What is the guage pressure of an object with a mass of 30 kg and 5ft height?

The charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electro

shusha [124]

Answer:

$1.6\times 10^{-18} C$

Explanation:

The fundamental charge of a single electron is $1.6\times 10^{-19} C$ .

If there are 10 excess electrons, the net charge that would be measured should be 10 times the fundamental charge of a single electron:

$Q=nq_e\\Q= 10\times 1.6\times 10^{-19} C\\Q= 1.6 \times 10^{-18} C$

3 0

3 years ago

Please answer ASAP for brainliest

OlgaM077 [116]

Answer:

Question #1- Scientists agree to a standard way of reporting measured quantities in which the number of reported digits reflects the precision in the measurement- more digits, more precision; less digits, less precision. You just studied 14 terms!

Question #2- Units are important because without proper measurement and units to express them, we can never express physical laws precisely just from qualitative reasoning. Units are incredibly important to physics. Two of the most important reasons are the following: (1) they help us. to avoid making mistakes in computation, and (2) they serve as a check on computations once they are completed. In the first case, you can avoid adding 3m and 25cm and coming up with the wrong answer.

Explanation: Hope this helps please mark brainliest!

4 0

2 years ago

Read 2 more answers

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$