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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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What is the external-internal pressure difference when the diver's lungs are at a depth of 6.1m (about 20ft)? assume that the di
Nesterboy [21]
Snorkel connects the diver's lungs with the air above the water. Because of this, the pressure inside the diver's lungs is the same as the atmospheric pressure.
The static pressure of a fluid is given with this equation:
P=\rho gh
The density of water is 1000kg/m^3.
P=1000\cdot 9.81 \cdot 6.1=59841$Pa
Standart atmospheric pressure is 101<span>325Pa. 
</span>The difference is:
\Delta P=101325-59841=41484 $Pa

8 0
3 years ago
A 100 W incandescent lightbulb emits about 5 W of visible light. (The other 95 W are emitted as infrared radiation or lost as he
frutty [35]

Answer:

n = 1.89 x 10¹⁹ photons/s

Explanation:

given,

Power of bulb = 100 W

Visible light = 5 W

wavelength of the visible light = 600 nm

number of photons emitted per second

using formula of wavelength

\lambda = \dfrac{c}{\nu}

\nu= \dfrac{c}{\lambda}

\nu= \dfrac{3 \times 10^8}{600 \times 10^{-9}}

\nu=5 \times 10^{14}\ Hz

energy of photon

U = h υ

U = 6.63 x 10⁻³⁴ x 4 x 10¹⁴

U = 2.65 x 10⁻¹⁹ J

number of photon

n = \dfrac{P}{U}

n = \dfrac{5}{2.65 \times 10^{-19}}

n = 1.89 x 10¹⁹ photons/s

6 0
3 years ago
Fifteen joules of heat are added to a cylinder with a piston. The system uses 7 joules of energy to raise the piston upward. By
Studentka2010 [4]
The first law of thermodynamics states that:
\Delta U = Q-L
where 
\Delta U is the variation of internal energy of the system
Q is the heat absorbed by the system
L is the work done by the system on the surrounding.

In this problem, the system absorbs 15 J of heat, so Q=+15 J (with positive sign, since it is heat absorbed by the system) while the work done by the system is L=+7 J (with positive sign, since it is work done by the system), so the variation of internal energy is
\Delta U= Q-L=(15 J)-(7 J)=+8 J
4 0
3 years ago
Read 2 more answers
A negative acceleration means that the speed of the object decreases. What makes this sentene incorrect?
Sophie [7]
It's not the speed , it's the rate of change of the speed decreases 
4 0
3 years ago
Read 2 more answers
A 50-cm-long spring is suspended from the ceiling. A 330g mass is connected to the end and held at rest with the spring unstretc
Anvisha [2.4K]

Explanation:

Given that,

Length of the spring, l = 50 cm

Mass, m = 330 g = 0.33 kg

(A) The mass is released and falls, stretching the spring by 28 cm before coming to rest at its lowest point. On applying second law of Newton at 14 cm below the lowest point we get :

kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.14}\\\\k=23.1\ N/m

(B) The amplitude of the oscillation is half of the total distance covered. So, amplitude is 14 cm.

(C) The frequency of the oscillation is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{23.1}{0.33}} \\\\f=1.33\ Hz

5 0
4 years ago
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