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3 months ago

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if m represents mass in kg, v represents speed in m/s, and r represents radius in m, show that the force F in the equation F=mv^

2 /r can be expressed in the unit kg×m/s^2

1 answer:

Keith_Richards [3.2K]3 months ago

4 0

This method is termed dimensional analysis, which focuses solely on the measurement units without considering their values. You need to perform calculations using variables while eliminating similar items from both the numerator and denominator. The outcome is as follows:

F = mv²/r = [kg][m/s]²/[m] = [kg][m²⁻¹][1/s²] = [kg·m/s²].

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What is the number N0 of 99mTc atoms that must be present to have an activity of 15mCi?

Keith_Richards [3271]

Based on my findings, within a period of 2 hours, there are certain atoms remaining. N = N0 * 2^(-t/6.020) = N = N0 * 2^-0.33223 = 0.7943 N0 Thus, the quantity of atoms that undergo disintegration is N0 - N = N0 * (1 - 0.79430) = 0.2057 N0 This must equate to 15 mCi = 15 * 3.7 * 10^7 = 5.55 * 10^8 atoms N0 = 5.55 * 10^8 / 0.2057 = 2.698 * 10^9 atoms Consequently, 2.698 * 10^9 atoms represents the value of N0.

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3 months ago

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Sav [3153]

Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

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Additional explanation

The Ideal Gas Law that should be remembered is:

$\large {\boxed {PV = nRT} }$

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Now, let’s proceed with the problem!

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Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

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Next, we will determine the work performed on the gas:

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

$\texttt{ }$

Step B:

By utilizing the methodology mentioned earlier:

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

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Next, we will ascertain the work completed on the gas:

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

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Ultimately, we can calculate the total work done and heat supplied as follows:

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

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Learn more

Minimum Coefficient of Static Friction:
The Pressure In A Sealed Plastic Container:
Effect of Earth’s Gravity on Objects:

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

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2 months ago

Which phrase describes an atom?

kicyunya [3294]

<span>an atom is described as having a negatively charged electron cloud surrounding a positively charged nucleus, which is the correct choice.</span><span>

The nucleus contains electrically neutral neutrons and positively charged protons, establishing its positive charge. In contrast, electrons carry a negative charge. The electromagnetic force keeps the atoms bound to the nucleus.
</span>

4 0

3 months ago

A 0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before

Ostrovityanka [3204]

Answer:

Explanation:

Provided:

mass of the steel ball $m=0.2\ kg$

initial velocity of the ball $u=10\ m/s$

Final velocity of the ball $v=-10\ m/s$ (moving upwards)

The impulse given is determined by the change in the momentum of the object.

<ptherefore the="" impulse="" j="" is="" defined="" by="">

$J=\Delta P$

$\Delta P=m(v-u)$

$\Delta P=0.2(-10-10)$

$\Delta =-4\ N-s$

Thus, the magnitude of the Impulse is 4 N-s.

</ptherefore>

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2 months ago

Read 2 more answers