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3 years ago

Drake lifted a couch a height of 0.2 meters in 1.3 seconds. He did 300 j of work. how much power did drake use?

Physics

1 answer:

Novosadov [1.4K]3 years ago

6 0

Answer:

c is correct

Explanation:

I hope this helps.

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A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at

yan [13]

Answer:

a) $s_{T} = 30\,m$ , b) $t = 5\,min$ , c) $\Delta t = 6.667\,s$ , d) $\Delta s_{R} = 33.333\,m$ , e) $t' = 11.667\,s$ , f) The rabbit won the race.

Explanation:

a) As turtle moves at constant speed, its position is determined by the following formula:

$s_{T} = v_{T}\cdot t$

Where:

$t$ - Time, measured in seconds.

$v_{T}$ - Velocity of the turtle, measured in meters per second.

$s_{T}$ - Position of the turtle, measured in meters.

Then, the position of the turtle when the rabbit starts to run is:

$s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)$

$s_{T} = 30\,m$

The position of the turtle when the rabbit starts to run is 30 meters.

b) The time needed for the turtle to finish the race is:

$t = \frac{s_{T}}{v_{T}}$

$t = \frac{150\,m}{0.5\,\frac{m}{s} }$

$t = 300\,s$

$t = 5\,min$

The time needed for the turtle to finish the race is 5 minutes.

c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:

$v_{R} = v_{o,R} + a_{R}\cdot \Delta t$

Where:

$v_{R}$ - Final velocity of the rabbit, measured in meters per second.

$v_{o,R}$ - Initial velocity of the rabbit, measured in meters per second.

$a_{R}$ - Acceleration of the rabbit, measured in $\frac{m}{s^{2}}$ .

$\Delta t$ - Running time, measured in second.

$\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}$

$\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }$

$\Delta t = 6.667\,s$

The time taken by the rabbit to reach maximum speed is 6.667 s.

d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:

$v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

Where $\Delta s_{R}$ is the travelled distance of the rabbit from rest to maximum speed.

$\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}$

$\Delta s_{R} = 33.333\,m$

The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.

e) The time required for the rabbit to finish the race can be determined by the following expression:

$t' = \frac{\Delta s_{R}}{v_{R}}$

$t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }$

$t' = 11.667\,s$

The time required for the rabbit from rest to maximum speed is 11.667 seconds.

f) The animal with the lowest time wins the race. Now, each running time is determined:

Turtle:

$t_{T} = 300\,s$

Rabbit:

$t_{R} = 60\,s + 6.667\,s + 11.667\,s$

$t_{R} = 78.334\,s$

The rabbit won the race as $t_{R} < t_{T}$ .

7 0

4 years ago

Vector A has a magnitude of 16 m and makes an angle of 44° with the positive x axis. Vector B also has a magnitude of 13 m and i

marshall27 [118]

Answer with explanation:

The given vectors in are reduced to their componednt form as shown

For vector A it can be written as

$\overrightarrow{v}_{a}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}$

Similarly vector B can be written as

$\overrightarrow{v}_{b}=-13\widehat{i}$

Hence The sum and difference is calculated as

$\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}+(-13\widehat{i})\\\\\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=(16cos(44^{o})-13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}+\overrightarrow{v}_{b}=-1.49\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}+\overrightarrow{v}_{b}|=\sqrt{(-1.49)^{2}+11.11^{2}}=11.21m$

The direction is given by

$\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{-1.49}=97.64^{o}$ with positive x axis.

Similarly

$\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}-(-13\widehat{i})\\\\\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=(16cos(44^{o})+13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}-\overrightarrow{v}_{b}=24.51\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}-\overrightarrow{v}_{b}|=\sqrt{(24.51)^{2}+11.11^{2}}=26.91m$

The direction is given by

$\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{24.51}=24.38^{o}$ with positive x axis.

3 0

3 years ago

Which is the best predictor of the radioactive nature of an isotope?

Luda [366]

The Neutron-to-Proton Ratio. Hope this helps Sorry if it's wrong I'm a little rusty on Physics.

6 0

3 years ago

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