Hii Happy New year can you all please answer this

A large heavy truck smashes into a brick wall totally crumpling the front end. what is the conclusion of this description

Oduvanchick [21]

The person driving the truck was killed
the wall was destroyed

8 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

An airplane is carrying 7575 passengers, which is 5656 of the capacity of the airplane. what is the capacity of the airplane?

BARSIC [14]

There is something wrong, 56 supposed to be 5/6 and 7575 is only 75 passengers.
So to solve for this problem:If an airplane carries 75 passengers and if it is 5/6 of the capacity which is x, then we can say that:

5/6 * x = 75

5x/6 = 75

5x = 75 x 6

5x = 450

5x/5 = 450/5

x = 90
The capacity of the airplane is 90

5 0

4 years ago

Read 2 more answers

1. Suppose that you have two samples with the

grandymaker [24]

the one with has a greater mass will have a higher density because the more particles/mass in a particular space, the more dense it is.

hope i helped you!

mark brainliest please! :D

4 0

3 years ago

Please help! A Boeing 747 aircraft has a landing speed of 72 m/s , and upon landing it is able to come to a stop in 2, 000m . As

Andre45 [30]

27.8 seconds is how long it would take to land

4 0

3 years ago

Hii Happy New year can you all please answer this​

Hii Happy New year can you all please answer this