2 years ago

What is the strength and direction of the electric field 3.560 cm?

1 answer:

8 0

Just treat the bead as a point charge.
E = kq/r^2, and E points away from the center of the sphere since the charge is positive.

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A box that has a mass of 80 kg slides down a ramp with a 30 degree angle. The free-body diagram shows the forces acting on the b

GaryK [48]

Answer:

4.9

Explanation:

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2 years ago

The airport has a new plane that can go from 0 miles per hour to 100 miles per hour in 10 seconds. What is being described about

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The answer is acceleration

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2 years ago

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The planet Krypton has a mass of 7.6 × 1023 kg and radius of 1.7 × 106 m. What is the acceleration of an object in free fall nea

olga55 [171]

Answer:

17.55 m/s²

Explanation:

Parameters given:

Mass of Krypton, M = 7.6 * 10^23 kg

Radius, R = 1.7 * 10^6 m

Gravitational constant, G = 6.6726 * 10^(-11) Nm²/kg²

Acceleration due to gravity of planet of mass M is given as:

g = GM/R²

Since the object is close to the surface of Krypton, we can say that the distance from the Centre of Krypton is the radius of the planet Krypton.

Therefore,

g = (6.6726 * 10^(-11) * 7.6 * 10^23)/(1.7 * 10^6)²

g = 17.55 m/s²

5 0

3 years ago

What is the force of gravity between two 50.0kg masses that are separated by 0.300m?3.71x10-8N5.59x10-7N2.78x104N1.85x10-6N

Varvara68 [4.7K]

We will have the following:

$\begin{gathered} F=G\frac{m_1m_2}{r^2}\Rightarrow F=\frac{(6.67\ast10^{-11}m^3\ast kg^{-1}\ast s^{-2})(50kg)(50kg)}{(0.3m)^2} \\ \\ \Rightarrow F=1.852777778...\ast10^{-6}N\Rightarrow F\approx1.85\ast10^{-6}N \end{gathered}$

So, the force is approximately 1.85*10^-6 N.

7 0

9 months ago

2 Points

yulyashka [42]

The force applied to lift the crate is 171 N

Explanation:

The lever works on the principle of equilibrium of moments, so we can write:

$F_i d_i = F_o d_o$

where

$F_i$ is the force in input

$d_i$ is the arm of the input force

$F_o$ is the output force

$d_o$ is the arm of the output force

For the lever in this problem, we have:

$d_i = 0.25 m$

$d_o = 0.19 m$

$F_i = 130 N$ (force applied)

Solving the equation for $F_o$ , we find the force applied to lift the crate:

$F_o = \frac{F_i d_i}{d_o}=\frac{(130)(0.25)}{0.19}=171 N$

Learn more about levers:

brainly.com/question/5352966

#LearnwithBrainly

6 0

3 years ago