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UkoKoshka [18]
2 years ago
5

A balloon will stick to a wooden wall if the balloon is charged

Physics
1 answer:
grigory [225]2 years ago
7 0
<h2>Answer: either way</h2>

The balloon contains neutral charge atoms, that is, it has the same number of electrons (negative charge), protons (positive charge) and neutrons (no charge).

Then, when two objects come into contact, the electrons of one of them can become part of the other.

Thus, by bringing the balloon closer to the wall, the wall, which is also made up of atoms, will reorder its charges in such a way that its electrons or protons become part of the balloon, charging it.

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an ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing the block to accelerate uni
stiv31 [10]
The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.

Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg

F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)

F = 83 N
3 0
2 years ago
A banana peel has lots of friction.<br> True or False
olga_2 [115]

Answer:

False

Explanation:

I learned it the hard way trust me T^T

3 0
2 years ago
If you took a white piece of plastic to a depth of 180 m, what color would it appear?.
SashulF [63]

Answer:

It would appear blue

Explanation:

3 0
2 years ago
How does the magnitude of the electrical force between a pair of charged particles change when they are brought to half their or
Ostrovityanka [42]

Answer:

5. Quadruple

Explanation:

The electrostatic force between two charged particles is given by:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

If the distance between the charges is reduced to half,

r' = \frac{r}{2}

So the new force will be

F'=k\frac{q_1 q_2}{(r/2)^2}=4(k\frac{q_1 q_2}{r^2})=4F

So, the force will quadruple.

4 0
3 years ago
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
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