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9 months ago

Review. (b) What is the mass difference?

Physics

1 answer:

nignag [31]9 months ago

8 0

The mass difference is in between the two is equivalent to the binding energy of the nucleus .

<h3>What is meant by mass difference? </h3>

Δm be the mass difference is defined as the difference between the mass of the nucleus and the total mass of the individual nucleons that comprise it.

Binding energy is the energy required to split the nucleus of an atom into neutrons and protons, while mass defect is the difference between the predicted mass and the actual mass of the nucleus of an atom .

The mass difference is difference between the mass of nucleus and the Total mass of the individual nucleons that make it up.

∆m = final mass - initial mass.

To learn more about mass refer to:

brainly.com/question/19694949

#SPJ4

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A ball is dropped from a height h and falls the last half of its distance in 4 seconds. How long does the ball fall? From what h

dlinn [17]

Answer:

How long does the ball fall is t_2 = 13.66 (s).

From what height is the ball originally dropped is h= 913.90 (m).

Explanation:

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2 years ago

A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the a

finlep [7]

Solution :

Let $m_1=m_2=4$ kg

$u_1 = 5$ m/s

Let $v_1$ and $v_2$ are the speeds of the disk $m_1$ and $m_2$ after the collision.

So applying conservation of momentum in the y-direction,

$0=m_1 .v_1_y -m_2 .v_2_y$

$v_1_y = v_2_y$

$v_1 . \sin 60=v_2. \sin 30$

$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$

$v_2=1.732 \times v_1$

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

$m_1.u_1=m_1.v_1_x+m_2.v_2_x$

$u_1=v_1_x+v_2_x$

$5=v_1. \cos 60 + v_2 . \cos 30$

$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$

$v_1 = 2.50$ m/s

So, $v_2 = 1.732 \times 2.5$

= 4.33 m/s

Therefore, speed of the disk 2 after collision is 4.33 m/s

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2 years ago

_________ is a layer of the earth that is classified not by composition

Sidana [21]

Answer:

Asthenosphere

Explanation:

The asthenosphere is a part of the upper mantle just below the lithosphere that is involved in plate tectonic movement and isostatic adjustments.

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2 years ago

En la figura los émbolos son de masa despreciable y están en reposo, como se ve en la figura, siendo M=300 kg. Determine: a) la

Leni [432]

The image mentioned is in the attachment

Answer: a) P = 2450 Pa;

b) P = 2940 Pa;

c) F = 4.9 N

Explanation:

a) Pressure is a force applied to a surface of an object or fluid per unit area.

The image shows a block applying pressure on the large side of the piston. The force applied is due to gravitation, so:

P = $\frac{F}{A}$

P = $\frac{m.g}{A}$

P = $\frac{300.9.8}{1.2}$

P = 2450 Pa

The pressure generated by the block is P = 2450 Pa.

b) A static liquid can also exert pressure and can be calculated as:

$P_{staticfluid} =$ ρ.g.h

where

ρ is the density of the fluid

h is the depth of the fluid

g is acceleration of gravity

$P_{staticfluid} =$ 600.9.8.0.5

$P_{staticfluid} =$ 2940 Pa

The pressure in the fluid at 50 cm deep is $P_{staticfluid} =$ 2940 Pa.

c) For the system to be in equilibrium both pressures, pressure on the left side and pressure on the right side, have to be the same:

$P_{s} = P_{b}$

$\frac{F}{A_s}$ = $\frac{F_b}{A_b}$

F = $\frac{F_b}{A_b}.A_s$

Adjusting the units, $A_{s}$ = 0.002 m².

F = $\frac{300.9.8.0.002}{1.2}$

F = 4.9 N

The force necessary to be equilibrium is F = 4.9 N.

4 0

2 years ago

A strings. A 50 gram bullet, traveling horzontally, strikes the block and becomes embedded inside the block. Immediately after t

vazorg [7]

Answer:

u= 200 m/s

Explanation:

Given that

Mass of bullet ,m= 50 gm

Assume that mass of block ,M= 1.2 kg

Lets take speed of the bullet before collision = u m/s

The speed of the system after collision ,v= 8 m/s

There is no any external force ,that is why linear momentum of the system will be conserve.

Linear momentum ,P = mass x velocity

m u = (M+m)v

0.05 x u = (1.2 + 0.05 ) x 8

u= 200 m/s

Therefore the speed of the bullet just before the collision is 200 m/s.

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3 years ago