2K]

Answer:

voltmeter

Explanation:

A voltmeter is the instrument used to measure a potential difference between two points in an electric circuit.

8 0

3 years ago

Dr. John Paul Stapp was U.S. Air Force officer whostudied the effects of extreme deceleration on thehuman body. On December 10,

Ira Lisetskai [31]

Answer:

(a) a = 56.4 m/s², his acceleration a, in multiples of gravity g, is 5.76 g

(b) a = -201.43 m/s², his deceleration -a, in multiples of gravity g, is -20.56 g

Explanation:

(a)

When moving upwards, the initial velocity, u = 0 (he accelerated from rest)

When moving upwards, the final velocity, v = 282 m/s

time of motion during this acceleration, t = 5 s

His acceleration is calculated as;

v = u + at

282 = 0 + 5a

a = 282 / 5

a = 56.4 m/s²

Ratio of his acceleration, a to gravity, g = a/g = 56.4 / 9.8 = 5.76

a = 5.76 g

(b)

When moving downwards, the initial velocity, u = 282 m/s

When moving downwards, the final velocity, v = 0 (he was brought to rest)

time of motion during this deceleration, t = 1.4 s

His deceleration is calculated as;

v = u + at

0 = 282 + 1.4a

1.4a = -282

a = -282 / 1.4

a = -201.43 m/s²

Ratio of his deceleration, -a to gravity, g = -a/g = 201.43 / 9.8 = 20.56

a = -20.56 g

3 0

3 years ago

What is meant by the statement "Motion is relativel"?

aleksklad [387]

Answer:

when an object move from its place is called an mation

4 0

3 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:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

2 years ago

Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through

Ad libitum [116K]

Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,

d l = d x

in this situation. We can therefore write

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

4 0

2 years ago