All categories

3 years ago

How does an electroscope detect negative static electricity

1 answer:

5 0

<span> An electroscope that might have a static charge is tested on a metal surface. From there, the charges move to the metal and straight to the foil leaves. If they repel, or move away from each other, that means they have identical charges. This applies for both positive and negative static electricity.

work cited: School of champions, Google
</span>

You might be interested in

How is force related to math

dybincka [34]

Answer:

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

Explanation:

8 0

3 years ago

Read 2 more answers

Se coloca una piedra de 600 g en una Honda de 50 cm y se la hace girar a una velocidad de 4 m/s. Dibuja la fuerza que ejerce la

Dima020 [189]

Answer:

Fc = 19.2 N

Explanation:

In this case, the force of the Honda over the rock, is a centripetal force. Then, you have:

$F_c=m\frac{v^2}{r}$

m: mass of the rock = 600g = 0.6 kg

v: tangential velocity of the Honda = 4m/s

r: radius of the Honda = 50cm = 0.5m

You replace the values of m, r and v in the equation for Fc:

$F_c=(0.6kg)\frac{(4m/s)^2}{0.5m}\\\\F_c=19.2N$

hence, the force has a magnitude of 19.2 N

If the rock would have more mass the centripetal force would be higher

4 0

3 years ago

A scientist bred fruit flies in two separate containers with different food sources for many generations. When she put the fruit

Nookie1986 [14]

Fruit flies prefer mates adapted to the same food source.

3 0

4 years ago

Read 2 more answers

How do Earth's and Jupiter's orbits compare?

shutvik [7]

Earth is smaller and have 1 moon so it rotates faster than jupiter and it have 6 moons
Also,earth is the earth is the earth and orbits the sun faster than jupiter the 7th

7 0

3 years ago

Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the

Svet_ta [14]

Answer:

b) $x_{cm}$ = 4.88 cm , c) $x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃) and d)

$x_{cm}$ ’= 1.88 cm

Explanation:

The definition of mass center is

$x_{cm}$ = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

D₂ = d₁ + d₂

D₂ = 1.1 + 1.9

D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

D₃ = D₂ + d₂

D₃ = 3.0 + 3.9

D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

M = m₁ + m₂ + m₃

M = 24 + 12 + 56

M = 92 g

$x_{cm}$ = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

$x_{cm}$ = 1/92 (448.8)

$x_{cm}$ = 4,878 cm

$x_{cm}$ = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

$x_{cm}$ ’= 1/M (m₁ d₁ + m₂ d₂ + m₃ d₃)

$x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

$x_{cm}$ ’= 1/92 ((24 (-1.9) +56 3.9)

$x_{cm}$ ’= 1/92 (172.8)

$x_{cm}$ ’= 1.88 cm

This distance is to the right of the central marble

3 0

3 years ago