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3 years ago

A charged balloon will stick to a neutral wall. which process is involved?

1 answer:

3 0

<span>The charged balloon will stick to a neutral wall because of the Static Electricity:
</span>
The matter is formed by atoms and these atoms are composed of electrons, protons and neutrons (the electrons have a negative charge, the protons have a positive charge and the neutrons have no charge).

As the balloon is charged (It gained electrons), and the charge of the same sign repel each other, when it approaches the wall, the electrons of this wall will move away, and the positive charges (protons) will remain in the nearest area to the balloon. As the charges of different signs are attracted, the balloon will be stuck to the wall.

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A circular wooden loop of mass m and radius R rest on a flat horizontal friction less surface. A bullet, also of mass m, and mov

bearhunter [10]

Answer:

w = vR/3

Explanation:

The centre of mass of the loop to bullet system is given by D / 4 from centre of loop, which is equivalent to R / 2 from its centre.

From the principle of conservation of linear momentum , we have

m*v = 2*m* Vcm

Where v = velocity of bullet, Vcm = velocity of wood

Hence, we have

Vcm = v2

Also, from the conservation of angular momentum about the centre of mass.

M*V*(R/2) = Ic*w - equation (I)

where Ic = moment of inertia and w = angular velocity

Ic for a ring is given by $mr^2 + m(r/2)^2$

Ic of a bullet is given by $m(r/2)^2$

Hence, the moment of inertia of the system is given by the summation of the two moments of inertia Ic(ring) + Ic(bullet) which gives

Ic(system) = $3*m*R^2/2$

Substituting back into equation (I), we have

$m*v*R^2=3*m*R^2*w/2$

Hence, we obtain w =vR/3

w=v3R

4 0

2 years ago

The Sun is a star that is in a stable state called the main sequence.

nlexa [21]

Astronomers believe some stars are moving away from earth by measuring the extent to which their light is “stretched” into the red part of the color spectrum, representing the lower frequencies on that spectrum.

In short, stars moving away from earth are “red-shift”, while if they were moving towards the earth they would take on a “blue-shift”.

5 0

1 year ago

An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 2530 km. When the astronaut jumps upwa

Natali [406]

Answer:

1.38*10^18 kg

Explanation:

According to the Newton's law of universal gravitation:

$F=G*\frac{m_a*m_p}{r^2}$

where:

G= Gravitational constant (6.674×10−11 N · (m/kg)2)

ma= mass of the astronaut

mp= mass of the planet

$F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2$

so:

$m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg$

7 0

3 years ago

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce

Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

$f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Frequency of pendulum is given by

$f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

Given in the question

$f_p=\dfrac{1}{2}f_s$

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m$

The unstretched length of the spring is 2.67 m

6 0

2 years ago

1) You slam on the brakes of your car in a panic, and skid a certain distance on a straight level road. If you had been travelin

aleksandr82 [10.1K]

Answer:

d = 4 d₀o

Explanation:

We can solve this exercise using the relationship between work and the variation of kinetic energy

W = ΔK

In that case as the car stops v_f = 0

the work is

W = -fr d

we substitute

- fr d₀ = 0 - ½ m v₀²

d₀ = ½ m v₀² / fr

now they indicate that the vehicle is coming at twice the speed

v = 2 v₀

using the same expressions we find

d = ½ m (2v₀)² / fr

d = 4 (½ m v₀² / fr)

d = 4 d₀o

3 0

2 years ago