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jeyben [28]
3 years ago
6

The strong nuclear force felt by a single proton in a large nucleus

Physics
1 answer:
Crank3 years ago
4 0
Your question: The strong nuclear force felt by a single proton in a large nucleus _______________________.

Answer: is about the same as that felt by a single proton in a small nucleus.
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Density=2g/mL and volume=20mL what is mass
leva [86]

Answer:

40g

Explanation:

Mass = density x volume

= 2 x 20

= 40g

8 0
3 years ago
Read 2 more answers
A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec
Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 8=50.24rad/sec

Velocity of the string wave is equal to v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec

Power of wave propagation is equal to P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt

So power of the wave will be equal to 5.464 watt

6 0
2 years ago
What is the best unit to use when measuring the mass of a mineral sample?
Ksivusya [100]
Measuring density: Measure the mass (in grams) of each mineral sample available to you. The mass of each sample is measured using a balance or electronic scale. Record mass on a chart.
8 0
2 years ago
Write an expression for the potential energy UD of a particle when it is at a distance D from the force center, in terms of B an
nadezda [96]

Answer:

E+mc+UD-3

Explanation:

7 0
3 years ago
A child’s toy that is made to shoot ping pong balls consists of a tube, a spring (k = 18 N/m) and a catch for the spring that ca
UkoKoshka [18]

Answer:

The height is 3.1m

Explanation:

Here we have a conservation of energy problem, we have a conversion form eslastic potencial  energy to gravitational potencial energy, so:

E_e=\frac{1}{2}K*x^2\\E_e=\frac{1}{2}18N/m*(9.5*10^{-2}m)^2\\E_e=0.081J

then we have only gravitational potencial energy when the ball is at its maximun height.

E_g=m*g*h

because all the energy was transformed Eg=Ee

h=\frac{0.081J}{9.8m/s^2*m}

searching the web, the mass of a ping pong ball is 2.7 gr in average. so:

h=\frac{0.081J}{9.8m/s^2*(2.7*10^{-3}kg)}\\h=3.1m

6 0
3 years ago
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