Answer:
I'll write it below
Explanation:
1) understand the parts.
2)read the scales
3)check the scale of your smallest divisions
4)clean the object you are measuring
5)If you have, unlock the screw
6)close the jaws
I hope this satisfies you sir.
If you have any questions related to this please feel free to ask me. I hope u will follow me and make this the brainliest answer.
Answer:
2.63 %.
Explanation:
Given that,
The calculated value of the specific heat of water is 4.29 J/g.C
Original value of specific heat of water is 4.18 J/g.C.
We need to find the student's percent error. The percentage error in any quantity is given by :
So, the student's percent error is 2.63 %.
Answer: Using fuel that has a lower-than-specified fuel rating.
Explanation:
Most likely, what causes the cylinder head temperature and engine oil temperature gauges to exceed their normal operating ranges is using fuel that has a lower-than-specified fuel rating. This can lead to detonation of the engine which is the tendency for the fuel to pre-ignite or auto-ignite in an engine's combustion chamber.The cylinder head and the engine oil are part of the automobile systems that helps in fuel combustion.
Answer:
A. 
B. 
C. ΔK
Explanation:
From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2
A. Since the two vehicles become entangled the final mass is:
From linear momentum we got that:
B. The change in velocity of both vehicles are:
For the car
For the truck
C. The change in kinetic energy is:
ΔK=
ΔK=
ΔK
Given Information:
KEa = 9520 eV
KEb = 7060 eV
Electric potential = Va = -55 V
Electric potential = Vb = +27 V
Required Information:
Charge of the particle = q = ?
Answer:
Charge of the particle = +4.8x10⁻¹⁸ C
Explanation:
From the law of conservation of energy, we have
ΔKE = -qΔV
KEb - KEa = -q(Vb - Va)
-q = KEb - KEa/Vb - Va
-q = 7060 - 9520/27 - (-55)
-q = 7060 - 9520/27 + 55
-q = -2460/82
minus sign cancels out
q = 2460/82
Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹
q = 2460(1.60x10⁻¹⁹)/82
q = +4.8x10⁻¹⁸ C
