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Effectus [21]
3 years ago
15

A car is traveling at 33.2 m/s. If the car has a kinetic energy of 688,900 J, what is the car's mass?​

Physics
2 answers:
wariber [46]3 years ago
7 0

Answer:1250kg

Explanation:

Kinetic energy(ke)=688900j

Velocity(v)=33.2m/s

Mass(m)=?

ke=(mv^2)/2

688900=(mx(33.2)^2)/2

688900x2=mx1102.24

1377800=mx1102.24

m=1377800/1102.24

m=1250kg

aivan3 [116]3 years ago
5 0

Answer:

1250kg

Explanation:

\frac{688900}{33.2\\^{2} } = 625

\frac{1}{2} m = 625

625 x 2 = 1250

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The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.

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The time that the dog will catch up with the rabbit is given as follows:

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The time taken will be the same T

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

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