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dexar [7]
4 years ago
11

The common value for the speed of light in a vacuum is:

Physics
1 answer:
OLEGan [10]4 years ago
5 0

The common value for “Speed of light in vacuum” is \bold{3 \times 10^{8}}  metre per second.

Answer: Option b

<u>Solution: </u>

Speed of light can be defined as the speed with which light waves propagate in different medium. In vacuum, speed of light is 186,282 miles per second or 299,792 km/s which is rounded off as 3 \times 10^{8} \mathrm{m} / \mathrm{s}.

“Speed of light in vacuum” is a universal constant and usually represented by ‘c’. Light waves travels at a speed of 3 \times 10^{8}  metre per second in vacuum.

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A quarter placed on a turntable has a centripetal acceleration of
icang [17]

Answer:

200

Explanation:

Because I know

4 0
3 years ago
a uniform metre rule of mass 100g balances at the 40cm mark when a mass X is placed at the 10cm mark what is the value of x.pls
Blababa [14]

Answer:

The fulcrum of the metre stick is at the 40 cm mark

100 g * 10 cm = 1000 g-cm clockwise torque

x * 30 cm = 1000 gm-cm = counterclockwise torque for balance

X = 1000 / (40 -10) = 1000 / 30 = 33.3 gm  at 10 cm to balance

3 0
3 years ago
Determine the acceleration that results when a 12 N net force is applied to a 3 kg object.
Vadim26 [7]

Heya!!

For calculate aceleration, let's applicate second law of Newton:

\boxed{F=ma}

⇒ Being:

→ F = Force = 12 N

→ m = Mass = 3 kg

→ a = aceleration = ?

Lets replace according formula and leave the "a" alone:

12\ N = 3\ kg * \textbf{a}

\textbf{a} = 12\ N / 3\ kg

\textbf{a} = 4\ m/s^{2}

Result:

The aceleration of the object is of <u>4 m/s²</u>

3 0
3 years ago
Please Help No Links
ale4655 [162]

Answer:

the moment of inertia

Explanation:

5 0
3 years ago
What is the magnitude of the force a 25 charge exerts on a 3 mc charge 35cm away?
lara31 [8.8K]
The electrostatic force between two charges is given by
F=k_e  \frac{q_1 q_2}{r^2}
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the charges

In our problem, 
q_1 = 25 \mu C=25 \cdot 10^{-6} C
q_2 = 3 mC = 3 \cdot 10^{-3} C
r=35 cm=0.35 m
Therefore the electrostatic force is
F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(25 \cdot 10^{-6}C)(3 \cdot 10^{-3}C)}{(0.35 m)^2}= 5510 N
4 0
4 years ago
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