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Hitman42 [59]
3 years ago
6

Anyone good with scientific notations? I sure am not

Physics
2 answers:
defon3 years ago
7 0
B is the answer that I know of.
algol133 years ago
5 0
I'm pretty sure it is b but don't take my word for it
You might be interested in
How many excess electrons are on a ball with a charge of -1.4944 10-16 C?
agasfer [191]

Answer:

There are 934 excess electrons on the ball with charge -1.4944*10^(-16).

Step-by-Step Explanation:

A ball will usually be neutral and have no charge on it.

If it is negatively charged, it means there are excess electrons on it.

If we know the charge on the ball, we can count the excess no. of electrons.

We know that the charge on the ball = c = -1.4944*10^-16

Charge of one electron = e = 1.6*10^-19

Charge of n electrons on the ball = ne = c

⇒ n = c/e

n = (-1.4944*10^-16)/(1.6*10^-19)

n = 934 electrons

7 0
3 years ago
You are handed a 5.00x10^-3kg coin and told that it is gold.
aksik [14]

Answer:

thats cool, gold is pretty

Explanation:

i dont get the question, though

7 0
3 years ago
1) Choose the answer choice that BEST completes the following sentence.
Lena [83]

1. one-Half

2. Apogee

3. Any object that revolves around another object

4. Venus's gravitation pull

5 0
3 years ago
Read 2 more answers
Pls help
belka [17]

Answer:

2156 J

Explanation:

From the question,

Work done = Combined mass of the bucket and water×height×gravity.

W = (M+m)hg............................. Equation 1

Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.

Given: M = 20 kg, m = 2 kg, h = 10 m

Constant: g = 9.8 m/s²

Substitute these  value into equation 1

W = (20+2)×10×9.8

W = 22×98

W = 2156 J

4 0
3 years ago
What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from
ElenaW [278]
When is at the end of the runway the velocity of the plane is given by the equation vf^{2}=0+2*a*s    where s=1800 m is the runway length. Thus
vf^{2}=2*5*1800=18000 (m/s)^{2}      
vf =134.164 (m/s)  

At half runway the velocity of the plane is
v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
 
v= \sqrt{9000}=94.87 ( \frac{m}{s})

Therefore at midpoint of runway the percentage of takeoff velocity is
‰P= \frac{v}{vf}=  \frac{94.87}{134.164}=0.707
6 0
3 years ago
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