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Marrrta [24]
3 years ago
8

11. A 3.8 kg object is lifted 12 meters. Approximately how much work is performed during the lifting?

Physics
1 answer:
Marianna [84]3 years ago
3 0
I found the answer for you if u need any help ask anytime!

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A student measures the pH of a solution to be 6.8. Which should the student add if she wants to decrease the pH of the solution?
zloy xaker [14]
The neutral pH is 7. Less than 7 indicates an acid and more than 7 indicates a base (up to 14).
<span> NaCl - it's a salt (we can't measure the pH)
H2O - it can be an acid but also a base  (the pH it is almost neutral,meaning close to 7 )
HF - it is a strong acid
</span><span> KOH  - it is a strong base (pH=14)
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He needs to use HF (Hydrogen fluoride) to decrease the pH.


7 0
3 years ago
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You leave the doctor's office after your annual checkup and recall that you weighed 688 N in her office. You then get into an el
Lapatulllka [165]

Answer:

a=0.5418\ m.s^{-2} upwards

a=1.283\ m.s^{-2} downwards

Explanation:

Given:

weight of the person, w=688\ N

So, the mass of the person:

m=\frac{w}{g}

m=\frac{688}{9.81}

m=70.132\ kg

  • Now if the apparent weight in the elevator, w_a= 726\ N

<u>Then the difference between the two weights is :</u>

\Delta w=w_a-w

\Delta w=726-688

\Delta w=38\ N is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

a=\frac{\Delta w}{m}

a=\frac{38}{70.132}

a=0.5418\ m.s^{-2} upwards, because the normal reaction that due to the weight of the body is increased here.

  • Now if the apparent weight in the elevator, w_a= 598\ N

<u>Then the difference between the two weights is :</u>

\Delta w=w-w_a

\Delta w=688-598

\Delta w=90\ N is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

a=\frac{\Delta w}{m}

a=\frac{90}{70.132}

a=1.283\ m.s^{-2} downwards, because the normal reaction that due to the weight of the body is decreased here.

7 0
3 years ago
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1.<br>Force that pulls toward the center of mass.<br><br>​
kondaur [170]
Gravity if I’m not mistaken
7 0
3 years ago
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A car starts moving from the position of rest with uniform acceleration of 8m/s^calculate the distance travelled by it during 10
ad-work [718]

We calculate the coordinates at t₁ = 9 min and t₂ = 10 min, since the 10th minute is between t₁ and t₂.

As it leaves from rest, it means that the initial speed is zero


t₁=9 min=540 s

t₂=10 min=600 s

x₁=at₁²/2=8*540²/2=4*291600=1166400 m

x₂=at₂²/2=8*600²/2=4*360000=1440000 m

Δx=x₂-x₁=1440000-1166400=273600 m represents the distance traveled by the car in the 10th minute of travel




8 0
3 years ago
If a car traveling at 60 km/h will skid 20m when its brakes lock, how far will it skid if it's traveling at 120 km/h when it's b
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