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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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How does the mass of the melted sundae compare to when it was first made? this for science
tatiyna
<h3><u>Answer:</u></h3>

When it was first made, it was heavier

<h3><u>Explanation:</u></h3>

So there was originally air in the Ice Cream because they fluff it up like that. When it melts, those are gone.

3 0
2 years ago
Which of the following represents a chemical change? (1 point)
fgiga [73]
Can I see a photo or can you comment the answer choices.
8 0
3 years ago
A diffraction grating is placed 1.00 m from a viewing screen. Light from a hydrogen lamp goes through the grating. A hydrogen sp
Colt1911 [192]

Answer:

λ = 396.7 nm

Explanation:

For this exercise we use the diffraction ratio of a grating

           d sin θ = m λ

in general the networks works in the first order m = 1

we can use trigonometry, remembering that in diffraction experiments the angles are small

           tan θ = y / L

           tan θ = \frac{sin \theta}{cos \theta} = sin θ

           sin θ = y / L

we substitute

          d \  \frac{y}{L} = m λ

with the initial data we look for the distance between the lines

           d = \frac{m \lambda \ L}{y}

           d = 1 656 10⁻⁹ 1.00 / 0.600

            d = 1.09 10⁻⁶ m

for the unknown lamp we look for the wavelength

           λ = d y / L m

           λ = 1.09 10⁻⁶ 0.364 / 1.00 1

           λ = 3.9676 10⁻⁷ m

           λ = 3.967 10⁻⁷ m

         

we reduce nm

           λ = 396.7 nm

8 0
2 years ago
A motion sensor emits sound, and detects an echo 0.0115 s after. A short time later, it again emits a sound, and hears an echo a
Mekhanik [1.2K]

Answer:

1.17 m

Explanation:

From the question,

s₁ = vt₁/2................ Equation 1

Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.

Given: v = 343 m/s, t = 0.0115 s

Substitute into equation 1

s₁ = (343×0.0115)/2

s₁ = 1.97 m.

Similarly,

s₂ = vt₂/2.................. Equation 2

Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo

Given: v = 343 m/s, t₂ = 0.0183 s

Substitute into equation 2

s₂ = (343×0.0183)/2

s₂ = 3.14 m

The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁

s₂-s₁ =  (3.14-1.97) m = 1.17 m

7 0
3 years ago
Read 2 more answers
What is energy?
mestny [16]

Answer:

Energy is the ability for an object to do work.

Kinetic energy is energy which a body possesses while in motion.

Stored Energy = Potential Energy.

Speed is a measurement of how quickly something is able to move or operate.

6 0
3 years ago
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