Multiples of the fundamental frequency created by plucking a string very quickly, several times in a row is called_________

A solid metal sphere with radius 0.430 m carries a net charge of 0.270 nC . Part A Find the magnitude of the electric field at a

rodikova [14]

Answer:

8.46 N/C

Explanation:

Using Gauss law

$E=\frac {kQ}{r^{2}}$

Gauss's Law states that the electric flux through a surface is proportional to the net charge in the surface, and that the electric field E of a point charge Q at a distance r from the charge

Here, K is Coulomb's constant whose value is $9\times 10^{9} Nm^{2}/C^{2}$

r = 0.43 + 0.106 = 0.536 m

$E=\frac {9\times 10^{9}\times 0.270\times 10^{-9}}{0.536^{2}}=8.4581755402094007\approx 8.46 N/C$

8 0

4 years ago

The vertical motion of mass A is defined by the relation x 5 10 sin 2t 1 15 cos 2t 1 100, where x and t are expressed in millime

nikklg [1K]

Explanation:

Given that,

The vertical motion of mass A is defined by the relation as :

$x=10\ sin2t+15\ cos2t+100$

At t = 1 s

$x=10\ sin2+15\ cos2+100$

x = 115.33 mm

(a) We know that,

Velocity, $v=\dfrac{dx}{dt}$

$v=\dfrac{d(10\ sin2t+15\ cos2t+100)}{dt}$

$v=20\ cos2t-30\ sin2t$

At t = 1 s

$v=20\ cos2-30\ sin2$

v = 18.94 mm/s

We know that,

Acceleration, $a=\dfrac{dv}{dt}$

$a=\dfrac{d(20\ cos2-30\ sin2)}{dt}$

$a=-40\ cos2t-60\ cos2t$

At t = 1 s

$v=-40\ cos2-60\ cos2$

$a=-99.93\ mm/s^2$

(b) For maximum velocity, $\dfrac{dv}{dt}=a=0$

$-40\ cos2t-60\ cos2t=0$

t = 45 seconds

For maximum acceleration, $\dfrac{da}{dt}=0$

$80\ sin2t+120\ cos2t=0$

t = 61.8 seconds

Hence, this is the required solution.

5 0

4 years ago

If all objects that have mass also have gravity, why doesn't your pencil get pulled towards you while it sits on your desk?

Lemur [1.5K]

Answer:

The pencil is not pulled towards a person due to a very small magnitude of force between them, due to lighter masses.

Explanation:

Let us apply Newton's Law of Gravitation between a person and pencil.

Average Mass of a Normal Pencil = m₁ = 10 g = 0.01 kg

Average Mass of a Person = m₂ = 80 kg

Distance between both = r = 1 cm = 0.01 m (Taking minimal distance)

Gravitational Constant = G = 6.67 x 10⁻¹¹ N.m²/kg²

So,

F = Gm₁m₂/r²

F = (6.67 x 10⁻¹¹ N.m²/kg²)(0.01 kg)(80 kg)/(0.01 m)²

This Force is very small in magnitude due to the light masses of both objects.

<u>Therefore, the pencil is not pulled towards a person due to a very small magnitude of force between them, due to lighter masses.</u>

5 0

3 years ago

A hammer thrower accelerates the hammer (mass = 7.90 kg ) from rest within four full turns (revolutions) and releases it at a sp

Dmitriy789 [7]

Answer: 1) a=5.98 rad/sec² 2) a tan= 8.97 m/s² 3) a rad= 450.7 m/s²

4) F= 3,560.5 N 5) theta = 180º

Explanation:

1) By definition, angular acceleration is equal to the change in angular velocity over time.

Assuming an constant angular acceleration, we can use one of the equivalent kinematic equations for circular movement, as follows:

ωf² - ω₀² = 2. Δθ.γ

and we know also that ω = v/r = 26.0 m/s / 1.50 m = 17.3 rad/s

As the hammer thrower accelerated from rest, ω₀ = 0, so replacing by the values, we get the angular acceleration, γ, as follows:

γ = ωf² / 2. Δθ = (17.3)² rad²/s² / 2. 8 π rad = 5.98 rad/s².

2) Tangential acceleration, has the same relationship with radius that angular velocity, so we can write the following:

at = γ . r = 5.98 rad/sec². 1.50 m = 8.97 m/s²

3) The centripetal acceleration, by definition, is the change in direction of the linear velocity vector, over time, is always directed towards the center of the circle, and her magnitude is as follows:

ac = v² / r

Just before release, the velocity has a magnitude of 26.0 m/s, so ac is as follows:

ac = (26.0)² m²/s² / 1.50 m = 450.7 m/s²

4) Ignoring gravity, the only force acting on the hammer, is the one exerted by the thrower, and this force is just the centripetal force, which is the product of the hammer mass times the centripetal acceleration, as follows:

Fc = m . ac = 7.9 Kg. 450. 7 m/s² = 3,560.5 N

5) Ignoring gravity, as the force exerted by the thrower is always along the radius of the circle, towards the centre, if we represent the radius as a vector with origin in the center, the force is always anti-parallel to it, so the angle of the force with respect to the radius of the circular motion is 180°.

7 0

3 years ago

PLEASE!!!!!!HELP!!!!!!!

REY [17]

DNA Polymerase proofreads and corrects bases added during replication.

6 0

3 years ago

Multiples of the fundamental frequency created by plucking a string very quickly, several times in a row is called__________.