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timama [110]
2 years ago
10

What keeps an inflated balloon from falling down if you rub it against your hair and place it against a wall? 1. rubbing distort

s the atoms inside the ballon and polarizes it. 2. rubbing leaves a balloon electrically charged; the charged balloon polarizes the wall. 3. when you rub the balloon against your hair, it will remove some m?
Physics
1 answer:
Keith_Richards [23]2 years ago
3 0
2 because when you are doing this it causes friction Which then cause the balloon to stick
~dany-ley
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A stationary particle of charge q = 2.1 × 10-8 c is placed in a laser beam (an electromagnetic wave) whose intensity is 2.9 × 10
alisha [4.7K]
(a) The intensity of the electromagnetic wave is related to the amplitude of the electric field by
I= \frac{1}{2} c \epsilon_0 E^2
where
I is the intensity
c is the speed of light
\epsilon_0 is the electric permittivity
E is the amplitude of the electric field

By substituting the numbers of the problem and re-arranging the equation, we can find E:
E= \frac{2 I}{c \epsilon_0} = \frac{2 ( 2.9 \cdot 10^3 Wm^{-2})}{(3 \cdot 10^8 m/s)(8.85 \cdot 10^{-12} Fm^{-1})} =2.2 \cdot 10^6 N/C

Now that we have the intensity of the electric field, we can calculate the electric force on the charge:
F=qE=(2.1 \cdot 10^{-8} C)(2.2 \cdot 10^6 N/C)=0.046 N

(b) We can calculate the amplitude of the magnetic field starting from the amplitude of the electric field:
B= \frac{E}{c}= \frac{2.2 \cdot 10^6 N/C}{3 \cdot 10^8 m/s}=7.3 \cdot 10^{-3} T

The magnetic force is given by
F=qvB \sin \theta
where v is the particle's speed, B the magnetic field intensity and \theta the angle between B and v.
In this case the charge is stationary, so v=0, and so the magnetic force is zero: F=0.

(c) The electric force has not changed compared to point (a), because it does not depend on the speed of the particle, so we have again F=0.046 N.

(d) This time, the particle is moving with speed v=3.7 \cdot 10^4 m/s, in a direction perpendicular to the magnetic field (so, the angle \theta is 90^{\circ}), and so by using the intensity of the magnetic field we found in point (b), we can calculate the magnetic force on the particle:
F=qvB \sin \theta = (2.1 \cdot 10^{-8}C)(3.7 \cdot 10^4 m/s)(7.3 \cdot 10^{-3} T)(\sin 90^{\circ} )=
=5.7 \cdot 10^{-6} N
5 0
2 years ago
Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy can be a
marysya [2.9K]

Answer:

The movable piston

Explanation:

Work is said to be done when a distance is been covered by a force . In this case kinetic energy will be change by an equal amount into work done.

Pushing the piston with a known mass of (m) and an accelarating rate from rest of ( a)   to cover a known distance of (d).The idea of work done is been achieved and can be mathematically represented by:

  • Work done = Force x distance (d)
  • Force = mass (m) x acceleration (a)
8 0
2 years ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
2 years ago
In a coordinate system in which the x-axis is east, for what range of angles is the x- component positive? For what range is it
trasher [3.6K]
If i remeber correctly when dealing with real world cordinate systems as you rotate around clockwise you move in a positive direction. but all the examples i have done said north was 0 degrees, so i may be wrong
6 0
2 years ago
In a photoelectric effect experiment, electromagnetic radiation containing a finite distribution of wavelengths shines on a meta
Sonja [21]

Answer with Explanation:

a.Intensity of radiation is directly proportional to the frequency of radiation

When the intensity of radiation  increases then the frequency of radiation increases and therefore, the number of photo-electrons emitted by the metal increases.

b.When all  of the wavelength in the radiation are increased by the same amount

We know that  

f=\frac{v}{\lambda}

Frequency is inversely proportional to the wavelength.

Therefore, the frequency decrease .

When the frequency decreases then the number of photo-electrons emitted by the metal decrease.

c.When the work function of the metal is increased  then the gain of kinetic energy decreases .

When energy decreases then the number of photo-electrons emitted by the metal decreases.

3 0
2 years ago
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