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Lunna [17]
2 years ago
15

What type of force slows you down as you slide across the floor?

Physics
2 answers:
Juli2301 [7.4K]2 years ago
6 0
Answer:
B-Friction

How?
Friction is when two surfaces rub against each other. This creates heat and resistance. It would only make sense that this force would slow you down will sliding across floors
timurjin [86]2 years ago
3 0

Friction can slows you down as you slide across the floor.

Answer: Option B

<u>Explanation:</u>

Every action has an equal and opposite reaction. So when we fall or slide upon a floor, just like resistance opposes the current flowing in the circuit, there is an external force that acts upon when we fall or slide away.

This external force is called as friction.  Friction opposes the object making the object slowly decrease its speed and lose energy. Gravity is the force that makes all the objects to return to the earth or ground when thrown upon.

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MA_775_DIABLO [31]

Answer:

Kinetic energy and potential energy.

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Give one example of something you use or make at home that is an example of solubility.
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The electric potential at the dot in the figure is 3160 V. What is charge q?
PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

V = \frac{kQ}{r}

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V

Lower left charge's potential:

V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V

Add the two, and subtract from the total EP at the point:

3160 + 1123.75 - 2247.5 = 2036.25


The remaining charge must have a potential of 2036.25 V, so:

2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}


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2 years ago
In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted?
agasfer [191]

Answer:

The frequency of the photon is 3.069\times10^{14}\ Hz.

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Given that,

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Where, \Delta E =  energy spacing

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\nu=\dfrac{4.07\times10^{-19}}{2h}

Put the value of h into the formula

\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}

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Hence, The frequency of the photon is 3.069\times10^{14}\ Hz.

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