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3 years ago

Which of the following nuclei is most stable based on its binding energy?

Physics

1 answer:

Anettt [7]3 years ago

4 0

We have that the most stable nuclei are the ones with the highest average binding energy. We see that Nitrogen has a mass number of 15 and that in this region of the graph average binding energy is low. Silver and Gold are along a line where there is a constant decline in average binding energy; silver has more than gold. However, we see that at the start of this decline, there is Fe 56. This region has the elements with the highest average binding energy; Nickel with a mass number of 58 is right there and thus it is the most stable nucleus out of the listed ones.

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On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo

ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

$\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i$

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

$\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i$

42 m/s, negative direction (to the west).

5 0

3 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= $\int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx$

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*0.8(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = $\frac{36}{11}$ kg/m

Mass of rope = weight of rope × change in distance

= $\frac{36}{11}$ kg/m × (11-x)m = 3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*3.27(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0

3 years ago

Question 2

Delvig [45]

Answer:

Approximately $73\; {\rm N}$ , assuming that the acceleration of this ball is constant during the descent.

Explanation:

Assume that the acceleration of this ball, $a$ , is constant during the entire descent.

Let $x$ denote the displacement of this ball and let $t$ denote the duration of the descent. The SUVAT equation $x = (1/2)\, a\, t^{2}$ would apply.

Rearrange this equation to find an expression for the acceleration, $a$ , of this ball:

$\begin{aligned} a &= \frac{2\, x}{t^{2}}\end{aligned}$ .

Note that $x = 11\; {\rm m}$ and $t = 1.5\; {\rm s}$ in this question. Thus:

$\begin{aligned} a &= \frac{2\, x}{t^{2}} \\ &= \frac{2 \times 11\; {\rm m}}{(1.5\; {\rm s})^{2}} \\ &\approx 9.78\; {\rm m \cdot s^{-2}}\end{aligned}$ .

Let $m$ denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is $a$ , the net external force on this ball would be $m\, a$ .

Since $m = 7.5\; {\rm kg}$ and $a \approx 9.78\; {\rm m\cdot s^{-2}}$ , the net external force on this ball would be:

$\begin{aligned} (\text{net force}) &= m\, a \\ &\approx 7.5\; {\rm kg} \times 9.78\; {\rm m\cdot s^{-2}} \\ &\approx 73\; {\rm kg \cdot m \cdot s^{-2} \\ &= 73\; {\rm N} && (1\; {\rm N} = 1\; {\rm kg \cdot m\cdot s^{-2}}) \end{aligned}$ .

4 0

2 years ago

Which of the following would decrease the magnetic field around a wire?

shusha [124]

The correct answer is
A. decreasing the current

In fact, the magnetic field produced by a current carrying wire is given by
 $B(r) = \frac{\mu_0 I}{2 \pi r}$
where
 $\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance from the wire at which the field is calculated

We see from the formula that the intensity of the field, B, is directly proportional to the current I, so if the current decreases, the magnetic field strength B decreases as well.

4 0

3 years ago

A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i

AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition