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3 years ago

Is electron positive

Physics

2 answers:

castortr0y [4]3 years ago

8 0

No the electron is a negative.

The proton is the positive.

Hope this helps!

Elena L [17]3 years ago

7 0

No it is negative a proton is postive

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The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

Circular Motion

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

3 years ago

What are the three subatomic particles of an atom?

Anestetic [448]

Protons, neutrons, and electrons are the three main subatomic particles found in an atom. Protons have apositive (+) charge. An easy way to remember this is to remember that both proton and positive start with theletter "P." Neutrons have no electrical charge.

5 0

3 years ago

Read 2 more answers

Three forces of magnitude 10N, 5N and 4N act on an object in the directions North, West and East respectively. Find the magnitud

Gekata [30.6K]

Answer:

19N to the south

Explanation:

F =10N + 5N + 4N

8 0

3 years ago

A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee

Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

$F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2} = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\$

where,

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Me = Mass of Earth = 5.97 x 10²⁴ kg

r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m

v = orbital speed = ?

Therefore,

$v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\$

v = 7.69 x 10³ m/s

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

$v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v = \frac{2\pi r}{T}\\\\$

where,

T = Time Period of Satellite = ?

Therefore,

$T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\$

T = 5500 s = 91.67 min = 1.53 h

7 0

3 years ago

The acceleration due to gravity, g , is constant at sea level on the Earth's surface. However, the acceleration decreases as an

blsea [12.9K]

Answer:

g = g₀ [1- 2 h / Re + 3 (h / Re)²]

Explanation:

The law of universal gravitation is

F = G m Me / Re²

Where g is the universal gravitational constant, m and Me are the mass of the body and the Earth, respectively and R is the distance between them

F = G Me /Re² m

We call gravity acceleration a

g₀ = G Me / Re².

When the body is at a height h above the surface the distance is

R = Re + h

Therefore the attractive force is

F = G Me m / (Re + h)²

Let's take Re's common factor

F = G Me / Re² m / (1+ h / Re)²

As Re has a value of 6.37 10⁶ m and the height of the body in general is less than 10⁴ m, the h / Re term is very small, so we can perform a series expansion

(1+ h / Re)⁻² = 1 -2 h / Re + 6/2 (h / Re) 2 + ...

Let's replace

F = G Me /Re² m [1- 2 h / Re + 3 (h / Re)²]

F = g₀ m [1- 2 h / Re + 3 (h / Re)²]

If we call the force of attraction at height

m g =g₀ m [1- 2 h / Re + 3 (h / Re)²]

g = g₀ [1- 2 h / Re + 3 (h / Re)²]

3 0

3 years ago