How do you find the average speed of an object?

A. Group of cross country runners decided to go on an hour and a half run. During the first hour, they ran a total of 13 kilomet

pshichka [43]

Answer:

The average speed for the entire run is 12 km/h.

Explanation:

The average speed is given by the following equation:

$\overline{v} = \frac{d_{T}}{t_{T}}$

Where:

$d_{T}$ : is the total distance

$t_{T}$ : is the total time

If during the first hour, they ran a total of 13 kilometers and then, they ran 5.0 kilometers during the next half an hour we have:

$d_{T} = 13 km + 5 km = 18 km$

$t_{T} = 1 h + \frac{1}{2} h = 1.5 h$

Hence, the average speed is:

$\overline{v} = \frac{d_{T}}{t_{T}} = \frac{18 km}{1.5 h} = 12 km/h$

Therefore, the average speed for the entire run is 12 km/h.

I hope it helps you!

3 0

3 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.60×106 N, one at an angle 13.0 west of north, an

Juliette [100K]

Answer: $W=1.93\times 10^9 J$

Explanation:

Given

Force $F=1.6\times 10^{6} N$

one at an angle of $13^{\circ}$ East of North and another at $13^{\circ}$ West of North

Net Force is in North Direction

$F_{net}=2F\cos 13$

Forces in horizontal direction will cancel out each other

thus Work done will be by north direction forces

$W=2F\cdot \cos 30\cdot s$

here $s=0.7 km$

$W=2\times 1.6\times 10^{6}\cdot \cos 30\cdot 700$

$W=1.93\times 10^9 J$

3 0

3 years ago

A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact

ArbitrLikvidat [17]

Answer:

0.09 % of the original radioactive nucllde its left after 10 half-lives
It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

$N ( t) \ = \ N_0 \ e^{ \ - \frac{t}{\tau}}$ ,

where N(t) its quantity of material at time t, $N_0$ its the initial quantity of material and $\tau$ its the mean lifetime of the radioactive element.

The half-life $t_{\frac{1}{2}}$ its the time at which the quantity of material its the half of the initial value, so, we can find:

$N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

so:

$\ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

$e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}$

$- \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )$

$t_{\frac{1}{2}}\ = \tau ln( 2 )$

So, after 10 half-lives, we got:

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ t_{\frac{1}{2}}}{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ \tau \ ln( 2 ) }{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - 10 \ \ ln( 2 ) }$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}$

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

$10 \ * \ 24,110 \ years \ = \ 241,100 \ years$

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

7 0

3 years ago

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A proton is accelerated through a potential difference of 250V. It then enters a uniform magnetic field and moves in a circular

3241004551 [841]

Answer: a) 7.52 ×10^-4 T, b) 4.72×10^-7s

Explanation: by using the work energy thereom, the work done on the electron by the voltage source (potential) equals the kinetic energy.

qV =mv^2/2

Where q = magnitude of electronic charge = 1.609×10^-19 c

V = potential difference = 250v

m = mass of an electronic charge = 9.11×10^-31 kg

v = velocity of electron =?

By substituting the parameters, we have that

1.609×10^-19 × 250 = 9.11×10^-31 × v^2/2

1.609×10^-19 × 250 ×2 = 9.11×10^-31 ×v^2

v^2 = 1.609×10^-19 × 250 ×2/ 9.11×10^-31

v^2 = 8.05×10^-17/ 9.11×10^-31

v^2 = 1.77×10^14

v = √1.77×10^14

v = 1.33×10^7 m/s

The centripetal force required for the motion of the electron is gotten from the magnetic force on the electron.

qvB = mv^2/r

By cancelling "v" on both sides of the equation, we have that

qB = mv/r

Where r = radius = 10cm = 0.1m

1.609×10^-19 × B = 9.11 ×10^-31 ×1.33×10^7/ 0.1

B = (9.11 ×10^-31 × ×1.33×10^7)/ (0.1 ×1.609×10^-19)

B = 1.21×10^-23/ 1.609×10^-20

B = 0.752×10^-3

B = 7.52 ×10^-4 T

To get the period of motion, we recall that

v = ωr

Where ω = angular frequency

1.33×10^7 = ω×0.1

ω = 1.33×10^7/ 0.1

ω = 13.3×10^7

ω = 1.33×10^6 rad/s

But the period (T) of a periodic motion is defined as

T = 2π/ω

T = 2 × 3.142 / 1.33×10^6

T = 4.72×10^-7s

6 0

3 years ago

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What would be the consequences of the discovery that life<br> independently took hold on mars?

Elina [12.6K]

Answer:

hope that helps

Explanation:

This implies that the discovery of life on Mars does not automatically mean the ... Hence, we hope that Mars may have been the site of an independent origin of life. ... The general view of the results of the Viking biology experiment is that there is ... subsurface may hold liquid water aquifers that support chemosynthetic life.

4 0

3 years ago