40 POINTS!!!

A net force of 3000 N is accelerating a 1200 kg elevator upward. If the elevator starts from rest, how long will it take to trav

alexdok [17]

F = net force acting on the elevator in upward direction = 3000 N

m = mass of the elevator = 1200 kg

a = acceleration of the elevator = ?

Acceleration of the elevator is given as

a = F/m

a = 3000/1200

a = 2.5 m/s²

v₀ = initial velocity of the elevator = 0 m/s

Y = displacement of the elevator = 15 m

t = time taken

Using the kinematics equation

Y = v₀ t + (0.5) a t²

15 = (0) t + (0.5) (2.5) t²

t = 3.5 sec

5 0

3 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

7 0

4 years ago

Instructions

RideAnS [48]

A group of students toured a limestone cave in northwest Georgia.
Which of these best explains how the limestone caves in Georgia were formed?
A. Plant roots split the rock.

B. Acidic water dissolved the rock.

C. Animals burrowed into the rock.

D. Ice formed and broke up the rock.

7 0

3 years ago

Help me bro, thanks all

bagirrra123 [75]

C beautiful yellow silk
D was sleeping

3 0

3 years ago

light waves are first transmitted through the ________ at the front of the eye and enter an opening called the ________ before s

viva [34]

The transmission of light waves is usually done through cornea of the eyes, then move through another opening which is regarded as pupil before it will get to the retina.

Light waves can be regarded as moving energy which contains microscopic particles known as photons.

The vision of the eye can be completed through the light wave passing through the components of the eyes and this process goes thus;

Light will move through the (cornea) which is situated at the front area of the eyes into lens.

Then both the cornea and the lens give room for the focusing of the light rays to the retina which is situated at the back of the eye .

Then through the help of the cells in the retina, the light will be absorbed and then be converted to electrochemical impulses and then transfer it to the brain as well as optic nerve.

Therefore, light wave are form of tiny microscopic particles.

brainly.com/question/19734585?referrer=searchResults

8 0

2 years ago