Which of the following ways is usable energy lost?

A basketball with mass of 0.8 kg is moving to the right with velocity 6 m/s and hits a volleyball with mass of 0.6 kg that stays

IceJOKER [234]

Answer:

26.67 m/s

Explanation:

From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.

p=mv where p is momentum, m is the mass of object and v is the speed of the object

Initial momentum

The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero

$p_i= m_bv_b+m_vv_v=8*6+0.6*0=48 Kg.m/s$

Final momentum

$p_f= m_bv_b+m_vv_v=8*4+0.6*v_v=32+0.6v Kg.m/s\\32+0.6v_v=48\\0.6v=16\\v_v=16/0.6=26.66666667\approx 26.67 m/s$

4 0

2 years ago

An apparatus is used to prepare an atomic beam by heating a collection of atoms to a temperature T and allowing the beam to emer

Zepler [3.9K]

Answer:

As shown in the attachment

Explanation:

The detailed steps and mathematical assumptions and manipulation is as shown in the attachment.

3 0

2 years ago

A valuable statuette from a Greek shipwreck lies at the bottom of the Mediterranean Sea. The statuette has a mass of 10,566 g an

leonid [27]

Answer:

A) W = 103.55 N

B) mass of displaced water = 4186 g

C) W_displaced water = 41.06 N

D) Buoyant force = 41.06 N.

E) ZERO

F) 62.54 N

Explanation:

We are given;

mass of statuette;m = 10,566 g = 10.566 kg

volume = 4,064 cm³

Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³

A) The dry weight of the statuette can be calculated as;

W = mg

So;

W = 10.556 × 9.81

W = 103.55 N

B) Mass of displaced water is calculated from;

Density = mass/volume

So, mass = Density × Volume

m = 1.03 × 4,064 = 4186 g

C) Weight of displaced water is given by;

W_displaced water = (m_displaced water) × g

W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N

D) The buoyant force is the same as the weight of the displaced water.

Thus, Buoyant force = 41.06 N.

E) The apparent weight of the statuette is calculated from;

Apparent weight = Dry weight - Weight of displaced water

Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.

F. From E above, The Force required to lift the statuette = 62.54 N

4 0

2 years ago

The system consists of a 20-lb disk A, 4-lb slender rod BC, and a 1-lb smooth collar C. If the disk rolls without slipping, dete

dlinn [17]

Answer:

The velocity of the collar will be 3.076 ft/s

Explanation:

Given data

weight of the disk, Wa = 20lb

weight of rod BC, Wbc = 4lb

weight of collar, Wc = 1lb

Considering the equation of equilibrium

Vb = 1.5Wbc

Wa = 1.875 Wbc

to calculate the velocity of the collar using energy conservation equation

T1 + V1 = T2 + V2

$0+4(1.5 \sin 45)+2(3 \sin 45)=\frac{1}{2}\left(\frac{1}{2}\left(\frac{20}{32.2}\right)(0.8)\right)$

=> $(1.875 W b c)+\frac{1}{2}\left(\frac{20}{32.2}\right)(1.5 W b c)+\frac{1}{2}\left(\frac{20}{32.2}\right)$

=> $(1.5 W b c) \frac{1}{2}\left\{\frac{1}{12}\left(\frac{4}{32.2}\right)(3)\right\}+\frac{1}{2}\left(\frac{1}{32.2}\right)$

=> $(2.598 W b c)+4(1.5 \sin 0)+2(3 \sin 0)$

Wbc = 1.18rad/sec

i.e.

$V _c=2.598 \times 1.18$

= 3.076 ft/ s

5 0

2 years ago

A spherical electron cloud surrounding an atomic nucleus would best represent

podryga [215]

Answer:

a S orbital

Explanation:

Atomic orbitals is the place where we are most likely to find at least one electron, this definition is based on the equation posed by Erwin Schrödinger.

It is said that each electron occupies an atomic orbital that is defined by a series of quantum numbers s, n, ml, ms. In any atom each orbital can contain two electrons. It is possible that thanks to the function of the orbitals, the appearance that atoms can have is that of a diffuse cloud.

The orbitals s (l = 0) have a spherical shape. The extent of this orbital depends on the value of the main quantum number, so a 3s orbital has the same shape but is larger than a 2s orbital.

The orbitals p (l = 1) are formed by two identical lobes that project along an axis. The junction zone of both lobes coincides with the atomic nucleus. There are three orbitals p (m = -1, m = 0 and m = + 1) in the same way, which differ only in their orientation along the x, y or z axes.

The orbitals d (l = 2) are also formed by lobes. There are five types of d orbitals (corresponding to m = -2, -1, 0, 1, 2)

6 0

2 years ago