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AleksandrR [38]
3 years ago
13

How many kindergarteners do you think you can take on a fight before getting tired or over powered?

Physics
2 answers:
NikAS [45]3 years ago
8 0

Answer:

ofc 10000000000000

Explanation:

just smart like that

yulyashka [42]3 years ago
4 0
Is this a serious question ?
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How to read a vernier calipers?
RideAnS [48]

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.

7 0
3 years ago
Read 2 more answers
A student measured the specific heat of water to be 4.29 J/g.Co. The
leonid [27]

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 :

P=\dfrac{|\text{original value-calculated value}|}{\text{original value}}\times 100\\\\P=\dfrac{4.29-4.18}{4.18}\times 100\\\\P=2.63\%

So, the student's percent error is 2.63 %.

7 0
3 years ago
Which would most likely cause the cylinder head temperature and engine oil temperature gauges to exceed their normal operating r
dusya [7]

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.

3 0
3 years ago
Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c
Mariulka [41]

Answer:

A. v_{3}=12.17m/s

B. v_{car}=6.3m/s\\v_{truck}=-6.3m/s

C. ΔK=-4.13x10^3J

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

v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg

A. Since the two vehicles become entangled the final mass is:

m_{3}=102kg+103kg=205kg

From linear momentum we got that:

p_{1}=p_{2}

m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}

v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}

v_{3}=12.17m/s

B. The change in velocity of both vehicles are:

For the car

v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s

For the truck

v_{truck}=12.17m/s-18.5m/s=-6.3m/s

C. The change in kinetic energy is:

ΔK=K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})

ΔK=\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J

ΔK=-4.13x10^{3}J

6 0
3 years ago
A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060 eV as the particl
Sati [7]

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

6 0
3 years ago
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