The symbol used for magnetic field is:

An electrostatic paint sprayer has a 0.200-m-diameter metal sphere at a potential of 25.0 kV that repels paint droplets onto a g

NeX [460]

The solution is in the attachment

3 0

3 years ago

What stands in for a "creator" in scientists view of the universe?

inessss [21]

In scientists view of the universe The creator is replace by: (B). The "Big Bang"

<h3>Meaning of Big Bang</h3>

The Big bang is a theory propagated by scientist and astronomers to replace the creator.

The Big Bang is said to be a reason for the universe and all the planets in it. Scientist say the universe started as just a single point with very high temperature and density, and then here comes the big bang.

In conclusion, In scientists view of the universe The creator is replace by: The "Big Bang"

Learn more about The Big Bang: brainly.com/question/10865002

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5 0

2 years ago

You are designing a system for moving aluminum cylinders from the ground to a loading dock. You use a sturdy wooden ramp that is

8_murik_8 [283]

Answer:

a = αR

Then We apply the Newton's second law of motion:

∑ F = ma

F - mg sinθ - f = ma.

F - mg sinθ - μmg cosθ = ma

Using given data we have:

F - 3217 = 470a ........................... (1)

Apply the Newton's law for rotation:

∑τ = I α

FR - (mg sinθ)R = $(\frac{mR^2}{2}+mR^2)\frac{a}{R}$

then: F - mg sinθ = 3ma / 2.

F - 2774 = 705a .......................... (2)

solving 1 and 2 we get:

F = 4103 N

a = 1.885 m/s²

b) In this case the time is given by:

t = $\sqrt{\frac{2d}{a}}$ it comes from: d = $v_ot+\frac{1}{2}at^2$ starting from rest vo = 0

t = $\sqrt{\frac{2*9}{1.885}$ = 3.09 s

8 0

3 years ago

A proton enters a uniform magnetic field of strength 2 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’

sattari [20]

Answer:

Magnetic force, $F=9.6\times 10^{-17}\ N$

Explanation:

It is given that,

Magnetic field, B = 2 T

Velocity of the proton, v = 300 m/s

Charge on the proton, $q=1.6\times 10^{-19}\ C$

The magnetic field is oriented perpendicular to the proton’s velocity. The magnetic force on the charged particle is given by :

$F=qvB\ sin\theta$

The magnetic field is oriented perpendicular to the proton’s velocity, $\theta=90^{\circ}$

$F=1.6\times 10^{-19}\times 300\times 2$

$F=9.6\times 10^{-17}\ N$

So, the magnitude of the force that the proton experiences while it moves through the magnetic field is $9.6\times 10^{-17}\ N$ . Hence, this is the required solution.