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BigorU [14]
3 years ago
10

For the potential , determine the number of bound states?A)6 b)4 c) 3 d) 1

Physics
1 answer:
s2008m [1.1K]3 years ago
7 0

Do you see that blank, open space after the word "potential ..." ?

There's supposed to be a number there that actually tells us the value of the potential.  Without that number ... and a lot more description of the whole scenario here ... there's no possible answer to the question.

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Which is likely to be more common in our Galaxy: white dwarfs or black holes? Why?
Nookie1986 [14]

Answer:

White dwarfs are likely to be much more common. The number of stars decreases with increasing mass, and only the most massive stars are likely to complete their lives as black holes. There are many more stars of the masses appropriate for evolution to a white dwarf.

4 0
3 years ago
Juan makes an adjustment to an electromagnet that causes the electromagnet to lose some of its strength. What did Juan most like
Assoli18 [71]
Reducing the amount of loops will cause a loss of strength, as the loops make the magnet stronger.
5 0
3 years ago
Read 2 more answers
Two children of about the same weight are playing at the playground. They both climb up to the top of a small tower. One slides
olya-2409 [2.1K]

Answer:

D

Explanation:

D) The overall work done by gravity is zero  

This statement is correct .

If m be the mass of each of the children and h be the height of tower

work done by gravity on the boys in going up = - mgh

it is so because force applied by gravity = mg downwards and displacement

is upwards

work done will be negative = - mgh

Work done by gravity on boys when they come down = + mgh because both force and displacement are downwards .

Hence total work done = - mgh + mgh = 0.

The children will have same kinetic energy as the inclined surface is friction-less so no energy will be dissipated hence addition of energy to boys in both the cases will be same.

4 0
3 years ago
A Pitot-static tube is mounted on a 2.5 cm pipe where oil (???? = 860 kg/m3, ???? = 0.0103 kg/m·s) is flowing. The Pitot tube is
emmasim [6.3K]

Answer:

\dot{V}=0.0733 \,m^3.s^{-1}

Explanation:

Given:

density, \rho=860\,kg.m^{-3}

diameter of the pipe, d=2.5\times 10^{-2}m

pressure difference, \Delta P=95.8\times 10^{5}\,Pa

In case of  pitot tube, the velocity is given by:

v=\sqrt{\frac{2.\Delta P}{\rho} }

v=\sqrt{\frac{2\times 95.8\times 10^{5}}{860} }

v=149.26\,m.s^{-1}

Now we know that volumetric flow rate is given as:

\dot{V}=a.v

where :

a= cross sectional area of the pipe

v= velocity of flow

\dot{V}=(\pi\times \frac{(2.5\times 10^{-2})^2}{4} ) \times 149.26

\dot{V}=0.0733 \,m^3.s^{-1}

4 0
3 years ago
An Atwood machine is constructed using two
Tomtit [17]

Answer:

0.47 m/s²

Explanation:

Assuming the string is inelastic, m₃ will accelerate downward at a rate of -a, and m₄ will accelerate upward at a rate of +a.

Draw a four free body diagrams, one for each hanging mass and one for each wheel.

For m₃, there are two forces: weight force m₃g pulling down, and tension force T₃ pulling up.  Sum of forces in the +y direction:

∑F = ma

T₃ − m₃g = m₃(-a)

For m₄, there are two forces: weight force m₄g pulling down, and tension force T₄ pulling up.  Sum of forces in the +y direction:

∑F = ma

T₄ − m₄g = m₄a

For m₁, there are two forces: tension force T₃ pulling down, and tension force T pulling right.  Sum of the torques in the counterclockwise direction:

∑τ = Iα

T₃r₃ − Tr₃ = (m₁r₃²) (a/r₃)

T₃ − T = m₁a

For m₂, there are two forces: tension force T₄ pulling down, and tension force T pulling left.  Sum of the torques in the counterclockwise direction:

∑τ = Iα

Tr₄ − T₄r₄ = (m₂r₄²) (a/r₄)

T − T₄ = m₂a

We now have 4 equations and 4 unknowns.  Let's add the third and fourth equations to eliminate T:

(T₃ − T) + (T − T₄) = m₁a + m₂a

T₃ − T₄ = (m₁ + m₂) a

Now let's subtract the second equation from the first:

(T₃ − m₃g) − (T₄ − m₄g) = m₃(-a) − m₄a

T₃ − m₃g − T₄ + m₄g = -(m₃ + m₄) a

T₃ − T₄ = (m₃ − m₄) g − (m₃ + m₄) a

Setting these two expressions equal:

(m₁ + m₂) a = (m₃ − m₄) g − (m₃ + m₄) a

(m₁ + m₂ + m₃ + m₄) a = (m₃ − m₄) g

a = (m₃ − m₄) g / (m₁ + m₂ + m₃ + m₄)

Plugging in values:

a = (1.64 kg − 1.27 kg) (9.8 m/s²) / (2.5 kg + 2.3 kg + 1.64 kg + 1.27 kg)

a = 0.47 m/s²

4 0
3 years ago
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