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Andreas93 [3]
3 years ago
12

What is the charge of a Sulfur atom if it has 17 electrons?

Physics
1 answer:
nadya68 [22]3 years ago
8 0

Answer:

-1

Explanation:

An atom contains a proton, electron and neutron as subatomic particles. The number of protons, which is the positively charged particle (+) is always equal to the number of electrons, which is the negatively charged particle (-) in A NEUTRALLY CHARGED ATOM i.e. no charge.

However, a decrease or increase in charge is dependent on the number of electrons and protons in that atom. Sulfur atom has an electron number of 16 in its neutral atom, however, when it gains one more electron to become 17, it means that the charge has become -1 because the charge = number of protons - number of electrons i.e. 16 - 17 = -1.

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Which body system remove carbon dioxide and waste
antiseptic1488 [7]
The excretory system

7 0
3 years ago
Read 2 more answers
A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra
lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

6 0
3 years ago
Number 3,4, 6, and 8
siniylev [52]

1. When the object is waiting to be released, it is storing a lot of potential energy. When it is released, the potential energy that was once stored is converted into kinetic energy.

3 0
3 years ago
Sean climbs a tower that is 71.3 m high to make a jump with a parachute. The mass of Sean plus the parachute is 81.4 kg. If U =
myrzilka [38]

Answer:

U = 56877.4 J

Explanation:

The potential energy of a body is that which it possesses because it is located at a certain height above the surface of the earth and can be calculated using the following formula:

U = mgh Formula (1)

Where:

U is the potential energy in Joules (J)

m is the mass of the body in kilograms (kg)

g is the acceleration due to gravity (m/s²)

h is the height at which the body is found from the surface of the earth in meters (m)

Data

m= 81.4 kg

g= 9.8 m/s²

h = 71.3 m

Potential energy of Sean and the parachute at the top of the tower

We replace data in the formula (1)

U = m*g*h

U = (81.4 kg)*(9.8 m/s²)*(71.3 m)

U = 56877.4 N*m

U = 56877.4 J

3 0
4 years ago
At what position or positions on the x-axis is the electric field zero?
ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

E=\dfrac{kq}{r^2}

The electric field vector due to charge one

\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})

The electric field vector due to charge second

\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})

We need to calculate the electric field

Using formula of net electric field

\vec{E}=\vec{E_{1}}+\vec{E_{2}}

\vec{E_{1}}+\vec{E_{2}}=0

Put the value into the formula

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})

(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}

\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}

Put the value into the formula

\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}

2.0+x=x

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0
4 years ago
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