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Alekssandra [29.7K]

3 years ago

6

A rocket sled is tested at "5 g" (5 times the acceleration due to gravity). If the sled starts from rest at position d o = 0.00,

how long does it take to travel 441 meters? t = _____ s
Please explain your answer and show the formula you used because I do not understand!
A 3.00
B. 4.24
C. 9.00
D. 18.0

1 answer:

vivado [14]3 years ago

7 0

Answer:

It will take 4.24 seconds to travel 441 meters.

Explanation:

We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case the velocity of body in vertical direction = 0 m/s, acceleration = 5g = 5 x 9.8 = 49 , we need to calculate time when s = 441 meter.

So it will take 4.24 seconds to travel 441 meters.

Hope this helps plz mark brainly:)

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4 years ago

Read 2 more answers

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:

$a=\omega ^{2}r$

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

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

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3 years ago

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Karolina [17]

Answer:

1.00 m/s^2

Explanation:

The problem can be solved by using Newton's second law:

$F=ma$

where

F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, F=35 N (the force applied by Amber) and

$m=32 kg+3 kg=35 kg$

is the mass of the wagon + the brother, so the acceleration of the wagon and the brother is

$a=\frac{F}{m}=\frac{35 N}{35 kg}=1.00 m/s^2$

5 0

4 years ago