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3 years ago

A little help please?

Physics

1 answer:

Tpy6a [65]3 years ago

7 0

A. Made of the marble.

the mass remains constant when you drop the marble but the rest of the variables change as the marble is dropped, therefore, the only constant variable is its mass.

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If you wanted the pitch of a horn to drop relative to an observer, which way would you move the horn, relative to where that obs

Vladimir [108]

We assume that horn releases sound of constant frequency. In order for observer to observe different frequency either horn or observer or both must move.

This happens due to Doppler effect. It states that when position of source of sound and observer relative to each other changes, the observed frequency also changes. If the source emits sound of constant frequency than observed frequency will be either higher or lower than original.

When distance between source and observer increases the observed frequency will be lower. This is because same number of sound waves must cover greater distance so they have greater wavelength.
When distance between source and observer decreases the observed frequency will be higher. This is because same number of sound waves must cover smaller distance so they have smaller wavelength.

Wavelength and frequency are inversely proportional meaning when one increases the other drecreases.

From this explanation we can find answer for our question. <span>If we wanted the pitch of a horn to drop relative to an observer we need to move horn away from an observer.</span>

3 0

2 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

What is the frequency heard by a person driving at 15 m/s toward a factory whistle emitting a

zmey [24]

Answer:

835.29 Hz

Explanation:

When moving towards the source of sound, frequency will be given by

f*=f(vd+v)/v

Where f is the freqiency of the source, vd is the driving speed, v is the speed of sound in air, f* is the inkown frequency when moving forward.

Substituting 800 Hz for f, 340 m/s for v and 15 m/s for vd then

f*=800(15+340)/340=835.29411764704 Hz

Rounded off, the frequency is approximately 835.29 Hz

4 0

3 years ago

What were formed after the big bang theory?

tatuchka [14]

Answer:

neutrons protons electrons

Explanation:

after the Big Bang, the universe was filled with neutrons, protons, electrons, anti-electrons, photons and neutrinos.

4 0

2 years ago

A 1.65 mol sample of an ideal gas for which Cv,m = 3R/2 undergoes the following two-step process:1) from an initial state of the

kirill [66]

Answer:

W = -1.97KJ, Q = 1.97KJ, Delta U = 0

Delta U = -1.03KJ, Q = -1.03KJ, Delta H = -1.72KJ

Explanation:

The deatiled step by step calculation using the ideal gas equation (Pv =nRT), The first law of thermodynamics ( dQ =dW + dU) as applied is as shown in the attached file.

7 0

3 years ago