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katen-ka-za [31]

3 years ago

11

two objects were lifted by a machine. One object had a mass of 2 kilograms, was lifted at a speed of 2m/sec . The other had a ma

ss of 4 kilograms and was lifted at 3m/sec. Which object had more kinetic energy while it was being lifted? Show all calculations

1 answer:

Lubov Fominskaja [6]3 years ago

5 0

Object 2 has more kinetic energy

Explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion, and it is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

In this problem, for object 1:

m = 2 kg

v = 2 m/s

So its kinetic energy is

$K_1 = \frac{1}{2}(2)(2)^2=4 J$

For object 2,

m = 4 kg

v = 3 m/s

So its kinetic energy is

$K_2 = \frac{1}{2}(4)(3)^2=18 J$

Therefore, object 2 has more kinetic energy.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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Answer:

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

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(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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Answer:

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2 years ago

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Answer:

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Explanation:

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Solution:

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