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weqwewe [10]
3 years ago
5

what does the density of an object tell you about the molecular arrangement of the atoms in an object

Physics
1 answer:
Andrews [41]3 years ago
4 0

Answer:

If the density of the object is high its molecular arrangement is compact while if the density is lows its molecular arrangement isnt that compact

You might be interested in
The solar system is estimated to be how old
weqwewe [10]
4.571 billion years
8 0
4 years ago
Calculate the nuclear binding energy per nucleon for 136^Ba if its nuclear mass is 135.905 amu.
kompoz [17]

Answer:

1.312 x 10⁻¹² J/nucleon

Explanation:

mass of ¹³⁶Ba = 135.905 amu

¹³⁶Ba contain 56 proton and 80 neutron

mass of proton = 1.00728 amu

mass of neutron = 1.00867 amu

mass of ¹³⁶Ba = 56 x  1.00728 amu + 80 x 1.00867 amu

                      = 137.10128 amu

mass defect = 137.10128 - 135.905

                    = 1.19628 amu

mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg

                     = 1.9858 x 10⁻²⁷ Kg

speed of light = 3 x 10⁸ m/s

binding energy,

E = mass defect x c²

E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²

E = 17.87 x 10⁻¹¹ J/atom

now,

binding energy per nucleon =\dfrac{17.87\times 10^{-11}}{136}

                                              = 0.1312 x 10⁻¹¹ J/nucleon

                                              = 1.312 x 10⁻¹² J/nucleon

4 0
3 years ago
Let's examine over the next few questions an NFL kick as described in the 3rd Law video. The announcer claimed that the ball is
hichkok12 [17]

Answer:

8451.62109367671 Newtons

Explanation:

1\ kg=1\times 9.8066500286389=9.8066500286389\ N

1 \lb=2.2046226218488\ kg

It is known that

1\ lbs=\dfrac{9.8066500286389}{2.2046226218488}=4.4482216282508\ N

So,

1900\ lbs=1900\times 4.4482216282508\\\Rightarrow 1900\ lbs=8451.6210936765\ N

The force in Newtons is 8451.6210936765 Newtons

4 0
4 years ago
A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made
anyanavicka [17]

Answer:

Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

Explanation:

Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2}  } }      ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

Current, I = 0.7 A

Length of wire, L = 0.62 m

Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

B=\frac{4\pi\times10^{-7}\times  0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2}  } }

B = 6.99 x 10⁻⁶ T

3 0
3 years ago
What can be the maximum value of the original kinetic energy of disk AA so as not to exceed the maximum allowed value of the the
timurjin [86]

The question is incomplete. The complete question is :

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

Solution :

Let M.I. of disk A = $I_0$

So, M.I. of disk B =  $3I_0$

Angular velocity of A = $\omega_0$

So the kinetic energy of the disk A = $\frac{1}{2}I_0\omega^2$

After coupling, the angular velocity of both the disks will be equal to ω.

Angular momentum will be conserved.

So,

$I_0\omega_0 = I_0 \omega + 3I_0 \omega$

$I_0\omega_0 = 4I_0 \omega$

$\omega = \frac{\omega_0}{4}$

Now,

$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$

$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$

$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$

$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$

$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4} $

$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$

$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J $

Therefore, the maximum initial K.E. = 3066.67 J

3 0
3 years ago
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