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Nostrana [21]
3 years ago
14

Mr. Nakamora wants to test the accuracy of an old balance in his classroom compared to a brand-new balance that he knows is accu

rate. He weighs an object on the old balance and finds it to have a mass of 15.6 grams. The same object weighs 15.9 grams on the brand-new balance
What is the percent error between the old and new balance?
Physics
2 answers:
Juli2301 [7.4K]3 years ago
8 0

the answer is d                     ..........................................

faust18 [17]3 years ago
7 0
It 5.009084564 and you probabably are correct

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Ex 10: My dog runs at 6 m/s for 18 meters. How long did she run for?
Hunter-Best [27]
She ran for 3s

Put 18/6 because in order to find how long she ran for you need to divide the distance by the meters ran, once you do that you will get 3.
7 0
3 years ago
Describe what happens when electrical impulses cross muscle fibers
kaheart [24]

Answer:

Sometimes may cause involuntary responses like twitching

Explanation:

8 0
3 years ago
What is the greatest distance an image can be located behind a convex spherical mirror?
Afina-wow [57]

Answer:

Maximum distance of image from mirror is equal to focal length of the mirror

Explanation:

As we know by the equation of mirror we have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

here we know for convex mirror

object position is always negative as it will be placed behind the mirror always

while the focal length of the convex mirror is always taken positive

So here we have

\frac{1}{d_i} + \frac{1}{-d_o} = \frac{1}{f}

\frac{1}{d_i} = \frac{1}{d_o} + \frac{1}{f}

so here maximum value of image distance is equal to focal length of the mirror

6 0
3 years ago
A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon r
dusya [7]

Answer:

28 degree C

Explanation:

We are given that

T_1=4.00^{\circ}

R_1=217.7 \Ohm

R_2=215.1\Ohm

\alpha=-5.00\times 10^{-4}C^{-1}

We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

\alpha(T_2-T_1)=\frac{R_2}{R_2}-1

Using the formula

-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1

-5\times 10^{-4}(T_2-4)=0.988-1=-0.012

T_2-4=\frac{0.012}{5\times 10^{-4}}=24

T_2=24+4=28^{\circ}C

Hence, the temperature  on a spring day 28 degree C.

7 0
3 years ago
PLEASE HELP ASAP!!!!!
Vadim26 [7]
Both waves would increase right? That seems correct since the water and air temp both equally changed.
4 0
3 years ago
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