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goldfiish [28.3K]

2 years ago

5

A deuterium atom is a hydrogen atom with a neutron added to its nucleus. Approximate the binding energy of this nucleus, given t

hat the mass of the deuterium atom is 2.014102 u and the masses of a hydrogen atom and a neutron are 1.007825 u and 1.008665 u, respectively.
a. 2 GeV.
b. 2 keV.
c. 2 MeV.
d. 2 eV.

1 answer:

Mademuasel [1]2 years ago

8 0

Answer:

c. 2 MeV.

Explanation:

The computation of the binding energy is shown below

$= [Zm_p + (A - Z)m_n - N]c^2\\\\=[(1) (1.007825u) + (2 - 1 ) ( 1.008665 u) - 2.014102 u]c^2\\\\= (0.002388u)c^2\\\\= (.002388) (931.5 MeV)\\\\=2.22 MeV$

= 2 MeV

As 1 MeV = (1 u) c^2

hence, the binding energy is 2 MeV

Therefore the correct option is c.

We simply applied the above formula so that the correct binding energy could come

And, the same is to be considered

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: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

e-lub [12.9K]

Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

$a_s = \dfrac{F}{m}$

$a_s = \dfrac{8.2}{12}$

$a_s = 0.68\ m/s^2$

$F = m a_m$

$a_m = \dfrac{F}{m}$

$a_m = \dfrac{8.2}{70}$

$a_m = 0.12\ m/s^2$

$x_1+x_2 = 25$

$\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25$

$(a_c+a_m)t^2=50$

$(0.12+0.68)t^2=50$

$t = \sqrt{\dfrac{50}{0.8}}$

t = 7.90 s

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$x_1 = 3.74m$

5 0

3 years ago

A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 s.

Stella [2.4K]

Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

$\omega =\frac{2\pi \times 469}{60}=49.11 rad/s$

t=16 s

Angular deceleration in 16 s

$\omega =\omega _0+\alpha \cdot t$

$\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2$

Moment of Inertia $I=mr^2=13\times 1.8^2=42.12 kg-m^2$

Change in kinetic energy =Work done

Change in kinetic Energy $=\frac{I\omega ^2}{2}-\frac{I\omega _0^2}{2}$

$\Delta KE=\frac{42.12\times 49.11^2}{2}=50,792.34 J$

(a)Work done =50.79 kJ

(b)Average Power

$P_{avg}=\frac{E}{t}=\frac{50.792}{16}=3.174 kW$

7 0

3 years ago

A water pump is a positive displacement-type pump true or false

Kay [80]

Answer: True

A water pump belong to a positive displacement pump that provides constant flow of water at fixed speed, regardless of changes in the counter pressure. The two main types of positive displacement pump are rotary pumps and reciprocating pumps.

Moreover, water pump is a reciprocating positive displacement pump that have an expanding cavity on the suction side and a decreasing cavity on the discharge side. In water pumps, the liquid flows into the pumps as the cavity on the suction side expands and then the liquid flows out of the discharge as the cavity collapses providing water in a pail.

6 0

3 years ago

17. A volleyball weighs about 300 grams.

atroni [7]

Answer:

PE = 44.1 J

Explanation:

Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)

We have:

1 kilograms = 1000 grams.

We convert it using a rule of 3, replacing, simplifying units and solving:

$\boxed{\bold{x=\frac{gr*1\ kg}{1000\ gr}=\frac{300\ gr*1\ kg}{1000\ gr}=\frac{300\ kg}{1000}=\boxed{\bold{0.3\ kg}}}}$

==================================================================

Earth's gravity is known to be 9.8 m/s², so we have:

Data:

m = 0.3 kg
g = 9.8 m/s²
h = 15 m
PE = ?

Use formula of potencial energy:

$\boxed{\bold{PE=m*g*h}}$

Replace and solve:

$\boxed{\bold{PE=0.3\ kg*9.8\frac{m}{s^{2}}*15\ m}}$

$\boxed{\boxed{\bold{PE=44.1\ J}}}$

Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.

The potential energy of the volleyball is <u>44.1 Joules.</u>

Greetings.

8 0

3 years ago

A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at 27 ∘ above t

ycow [4]

Answer:

Explanation:

Applied force, F = 18 N

Coefficient of static friction, μs = 0.4

Coefficient of kinetic friction, μs = 0.3

θ = 27°

Let N be the normal reaction of the wall acting on the block and m be the mass of block.

Resolve the components of force F.

As the block is in the horizontal equilibrium, so

F Cos 27° = N

N = 18 Cos 27° = 16.04 N

As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .

The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)

The vertically downward force acting on the block is mg - F Sin 27°

= mg - 18 Sin 27° = mg - 8.172 ... (2)

Now by equating the forces from equation (1) and (2), we get

mg - 8.172 = 6.42

mg = 14.592

m x 9.8 = 14.592

m = 1.49 kg

Thus, the mass of block is 1.5 kg.

6 0

3 years ago