Activity: Lab safety and Equipment Puzzle

A current of 5.0 a flows through an electrical device for 10 seconds. how many electrons flow through this device during this ti

melisa1 [442]

1 Amp = 1 Coulomb/sec
1 Coulomb/sec = 6.25*10^18 electrons/sec

Therefore,
5.0 A = 5 C/s = 5*6.25*10^18 = 3.125*10^19 e/s

In 10 second, number of electrons are calculated as;
Number of electrons through the device = 3.125*10^19*10 = 3.125*10^20 electrons

4 0

2 years ago

In some cases fixture wires may be used for

zalisa [80]

You can use fixture wires: For installation in luminaires where they are enclosed and protected and not subject to bending and twisting and also can be used to connect luminaires to their branch circuit conductors.

<h3>What are some uses of fixture wires?</h3>

Fixture wires are flexible conductors that are used for wiring fixtures and control circuits. There are some special uses and requirements for fixture wires and no fixture can be smaller than 18 AWG

In modern fixtures, neutral wire is white and the hot wire is red or black. In some types of fixtures, both wires will be of the same color.

To know more about fixture wires, refer

brainly.com/question/26098282

#SPJ4

3 0

8 months ago

A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is

goblinko [34]

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, $v_i$ = 1.4 m/s

final velocity of the child, $v_f$ = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

total mass, m = 48 kg

frictional force, $f_k$ = 41 N

The work done by the child is calculated as;

$\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech} + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96 \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J$

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

5 0

2 years ago

In a certain two slit diffraction experiment, two slits 0.02mm wide are spaced0.2mm between centers.(a) How many fringes appear

Kamila [148]

Answer:

a) m = 10 and b) λ = 3.119 10⁻⁷ m

Explanation:

In the diffraction experiments the maximums appear due to the interference phenomenon modulated by the envelope of the diffraction phenomenon, for which to find the number of lines within the maximum diffraction center we must relate the equations of the two phenomena.

Interference equation d sin θ = m λ

Diffraction equation a sin θ = n λ

Where d is the width between slits (d = 0.2 mm), a is the width of each slit (a = 0.02 mm). θ is the angle, λ the wavelength, m and n are an integer.

Let's find the relationship of these two equations

d sin θ / a sin θ = m Lam / n Lam

The first maximum diffraction (envelope) occurs for n = 1, let's simplify

d / a = m

Let's calculate

m = 0.2 / 0.02

m = 10

This means that 10 interference lines appear within the first maximum diffraction.

b) let's use the interference equation, remember that the angles must be given in radians

θ = 0.17 ° (π rad / 180 °) = 2.97 10⁻³ rad

d sin θ = m λ

λ = d sin θ / m

λ = 0.2 10⁻³ sin (2.97 10⁻³) / 2

λ = 3.119 10⁻⁷ m

8 0

2 years ago

Anyone want to be Friends ?

In-s [12.5K]

i would like a friend

3 0

2 years ago

Read 2 more answers