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Mice21 [21]
2 years ago
6

Which of the following is true of our current knowledge of electrons?

Physics
1 answer:
frosja888 [35]2 years ago
8 0
They have a negative charge and rotate around the nucleus
You might be interested in
A particle moving along the y-axis has the potential energy u =4y3j, where y is in m. what is the y-component of the force on th
Ivenika [448]

Potential energy is given as

U = 4y^3

now as we know that force is related by potential energy by the formula

F = - \frac{dU}{dy}

So it is gradient of energy with position in Y

F = - \fracd(4y^3}{dy}

F = -12y^2 \hat j

Now at y = 0

F = 0 N

at y = 1

F = - 12*1^2

F = - 12 N

at y = 2

F = - 12*2^2

F = - 48 N

so above is the forces at given positions

7 0
2 years ago
is the science that manages the database, digital tools, and software used by both medicine and forensic science to store and an
Ne4ueva [31]

Explanation :

Bioinformatics is the science that manages the database, digital tools and software used by both medicine and forensic science to store and analyze DNA.

It involves the use of computers to collect all the data and organize data. It also develops methods and computational tools for understanding biological data.

It helps in tracing the evolution of organism using DNA and builds a computational model.

8 0
3 years ago
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

5 0
3 years ago
Technician A says that current is created in the stator windings of an alternator. Technician B says that the voltage regulator
drek231 [11]

Answer:

Technician A and Technician B both are right.

Explanation:

In an AC alternator,  there are two windings

1. Stator winding (stationary)

2. Rotor winding (rotating)

The current is induced in the stationary coils due to the magnetic field produced by the rotor. The DC suppy is provided to the rotor winding via slip rings and brushes and a voltage regulator precisely controls this supply to control the current flow through the rotor.

Therefore, both technicians are right.

4 0
2 years ago
During each cycle, a refrigerator ejects 610 kJ of energy to a high-temperature reservoir, and takes in 505 kJ of energy from a
jenyasd209 [6]

Answer

A. the work done on the refrigerant in each cycle is 105kJ

B the coefficient of performance of the refrigerator is 4.8

Explanation

Given data

Work done at high temperature T2 Qh=610kJ

Work done at low temperature T1 Ql=505kJ

We know that the net work done by the refrigerator is expressed as

Wnet= Qh-Ql

=610-505

=105kJ

Also we know that the coefficient of performance is expressed as

COP= Ql/Wnet

COP= 505/105

= 4.8

8 0
2 years ago
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