Are electric files lines due to an isolated positive point charge emanate radially away from the charge?

Think of the last five things you read. Now think of a type of reading you havent done recenty

GarryVolchara [31]

Answer:

1) Half Girlfriend

2) I am Malala

3) Diamond fire

4)I too had a love story

5)Your Dream are Now Mine

Explanation:

I don't have a same pattern to read but mostly of romantic and fiction. I read less autobiographies but when I read I am Malala , it was an inspiring one.My favorite type of reading is romantic types and least favorite is non fiction especially History

6 0

3 years ago

A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal

Gemiola [76]

Answer:

Time taken, $T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

$T\ cos\theta-mg=0$

$T=\dfrac{mg}{cos\theta}$

Sum of forces in x direction,

$T\ sin\theta=\dfrac{mv^2}{r}$

$mg\ tan\theta=\dfrac{mv^2}{r}$ .............(1)

Also, $r=l\ sin\theta$

Equation (1) becomes :

$mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}$

$v=\sqrt{gl\ tan\theta.sin\theta}$ ...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

$T=\dfrac{2\pi r}{v}$

Put the value of T from equation (2) to the above expression:

$T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}$

$T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}$

On solving above equation :

$T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Hence, this is the required solution.

4 0

3 years ago

An green hoop with mass mh = 2.8 kg and radius rh = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass

vladimir2022 [97]

The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2

6 0

2 years ago

A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w

amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

10 x 3 + 25 x 0 = v(10 + 25)

30 = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

$K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\$

the final kinetic energy of the package;

$K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J$

the fraction of the initial energy lost;

$= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71$

7 0

3 years ago

Which undergoes greater acceleration: an airplane that goes from 1000 km/h to 1005 km/h in 10 seconds or a skateboard that

Alex_Xolod [135]

Answer:

Skateboard

Explanation:

Acceleration is change in velocity over time.

a = Δv / Δt

The airplane's acceleration is:

a = (1005 km/h − 1000 km/h) / 10 s

a = 0.5 km/h/s

The skateboard's acceleration is:

a = (5 km/h − 0 km/h) / 1 s

a = 5 km/h/s

6 0

2 years ago