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Katena32 [7]
3 years ago
11

Please help need answers asap ​

Physics
1 answer:
Soloha48 [4]3 years ago
4 0
Answer- Helium
You can by the number of protons, and if you look at a periodic table the atomic number of helium is the same as the number of protons
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Scientific notation of 86400
Mandarinka [93]
8.64×10^4

this is 86400 in scientific notation

3 0
3 years ago
Read 2 more answers
3. A fighter jet covers a distance of 895m while accelerating from 22m/s to 35m/s. How long (in
larisa86 [58]

Answer:

31.404 seconds

Explanation:

To answer this equation, SUVAT is your best option utilizing and rearranging the known values to solve for the unknown.

here we have the values for

s=895

u=22

v=35

t= the unknown value

in this instant the equation s=0.5 x (u+v)t is the best equation to use

so we sub in the known values

895=0.5 x (22+35)t

rearrange to solve for t

895=28.5t

895/28.5=t

t=31.404 seconds (rounded to 3 decimal places)

6 0
2 years ago
What is the ability to react with air
Bond [772]
Reactivity is the answer
3 0
3 years ago
Read 2 more answers
A phone cord is 4.89 m long. The cord has a mass of 0.212 kg. A transverse wave pulse is produced by plucking one end of the tau
slega [8]

Answer:

Tension in Cord=174 N

Explanation:

Given Data

L (Phone Cord Length)=4.89 m

m (Cord Mass)=0.212 Kg

T (Time for four trips)=0.617 s

Tension=?

Solution

V=λ×f

V=\frac{8*4.89}{0.617}\\ V=63.4m/s

Sigma=\frac{mass}{length}\\ Sigma=\frac{0.212}{4.89}\\ Sigma=0.0433 \frac{kg}{m}

Wave Speed=\sqrt{\frac{Tension}{Sigma} }\\ \\V=\sqrt{\frac{T}{Sigma} }\\ V^{2}=\frac{T}{Sigma}\\  T=V^{2}*Sigma\\ T=(63.4)^{2}*(0.0433)\\ T=174 N

5 0
3 years ago
Situation 6.1 A 13.5-kg box slides over a rough patch 1.75 m long on a horizontal floor. Just before entering the rough patch, t
GenaCL600 [577]
B) 14.0 N    

The way to solve this problem is to determine the kinetic energy the box had before and after the rough patch of floor. The equation for kinetic energy is: 

 E = 0.5 M V^2 

 where 

 E = Energy 

 M = Mass 

 V = velocity   

 Substituting the known values, let's calculate the before and after energy. 

 Before: 

 E = 0.5 M V^2 

 E = 0.5 13.5kg (2.25 m/s)^2 

 E = 6.75 kg 5.0625 m^2/s^2 

 E = 34.17188 kg*m^2/s^2 = 34.17188 joules   

 After: 

 E = 0.5 M V^2 

 E = 0.5 13.5kg (1.2 m/s)^2 

 E = 6.75 kg 1.44 m^2/s^2 

 E = 9.72 kg*m^2/s^2 = 9.72 Joules   

 So the box lost 34.17188 J - 9.72 J = 24.451875 J of energy over a distance of 1.75 meters. Let's calculate the loss per meter by dividing the loss by the distance.   

 24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N   

 Rounding to 1 decimal place gives 14.0 N which matches option "B".
8 0
2 years ago
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