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3 years ago

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What is the Coulomb's Law?

1 answer:

bezimeni [28]3 years ago

6 0

Coulomb's law states that: The magnitude of the electrostatic force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distance between them. The force is along the straight line joining them.

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A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10

sashaice [31]

Answer:

Time interval;Δt ≈ 37 seconds

Explanation:

We are given;

Angular deceleration;α = -1.6 rad/s²

Initial angular velocity;ω_i = 59 rad/s

Final angular velocity;ω_f = 0 rad/s

Now, the formula to calculate the acceleration would be gotten from;

α = Change in angular velocity/time interval

Thus; α = Δω/Δt = (ω_f - ω_i)/Δt

So, α = (ω_f - ω_i)/Δt

Making Δt the subject, we have;

Δt = (ω_f - ω_i)/α

Plugging in the relevant values to obtain;

Δt = (0 - 59)/(-1.6)

Δt = -59/-1.6

Δt = 36.875 seconds ≈ 37 seconds

6 0

4 years ago

1 point<br> You throw a ball up in the air with a velocity of 30 m/s. How high does it<br> go?

choli [55]

Answer:

Explanation:

2as=vf^2-Vi^2

vf=30 m/s

vi= 0 m/s

a=g=9.8 m/s^2

s=vf^2-Vi^2/2a

s=(30)²-(0)²/2*9.8

s=900-0/19.6

s=45.9=46 m

4 0

3 years ago

Remember KE=M*V2/2…  A model airplane moves twice as fast as another identical model airplane. Compared with the kinetic energy

pishuonlain [190]

Answer: 4 times as much

Explanation:

4 0

2 years ago

I need help with this

lisabon 2012 [21]

A. Condensation

Hope this helps!!!

5 0

2 years ago

A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800

Arada [10]

Explanation:

The given data is as follows.

m = 5000 kg, h = 800 km = $0.8 \times 10^{6} m$

$R_{e} = 6.37 \times 10^{6} m$ , r = R + h = $7.17 \times 10^{6} m$

$M_{e} = 5.98 \times 10^{24}$ kg, G = $6.67 \times 10^{-11} Nm^{2}/kg^{2}$

As we know that,

$\frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}$

v = $\sqrt{\frac{GM_{e}}{r^{2}}}$

And, it is known that formula to calculate angular velocity is as follows.

$\omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}$

v = $\sqrt{\frac{GM_{e}}{r^{3}}}$

= $\sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}$

= $1.0402 \times 10^{-3} rad/s$

Thus, we can conclude that speed of the satellite is $1.0402 \times 10^{-3} rad/s$ .

6 0

3 years ago