All categories

3 years ago

What is 16.558 m/s rounded to three significant figures?

Physics

1 answer:

In-s [12.5K]3 years ago

3 0

Answer:

Option C. 16.6 m/s

Explanation:

To round this 16.558 m/s to 3sf, we need to count the number beginning from 1. When we get to the 3rd number( ie 5), we'll examine the fourth number(i.e 5)to see if it less than five or greater. If it less than five, then we'll discard it. But if it five or greater, we'll approximate it and add it to the 3rd number.

So.

16.558 m/s = 16.6m/s to 3sf

You might be interested in

When there are more coils added to the wire of an electromagnet what happens to

fiasKO [112]

In technical terms, every coil of wire increases the "magnetic flux density" (strength) of your magnet.

So it's A (magnetic field increase)

5 0

3 years ago

Strontium has density of 2.64 g/cm3 and crystallizes with the face-centered cubic unit cell. Calculate the radius of a strontium

pantera1 [17]

Answer:

The radius of strontium atom is

2.14 E-8cm

Explanation:

Go through the attached file for a comprehensive detailed explanation.

6 0

3 years ago

What force is required to accelerate a 385 kg couch at 0.2 m/s^2 ?

Juliette [100K]

Answer:

It takes 77 N

Explanation:

Using Newton's second law of motion, F=ma (Force equals mass times acceleration. Since the mass of the couch is 385 kg and the target acceleration is 0.2 m/s, you simply multiply mass times acceleration (ma) to get the total force, or 77 N.

6 0

2 years ago

In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere

muminat

Answer:

$2.1\cdot 10^{21}$ electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

$E=\frac{kQ}{r^2}$

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a charged sphere, so we have:

$E_b=3.00\cdot 10^6 N/C$ is the maximum electric field strength tolerated by the air before breakdown occurs

$r=1.00 km = 1000 m$ is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

$Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C$

Assuming that the cloud is negatively charged, then

$Q=-333.3 C$

And since the charge of one electron is

$e=-1.6\cdot 10^{-19}C$

The number of excess electrons on the cloud is

$N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}$

5 0

3 years ago

During a clinic visit a4 year old girl suddenly yells don't sit on erin the parents wispers that erin is an imaginary friend wha

igor_vitrenko [27]

Try and calm the child and say that you won't sit on Erin. If you tell the young girl that Erin isn't real at that age, they may have social issues later.

6 0

3 years ago