Which combination of elements below will bond ionically and in the same

A car starts from rest at a stop sign. It accelerates at 4.0 m/s^2 for 3 seconds, coasts for 2 s, and then slows down at a rate

zubka84 [21]

FIRST SECTION You should use the formula for uniformly accelerated linear movement. Initial speed is 0 because it starts from rest. d=(1/2)*a*t^2+vo*t =(1/2)*(4.0 m/s^2)*(3s)^2+0*3s=(1/2)*(4.0 m/s^2)*3^2*s^2+0=2.0 m*9=18m You can calculate the final speed with the other formula: v=a*t+vo=(4.0 m/s^2)*(3s)+0=(4.0 m/s)*(3)=12m/s SECOND SECTION You should use the formula for uniform linear movement. Velocity is a constant: it remains in 12m/s. d=v*t=12m/s*2s=12m*2=24m THIRD SECTION We should use the same formulas as the first section, but with different numbers. Initial velocity will be 12m/s, and then velocity will start to decrease until it gets to 0. We don’t know what the time is for this section. Acceleration is negative, because it’s slowing down. v=a*t+vo 0=-3.0 m/s^2*t+12m/s 3.0 m/s^2*t=12m/s t=(12m/s)/(3.0 m/s^2)=4(1/s)/(1/s^2)=4s^2/s=4s Now let’s use that time in the other formula: d=(1/2)*a*t^2+vo*t =(1/2)*(-3.0 m/s^2)*(4s)^2+(12m/s)*3s=(-1.5 m/s^2)*4^2*s^2+12*3m*s/s=-1.5 m*4^2+36m=-1.5*16m+36m=-24m+36m=12m Now let’s add the 3 stages: d=18m+24m+12m=54m

5 0

3 years ago

Read 2 more answers

a 25 newton force applied on an object moves it 50 meters. the angle between the force and displacement is 40.0°. what is the va

LiRa [457]

Work = force × distance × cos(angle)

work = (25)(50)(cos (40))

work = 957.56 Joules

= 9.6x10^2 Joules

3 0

3 years ago

Which one will it be

liq [111]

Answer:

none

Explanation:

it's to high up to be affected by the gravity

3 0

2 years ago

Read 2 more answers

Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration

eimsori [14]

The complete text of the problem is:

"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "

Solution:

1) Acceleration: $-2336 m/s^2$ on the hardwood floor, $-382 m/s^2$ on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

$v^2 - u^2 = 2ad$

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

a = g = 9.8 m/s^2 is the acceleration of gravity

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

$v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s$

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

$d = 1.8 mm = 0.0018 m$

So we can find the acceleration by using again the equation

$v'^2 - v^2 = 2ad$

Solving for a,

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2$

For the carpeted floor instead,

$d=1.1 cm = 0.011 m$

therefore the acceleration is

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2$

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

$a=\frac{v'-v}{t}$

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

$a=-2336 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms$

For the carpeted floor,

$a=-382 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms$

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

8 0

3 years ago

which type of lightening device would you use for each of the following needs: an economical light source in a manufacturing pla

Yakvenalex [24]

Btw only someone who is nice will answer tour question. You can't expect for explanition when the question is only worth 5 points. Not trying to be mean sorry if i am being mean

7 0

3 years ago