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erik [133]
3 years ago
15

A ball rolls along the floor with a constant velocity of 6m/s. How far will it go in 10 sec?

Physics
1 answer:
Nady [450]3 years ago
5 0

Answer:

60m

Explanation:

.

X = V × t

X = 6 × 10 = 60m

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The peak of intensity for our Sun on a black body curve would be in which of the following wavelength ranges?0.01 – 10nm
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The answer is,

560- 600nm


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The electric potential is 40 V at point A near a uniformly charged sphere. At point B, 2.0 μm farther away from the sphere, the
KatRina [158]

Answer:

Explanation:

Potential at a point near a charge = Q / 4πε₀ R

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= .16 x 10⁻³ / 2 x 10⁻⁶

= .08 x 10³

= 80 N/C

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80 =  Q / 4πε₀ R x R

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4 0
3 years ago
Select the correct expression that gives the block's acceleration at a displacement x from the equilibrium position. Note that x
Tcecarenko [31]

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the correct is d

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Explanation:

I got it correct

6 0
4 years ago
Ohms law? explanation.​
shusha [124]

Answer:

\boxed{ \bold{ \sf{see \: below}}}

Explanation:

\boxed{ \bold{ \underline{ \huge{  \boxed{ \sf{ohm's \: law}}}}}}

The relation between current through a metallic conductor and potential difference across its ends was studied systematically by a German physicist , <u>George </u><u>Simon </u><u>Ohm </u> in <u>1</u><u>8</u><u>2</u><u>6</u><u> </u><u>AD </u>. This relation is now known as Ohm's law. It states that <u>the </u><u>electric </u><u>current </u><u>passing </u><u>through</u><u> </u><u>a </u><u>conductor </u><u>is </u><u>directly </u><u>proportional</u><u> </u><u>to </u><u>the </u><u>potential </u><u>difference</u><u> </u><u>across </u><u>its </u><u>two </u><u>ends </u><u>at </u><u>a </u><u>constant </u><u>physical </u><u>condition</u><u> </u><u>[</u><u> </u><u>Temperature</u><u> </u><u>,</u><u> </u><u>cross </u><u>-</u><u> </u><u>sectional </u><u>area </u><u>,</u><u> </u><u>length </u><u>,</u><u> </u><u>nature </u><u>of </u><u>material </u><u>etc </u><u>]</u>

Hope I helped!

Best regards!!

4 0
4 years ago
Predict the gravitational force between two 15-kg balls whose centers are 35 m apart.
ArbitrLikvidat [17]

F=G\frac{m_1-m_2}{r^2}

=\frac{(6.67\times10^{-11}N*m/kg^2)(15kg) }{(0.35m)^2}

=1.2\times10^{-7} N

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5 0
2 years ago
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