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Oksana_A [137]
4 years ago
10

What is the name of the element with 83 protons and 80 electrons?

Physics
1 answer:
dmitriy555 [2]4 years ago
8 0
An element with 83 protons has the atomic number 83. If you look at a periodic table you will see Bismuth is the element with the atomic number of 83. It will have a positive charge of +3 because it has only 80 electrons and 83protons. 83-80=+3. 
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3 years ago
Prečo je omnoho ľahšie udržať sa na hladine vody v mori ako v rieke
Vika [28.1K]

Answer:

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6 0
3 years ago
A _____ magnetic field can create an electric current.
bezimeni [28]

Answer:

D.Strong

Explanation:

4 0
3 years ago
Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and t
BabaBlast [244]

Answer:

-\frac{8\pi}{3}rad/s^2

Explanation:

To solve this problem we need to apply the concept related to Angular Acceleration. We can find it through the equation

\omega_f^2-\omega_i^2=2\alpha\theta

Where for definition,

\omega_i = \frac{\theta}{t}

The number of revolution (\theta)was given by 20 times, then

\omega_i = \frac{20*2pi}{5}

\omega = 8\pi rad/s

We know as well that the salad rotates 6 more times, therefore in angle measurements that is

\theta = 6*2\pi rad = 12\pi rad

The cook at the end stop to spin, then using our first equation,

0-8\pi = 2\alpha (12\pi)

re-arrange to solve\alpha ,

\alpha = \frac{-8\pi}{2*12\pi}

\alpha = -\frac{8\pi}{3}rad/s^2

We can know find the required time,

\omega_f-\omega_i = \alpha t

Re-arrange to find t, and considering that \omega_f=0

t= \frac{\omega_i}{\alpha}

t=\frac{-8\pi}{-8\pi/3}

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6 0
3 years ago
Read 2 more answers
A test charge of -1.4 x 10-7 coulombs experiences a force of 5.4 x 10-1 newtons. Calculate the magnitude of the electric field c
VARVARA [1.3K]

Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F  = Force, q = charge.

From the question,

Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs

Substitute these values into equation 1

E = 5.4×10⁻¹/ -1.4×10⁻⁷

E = -3.86×10⁶ Newtons/coulombs

Hence the magnitude of the electric field created by the

negative test charge is 3.86×10⁶ Newton/coulombs

5 0
3 years ago
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