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2 years ago

Use the periodic table and your knowledge of isotopes to complete these statements.

2 answers:

7 0

Use the periodic table and your knowledge of isotopes to complete these statements.

When polonium-210 emits an alpha particle, the child isotope has an atomic mass of ⇒ 206.

I-131 undergoes beta-minus decay. The chemical symbol for the new element is ⇒ Xe.

Fluorine-18 undergoes beta-plus decay. The child isotope has an atomic mass of ⇒ 18.

sleet_krkn [62]2 years ago

7 0

Answer:

206,Xe,18

Explanation:

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A 250 Hz tuning fork is struck and the intensity at the source is I1 at a distance of one meter from the source. (a) What is the

Zina [86]

Answer:

a) 0.0625 I_1

b) 3.16 m

Explanation:

Concepts and Principles

The intensity at a distance r from a point source that emits waves of power P is given as:

I=P/4π*r^2 (1)

Given Data

f (frequency of the tuning fork) = 250 Hz

I_1 is the intensity at the source a distance r_1 = I m from the source.

Required Data

- In part (a), we are asked to determine the intensity I_2 a distance r_2 = 4 in from the source.

- In part (b), we are asked to determine the distance from the tuning fork at which the intensity is a tenth of the intensity at the source.

solution:

(a)

According to Equation (1), the intensity a distance r is inversely proportional to the distance from the source squared:

I∝1/r^2

Set the proportionality:

I_1/I_2=(r_2/r_1)^2 (2)

Solve for I_2 :

I_2=I_1(r_2/r_1)^2

I_2=0.0625 I_1

(b)

Solve Equation (2) for r_2:

r_2=(√I_1/I_2)*r_1

where I_2 = (1/10)*I_1:

r_2=(√I_1/1/10*I_1)*r_1

=3.16 m

3 0

2 years ago

The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2

masha68 [24]

Answer:

the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

Explanation:

Given that:

Modulus for elasticity for concrete post $E_c$ = $3.6 *10^6$ psi

Thermal coefficient for concrete post $\alpha _c$ = $5.5 *10^{-6}/^0F$

Modulus for elasticity of steel bar $E_s$ = $29*10^6psi$

Thermal coefficient of steel bar $\alpha _2$ = $6.5*10^{-6}/^0F$

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod $A_s$ = $6(\frac{ \pi}{4} )(d_s)^2$

= $6(\frac{ \pi}{4} )(\frac{7}{8} )^2$

= 3.61 in²

Area of concrete parts $A_c$ = (10)(10) - $A_s$

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= $[\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)$

= $[\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)$

= $[(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}$

= $1.234*10^{-8}P = 8.0*10^{-5}$

$P = \frac {8.0*10^{-5}}{1.234*10^{-8}}$

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

$\sigma_c =\frac{P}{A_c}$

$\sigma_c = \frac{6482.98}{96.39}$

$\sigma_c = 67.26 psi$

Thus, the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

Also; the normal stress in the steel bars induced as a result of temperature rise is as follows:

$\sigma_s = \frac{-P}{A_s}$

$\sigma_s =\frac{-6482.98}{3.61}$

$\sigma_s = - 1795.84 psi$

Thus, the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

6 0

2 years ago

Which points are most efficient for the utilization of resources on a production possibilities diagram?

KonstantinChe [14]

Answer: most effective way is to practice reduce reuse and recycle for utilisation of resources

3 0

1 year ago

3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag

aleksandr82 [10.1K]

Answer:

(a) $W=-19.25J$

(b) $W=-52.8J$

Explanation:

Hello.

(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:

$W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J$

Which is negative as it expands against the given pressure.

(b) Moreover, of the process is carried out reversibly, the pressure can change, therefore, we need to compute the work via:

$W=nRTln(\frac{V_1}{V_2} )$

Whereas the moles are computed from the given mass of argon:

$n=6.56g*\frac{1mol}{39.95g}=0.164mol$

Thus, the work is:

$W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J$

Regards.

4 0

2 years ago

A baseball is hit with a speed of 33.6 m/s. Calculate its height and the distance traveled if it is hit at angles of 30.0°, 45.0

Solnce55 [7]

Answer:

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

Explanation:

Let ∝ be the angle with respect to the horizontal that the baseball is hit with.

The horizontal component of the velocity is vcos(∝) and the vertical component of the velocity is vsin(∝).

Ignore air resistant, only gravitational acceleration g = -9.81 m/s2 affect the ball vertically. We can use the following equation to calculate the time it takes to reach maximum height (at 0 speed)

$vsin(\alpha) + gt = 0$