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fomenos
2 years ago
10

Once again, move the balloon to the right and let it go. Note how fast the balloon moves. Next, brush the balloon against the en

tire sweater. Allow all the electrons to transfer. Again, move the balloon all the way to the right, let it go, and note how fast it moves. Is there a difference in how fast the balloon moves when the balloon has more electrons and the sweater has fewer electrons?
Physics
2 answers:
lilavasa [31]2 years ago
8 0

Answer:

Yes. The balloon moves faster when it has more electrons and the sweater has fewer electrons

Explanation:

From Plato. Hope this helps!

olya-2409 [2.1K]2 years ago
6 0

Answer:

Yes. The balloon moves faster when it has more electrons and the sweater has fewer electrons

Explanation:

Edmentum

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The instantaneous emf resulting from magnetic induction equals the rate of change of flux is
katen-ka-za [31]

Lenz's Law: The polarity of the induced emf is such that it produces a current whose magnetic field opposes the change in magnetic flux through the loop.

7 0
3 years ago
What is the coefficient of friction between the skates and the ice?
natta225 [31]

The coefficient of friction is 0.051

Explanation:

The motion of the skater is a uniformly accelerated motion, therefore we can use the following suvat equation:

v^2 - u^2 = 2as

where:

v = 0 is the final velocity of the skater (he comes to a stop)

u = 10.0 m/s is his initial velocity

a is the acceleration

s=1.0\cdot 10^2 m = 100 m is the distance he travels before stopping

Solving for a, we find the acceleration of the skater:

a=\frac{v^2-u^2}{2s}=\frac{0-10.0^2}{2(100)}=-0.5 m/s^2

We also know that the net force acting on the skater is the force of friction, therefore we can write (Newton's second law of motion):

F= ma = -\mu mg

where

-\mu mg is the force of friction

m is the mass of the skater

\mu is the coefficient of friction

a=-0.5 m/s^2 is the acceleration

g=9.8 m/s^2 is the acceleration of gravity

Solving for \mu, we find the coefficient of friction:

\mu = -\frac{a}{g}=-\frac{-0.5}{9.8}=0.051

Learn more about friction:

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8 0
3 years ago
Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00 ✕ 105 kg on springs that ha
jasenka [17]

Explanation:

It is given that,

Mass of an object, m=4\times 10^5\ kg

(a) Time period of oscillation, T = 2.4 s

The formula for the time period of spring is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

Where

k is the force constant

k=\dfrac{4\pi ^2 m}{T^2}

k=\dfrac{4\pi ^2 \times 4\times 10^5}{(2.4)^2}

k=2.74\times 10^6\ N/m

(b) Displacement in the spring, x = 2.2 m

Energy stored in the spring is given by :

U=\dfrac{1}{2}kx^2

U=\dfrac{1}{2}\times 2.74\times 10^6\ N/m\times (2.2\ m)^2

U=6.63\times 10^6\ J

Hence, this is the required solution.

7 0
3 years ago
What kind of waves can cause the most damage on the earth's surface?
Karo-lina-s [1.5K]

Answer:

Surface waves, in contrast to body waves can only move along the surface. They arrive after the main P and S waves and are confined to the outer layers of the Earth. They cause the most surface destruction. Earthquake surface waves are divided into two different categories: Love and Rayleigh.

Explanation:

Hope this helped Mark BRAINLIEST!!!

4 0
3 years ago
Read 2 more answers
The mass of Earth is 5.972 x 1024 kg and its orbital radius is an average of 1.496 x 1011 m. Calculate its linear momentum, give
Anna007 [38]

Answer: its linear momentum is 1.78 × 10²⁹ kg.m/s

Explanation:

Given that;

mass of Earth m =  5.972 x 10²⁴ kg

radius r = 1.496 x 10¹¹ m

period t = 3.15 x 10⁷ s

now we know that Earth rotates in a circular path so the distance travelled per rotation is;

d = 2πr we substitute

d = 2π × 1.496 x 10¹¹ m

= 9.4 × 10¹¹ m

Now formula for speed v is;

v = d/t

we substitute

v = 9.4 × 10¹¹ m / 3.15 x 10⁷ s

v = 2.98 × 10⁴ m/s

now we determine the linear momentum p

linear momentum p = mv

we substitute

p = (5.972 x 10²⁴ kg) × (2.98 × 10⁴ m/s)

p = 1.78 × 10²⁹ kg.m/s

Therefore its linear momentum is 1.78 × 10²⁹ kg.m/s

8 0
3 years ago
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