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3 years ago

Electrons are (2 points)

2 answers:

8 0

A) able to move from one atom to another; the atom that loses an electron becomes positively charged and the atom that gains an electron becomes negatively charged.

ddd [48]3 years ago

3 0

electrons are able to move from one atom to another; the atom that loses an electron becomes positively charged and the atom that gains an electron becomes negatively charged.

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A newton-meter is a measure of work also known as the _____.

MatroZZZ [7]

1 newton-meter is 1 Joule, the unit of work and energy.

4 0

3 years ago

b. A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg  m2. If the string is pulled with a force

Oduvanchick [21]

Answer:

f = 8 N

Explanation:

Data provided in the question

Radius of the pulley = r = 0.05 m

Moment of inertia = (I) = 0.2 kg.m^{2}

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

Torque is

$= force \times radius$

$= f \times r$

And,

Torque is also

$= moment\ of\ inertia \times angular\ acceleration$

$= I \times \alpha$

So,

We can say that

$f \times r = I \times \alpha$

$f \times 0.05 = 0.2 \times 2$

0.05f = 0.4

f = 8 N

We simply applied the above formulas

8 0

3 years ago

One of your delivery trucks traveled 1,200 miles on 55 gallons of gas. How many miles per gallon did the truck get? (Round off y

Aleksandr-060686 [28]

The word "Per" means divide

"miles per gallon" is the same as "miles / gallon"

The truck went 1,200 miles
on 55 gallons

1,200 ÷ 55 = 21.81

7 0

3 years ago

Read 2 more answers

Describe the Compton Effect

Nana76 [90]

<h3>Compton scattering, discovered by Arthur Holly Compton, is the scattering of a photon after an interaction with a charged particle, usually an electron. If it results in a decrease in energy of the photon, it is called the Compton effect. Part of the energy of the photon is transferred to the recoiling electron.</h3>

7 0

3 years ago

Read 2 more answers

A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =

natali 33 [55]

With acceleration

$\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j$

and initial velocity

$\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i$

the velocity at time t (b) is given by

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

We can get the position at time t (a) by integrating the velocity:

$\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du$

The particle starts at the origin, so $\mathbf x(0)=\mathbf0$ .

$\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du$

$\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}$

$\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j$

Get the coordinates at t = 8.00 s by evaluating $\mathbf x(t)$ at this time:

$\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j$

$\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j$

so the particle is located at (x, y) = (64.0, 64.0).

Get the speed at t = 8.00 s by evaluating $\mathbf v(t)$ at the same time:

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j$

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

This is the velocity at t = 8.00 s. Get the speed by computing the magnitude of this vector:

$\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}$

5 0

3 years ago