With a speed of 75 m sl. Determine

One mole of iron (6 x 10^23 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-

choli [55]

Answer:

Explanation:

Given that:

length l = 2.3 m

a = 0.12 cm = $0.12 \times 10^{-2} \ m$

$x = 1.17 \ cm = 1.17 \times 10^{-2}\ m$

m = 149 kg

$\delta = 7.87 \ g/cm^3$

$da = 2.28 \times 10^{-10}\ m$

$F_{net} = F-mg\\ \\0 = F - mg \\ \\ F = mg \\ \\ k_sx = mg \\ \\$

∴

$k_s = \dfrac{149(9.8)}{1.17 \times 10^{-2}} \\ \\ k_s = 124803.42 \ N /m$

$N_{chain} = \dfrac{A_{wire}}{A_{atom}} = \dfrac{A_w}{da^2}$

$N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2$

$N_{chain} = (\dfrac{0.12 \times 10^{-2} }{2.28 \times 10^{-10}})^2$

$N_{chain} = 2.77 \times 10^{13}$

$N_{bond} = \dfrac{L}{da} \\ \\ = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}$

$\text{Finally; the stiffness of a single interatomic spring is:}$

$k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s$

$k_{si} =\dfrac{(1.009 \times 10^{10})}{2.77*10^{13}}}\times (124803.42)$

$\mathbf{k_{si} =45.46 \ N/m}$

4 0

2 years ago

Two point charges each carrying a charge of + 4.5 E - 6 C are located 4.5 meters away from each other. How strong is the electro

BARSIC [14]

Answer:

0.009 N, repulsive

Explanation:

The electrostatic force between two electric charges is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, we have

$q_1 =q_2 = +4.5\cdot 10^{-6}C$ are the two charges

r = 4.5 m is their separation

Substituting into the equation, we find

$F=(9\cdot 10^9 Nm^2 C^{-2})\frac{(+4.5\cdot 10^{-6} C)(4.5\cdot 10^{-6} C)}{(4.5 m)^2}=0.009 N$

Moreover, the force is repulsive. In fact, the following rules apply:

- When two charges have same sign, they repel each other

- When two charges have opposite signs, they attract each other

7 0

2 years ago

Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application

VARVARA [1.3K]

Answer:

The vertical trajectory is governed by Ordinary Differential Equation.

Time derivatives of each state variables.

d(d)/dt = v, d(m)/dt = -d(m-fuel)/dt, d(v)/dt = F/m.

Where V is velocity positive upwards, t is time, m is mass, m-fuel is fuel mass, F is Total force, positive upwards.

Therefore,

F = -mg - D + T, If V is positive and

F = -mg + D - T, If T is negative.

D is drag and the questions gave it as zero.

Explanation:

The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p

8 0

3 years ago

Small paragraph explaining how how potential and kinetic energy are related

weeeeeb [17]

Answer:

Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or state. While kinetic energy of an object is relative to the state of other objects in its environment, potential energy is completely independent of its environment.

Explanation:

4 0

2 years ago

Read 2 more answers

A car has uniformly accelerated from rest to a speed of 25m/s after traveling 75m. What is its acceleration in m/s^2

Roman55 [17]

<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’): </em></h2><h2><em> </em></h2><h2><em>The car’s speed is 25m/s </em></h2><h2><em>The distance travelled is 75m </em></h2><h2><em>Then we have the formulas for speed and distance: </em></h2><h2><em> </em></h2><h2><em>v = a x t -> 25 = a x t </em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2 </em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say: </em></h2><h2><em> </em></h2><h2><em>t = 25 / a </em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a </em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1): </em></h2><h2><em> </em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2 </em></h2><h2><em>t^2 = 150 / a </em></h2><h2><em>Since t has the same value for both truths we can say: </em></h2><h2><em> </em></h2><h2><em>625 / a^2 = 150 / a </em></h2><h2><em> </em></h2><h2><em>Thus multiply both sides with a^2: </em></h2><h2><em> </em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17 </em></h2><h2><em> </em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>

4 0

3 years ago