Ask question

Ask question

All categories

3 years ago

5

Greg and Tabitha are jogging together. At time t = 0, they are both traveling at a speed of 4 m/s, but Greg is 8 m ahead of Tabi

tha. Tabitha accelerates at a rate of 1.5 m/s2 and Greg accelerates at a rate of 1.0 m/s2.
a. Write an expression for Greg's position. Plug in all the values you can to simplify the expression. (1 point)

b. Write an expression for Tabitha's position. Plug in all the values you can to simplify the expression. (1 point)

c. How long does it take Tabitha to catch up to Greg? (1 point)

d. How far has Tabitha run while catching up to Greg? (1 point)

1 answer:

Ray Of Light [21]3 years ago

8 0

Answer:

The answer to question is below

Explanation:

Data

t1 = 0

vo = 4 m/s

a)

d = vot + $\frac{1}{2} at^{2}$

d = 4t + $\frac{1}{2} (1)t^{2} + 8$

$d = 8 + 4t + \frac{t^{2} }{2}$

b)

$d = 4t + \frac{1}{2} (1.5)t^{2}$

$d = 4t + 0.75t^{2}$

c)

$8 + 4t + \frac{t^{2} }{2}$ = $4t + 0.75t^{2}$

$4t + \frac{1}{2} t^{2} + 8 = 4t + \frac{1}{2} (1.5)t^{2} \\$

$0.75t^{2} - 0.5t^{2} = 8$

$0.25t^{2} = 8$

t² = 32

t = 5.66 s

d)

d = 4(5.66) + 0.75(5.66)²

d = 22.64 + 24.02

d = 46.67 m

You might be interested in

¿Que es un átomo???????

disa [49]

Un átomo es una porción material menor de un elemento químico que interviene en las reacciones químicas y posee las propiedades características de dicho elemento.

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Chuge%5Cmathcal%5Ccolorbox%7Bred%7D%7B%5Ccolor%7Bpink%7D%7B%E3%82%AF%E3%82%A8%E3%82%B9%E3

nirvana33 [79]

Answer:

The theory of Relativity state that :<u>Special relativity applies to all physical phenomena in the absence of gravity and General relativity explains the law of gravitation and its relation to other forces of nature. </u>

3 0

3 years ago

Determine the shortest frequency of light required to remove an electron from a sample of ti metal if the binding energy of tita

Mama L [17]

The correct answer is: shortest frequency = 7.86*10^15 $s^{-1}$

Explanation:
The binding energy of titanium = 3.14*10^6<span> J/mol
</span>The energy required to remove an electron = (3.14*10^6) /(6.023*10^23) = 5.213*10^-18 J<span>
Where 6.023*10^</span>23 = Avagadro number

Since E = hv

Frequency = v = E/h
E = Energy = 5.213*10^-18
h = Planck's constant = 6.626*10^-34
v = (5.213*10^-18) / 6.626*10^-34<span>)
</span><span>v = </span><span>7.86*10^15 </span> $s^{-1}$ (shortest frequency)

7 0

4 years ago

Mrs. Stark weighs 70kg. How much Potential Energy does she have if she stands on the roof of an 80m tall building?

sesenic [268]

70kg

1 earth gravity

80 M of height

---------------------------------

54917.2 Joules of energy

Mrs. Stark has 54,917.2 Joules of energy.

3 0

3 years ago

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

tia_tia [17]

Answer:

a) $a=33.73mm/s^{2}$

b) mg>N

c) $\%_{change}=0.343\%$

d) $a=24.07mm/s^{2}$

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

$a_{c}=\omega ^{2}r$

where:

$\omega=\frac{2\pi}{T}$

we know the period of rotation of the earth is about 24 hours, so:

$T=24hr*\frac{3600s}{1hr}=86400s$

so we can now find the angular speed:

$\omega=\frac{2\pi}{86400s}$

$\omega=72.72x10^{-6} rad/s^{2}$

So the centripetal acceleration will be:

$a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)$

which yields:

$a_{c}=33.73mm/s^{2}$

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

$\sum F=0$

so we get that:

$N-mg+ma_{c} = 0$

and solve for the normal force:

$N=mg-ma_{c}$

In this case, we can clearly see that:

$mg>mg-ma_{c}$

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$

and let's calculate the normal force:

$N=m(g-a_{c})$

$N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})$

N=586.58N

so now we can calculate the percentage change:

$\%_{change} = \frac{mg-N}{mg}x100\%$

so we get:

$\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%$

$\%_{change}=0.343\%$

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

$cos \theta = \frac{AS}{h}$

In this case:

$cos \theta = \frac{r}{R_{E}}$

so we can solve for r, so we get:

$r= R_{E}cos \theta$

in this case we'll use the average radius of earch which is 6,371 km, so we get:

$r = (6371x10^{3}m)cos (44.4^{o})$

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

$a=\omega ^{2}r$

$a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m$

$a=24.07 mm/s^{2}$

5 0

3 years ago