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photoshop1234 [79]

3 years ago

6

According to Coulomb’s Law, the force between two charged objects is related to _____.

1 answer:

Natasha_Volkova [10]3 years ago

5 0

Answer:

A.) the inverse of the square of the distance separating them

Explanation:

Coulombs law states that "the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them."

Mathematically, F = kq1q2/r²

Where q1 and q2 are the charges

r is the distance between the charges.

According to the law, the force between two charged objects is related to the inverse of the square of the distance separating them.

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Describe what a convection current is in how it was material than the earth

Leona [35]

The movement of fluid as a result of differential heating or convection. Earth convection currents refer to the motion of molten rock in the mantle as radioactive decay heats up magma, causing it to rise and driving the global scale flow of magma.

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2 years ago

What is the de Broglie wavelength for a proton with energy 50 keV? Due to the limitations of Canvas, please give the wavelength

ad-work [718]

Answer:

1.2826 x 10^-13 m

Explanation:

$\lambda = \frac{h}{\sqrt{2 m K}}$

Here, k be the kinetic energy and m be the mass

K = 50 KeV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J

m = 1.67 x 10^-27 kg

$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times 80\times 10^{-16}}}$

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2 years ago

How much does a 60 kg table weigh

soldier1979 [14.2K]

Answer:

588 N

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2 years ago

What makes a good scientific question? (1 point)

liraira [26]

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7 0

2 years ago

A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch

Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

$x=V_{o}cos \theta t$ ----------------------(1)

$y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}$ ---------(2)

$V=V_{o}-gt$ ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

$V_{o}$ is the golf ball's initial velocity

$\theta=0\°$ is the angle (it was a horizontal shot)

t is the time

y is the final height of the ball

$y_{o}$ is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

$t=\sqrt{\frac{2 y_{o}}{g}}$

$t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}$

$t=1.597$ s

Substituting (6) in (1):

$67.1 =V_{o} cos(0\°) 1.597$ -------------------(4)

Step 2: Finding $V_{o}$ :

From equation(4)

$67.1 =V_{o}(1) 1.597$

$V_0 = \frac{6.71}{1.597}$

$V_{o}=42.01$ m/s (8)

Substituting $V_{o}$ in (3):

$V=42.01 -(9.8)(1.597)$

v =42 .01 - 15.3566

V=26.359 m/s

5 0

2 years ago