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3 years ago

How does charging by conduction occur?

Physics

2 answers:

Aloiza [94]3 years ago

7 0

Charging by conduction involves the contact of a charged object to a neutral object.

C) A charged object touches a neutral object

tia_tia [17]3 years ago

4 0

Answer:

Option (C)

Explanation:

When a charged body is brought in contact to uncharged body, the process of conduction takes place.

For example, there are two identical bodies one having charge 10 C and another is uncharged, when they kept in contact to each other the charges will be distributed.

It means after conduction the charge on both the bodies will be 5 C.

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Define reversible change

inysia [295]

A reversible change is a change that can be undone or reversed

8 0

2 years ago

How much speed is needed for an object to travel 250 km in 1.5 hrs? *<br> Your answer

NemiM [27]

Answer: 166.67km/hr

Explanation:

Given the following :

Distance traveled = 250km

Time taken = 1.5 hours

Recall :

Speed = Distance traveled / time taken

Speed = 250 km / 1.5 hours

Speed = 166. 67 km/hr

Speed in m/s:

166.67km/hr = (166.67 × 1000)m / 3600 s

= 166670m / 3600s

= 46.3m/s

7 0

3 years ago

For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same.

andrey2020 [161]

Answer:

he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

Explanation:

In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values where some natural frequency of the system coincides with the excitation frequency.

In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.

From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

7 0

2 years ago

A race car circles 10 times around a circular 8.0-km track in 20 min.

Otrada [13]

Find the average speed and the average velocity.

Average speed = distance / time

distance = 10 x 8000 m = 80,000 m
time = 20 min * 60 s/min = 1200 s

Average speed = 80,000 m / 1200 s = 66.67 m/s

Average velocity = displacement / time

Given that the race car made complete circles the final poin is the same initial point, then its displacement is zero and the average velocity is zero too.

6 0

3 years ago

Read 2 more answers

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

2 years ago