Which quantity is the vector? a) weight b) power c) kinetiic energy d) speed

What physical property of nylon and leather make them get materials to use for shoelaces

Ulleksa [173]

What physical properties of nylon and leather make them good materials to use for shoelaces? A. high density and low conductivity B. durability and flexibility C. hardness and durability D. low viscosity and flexibility.

4 0

3 years ago

Read 2 more answers

A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the

Tanya [424]

Answer:

Afterward the magnitude of the acceleration of the stone is four times as great

Explanation:

A stone tied to a string and whirled at a constant speed in a horizontal circle will have a centripetal acceleration given by

$a = \frac{v^{2} }{r}$

Where $a$ is the centripetal acceleration

$v$ is the speed

and $r$ is the radius of the circle

Here, the radius of the circle is the length of the string.

Now, from the question

The speed is then doubled without changing the length of the string,

Let the new speed be $v_{2}$ , that is

$v_{2} = 2v$

and without changing the length of the string means radius $r$ is constant.

To determine the magnitude of the acceleration of the stone afterwards,

Let the new acceleration be $a_{2}$ .

Then we can write that

$a_{2} = \frac{v_{2}^{2} }{r}$

From

$a = \frac{v^{2} }{r}$

$v = \sqrt{ar}$

Recall that

$v_{2} = 2v$

∴ $v_{2} = 2\sqrt{ar}$

Now, we will put the value of $v_{2}$ into

$a_{2} = \frac{v_{2}^{2} }{r}$

Then,

$a_{2} = \frac{(2\sqrt{ar}) ^{2} }{r}$

$a_{2} = \frac{4ar }{r}$

$a_{2} = 4a$

The new magnitude of the acceleration of the stone is four times the initial acceleration.

Hence,

Afterward the magnitude of the acceleration of the stone is four times as great

3 0

3 years ago

A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius

Nostrana [21]

Explanation:

Gauss Law relates the distribution of electric charge to the resulting electric field.

Applying Gauss's Law,

EA = Q / ε₀

Where:

E is the magnitude of the electric field,

A is the cross-sectional area of the conducting sphere,

Q is the positive charge

ε₀ is the permittivity

We be considering cases for the specified regions.

Case 1: When r < R

The electric field is zero, since the enclosed charge is equal to zero

E(r) = 0

Case 2: When R < r < 2R

The enclosed charge equals to Q, then the electric field equals;

E(4πr²) = Q / ε₀

E = Q / 4πε₀r²

E = KQ /r²

Constant K = 1 / 4πε₀ = 9.0 × 10⁹ Nm²/C²

Case 3: When r > 2R

The enclosed charge equals to Q, then the electric field equals;

E(4πr²) = 2Q / ε₀

E = 2Q / 4πε₀r²

E = 2KQ /r²

4 0

3 years ago

Why quantity refractive index dooesn't have unit

ASHA 777 [7]

Refractive Index is a ratio of two similar physical quantity which is dimension less
refractive index = sin I / sin r

therefore it doesn't have a unit.

5 0

3 years ago

Read 2 more answers

A 0.9 kg ball attached to a cord is whirled in a vertical circle of radius 2.5 m. Find the minimum speed needed at the top of th

lapo4ka [179]

Answer:

The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.

Explanation:

By the Principle of Energy Conservation we understand that the minimum speed needed by the ball is that speed such that maximum height reached is equal to the diameter of the vertical circle, that is:

$K =U_{g}$ (1)

Where:

$K$ - Translational kinetic energy, measured in joules.

$U_{g}$ - Gravitational potential energy, measured in joules.

By definitions of translational kinetic and gravitational potential energies, we expand the equation above and clear the initial speed of the ball:

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g\cdot h$

$v = \sqrt{2\cdot g\cdot h}$ (2)

Where:

$m$ - Mass, measured in kilograms.

$v$ - Initial speed, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

$h$ - Maximum height of the ball, measured in meters.

If we know that $g = 9.807\,\frac{m}{s^{2}}$ and $h = 5\,m$ , then the initial speed of the ball is:

$v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}$

$v\approx 9.903\,\frac{m}{s}$

The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.

3 0

4 years ago