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4 years ago

In a 2-dimensional coordinate system, the x- and y-axes are typically _____.

Physics

1 answer:

dem82 [27]4 years ago

7 0

<span>In a 2-dimensional coordinate system, the x- and y-axes
are typically perpendicular to each other. (C) </span>

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What forces must be balanced in the Bohr model?

Solnce55 [7]

The correct answer is:
<span>A. electron attraction between electron and nucleus = centripetal force of the orbiting electron

In fact, Bohr model depicts the atom as a nucleus surrounded by electrons in circular orbit around it, similar to the planets around the Sun. The centripetal force that keeps the electrons in circular motion around the nucles is provided by the electrostatic force between the electrons and the nucleus, so option A is the correct one.</span>

4 0

3 years ago

Read 2 more answers

Q|C A string with linear density 0.500 g/m is held under tension 20.0 N. As a transverse sinusoidal wave propagates on the strin

julia-pushkina [17]

A string with linear density 0.500 g/m.

Tension 20.0 N.

The maximum speed $v_{y, max}$

The energy contained in a section of string 3.00 m long as a function of $v_{y, max}$ .

We are given following data for string with linear density held under tension :

μ = 0.5 $\frac{g}{m}$

= 0.5 x 10⁻³ $\frac{kg}{m}$

T = 20 N

If string is L = 3m long, total energy as a function of $v_{y, max}$ is given by:

E = 1/2 x μ x L x ω² x A²

= 1/2 x μ x L x $v^{2} _{y, max}$

= 7.5 x 10⁻⁴ $v^{2} _{y, max}$

So, The total energy as a function of $v^{2} _{y, max}$ = 7.5 x 10⁻⁴ $v^{2} _{y, max}$

Learn more about linear density problem here:

brainly.com/question/17190616

#SPJ4

3 0

2 years ago

Up-regulation involves the loss of receptors and prevents the target cells from overreacting to persistently high hormone levels

kotegsom [21]

Answer:

False.

Explanation:

Up regulation involves the increase of receptors and it makes the cells more sensitive to the hormones.It's the down regulation that involves the loss of receptors and prevents the target cells from overreacting to persistently high hormone levels.

5 0

3 years ago

A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the c

icang [17]

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

$W = F x = \mu N x = \mu\ m\ g\ x$

$W = 150 \times 6$

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

$\mu = \frac {F}{m\timesg}$

$= \frac{150}{40\times 9.8}$

= .383

We simply applied the above formulas so that each one part could calculate

3 0

4 years ago

Determine the average acceleration for x(t)=19t^2+7t^3 for a time interval between 3 and 9 seconds

AleksandrR [38]

Answer:

290

Explanation:

Average acceleration is the change in velocity over change in time.

First, find the velocity by taking the derivative of position.

v(t) = dx/dt

v(t) = 38t + 21t²

At t = 3 and t = 9:

v(3) = 303

v(9) = 2043

So the average acceleration is:

a = Δv / Δt

a = (2043 − 303) / (9 − 3)

a = 290

Use appropriate units.

7 0

3 years ago