Calculate the speed for a car that went a distance of 125 kilometers in 2 hours time.

A racecar driver steps on the gas, and his racecar travels 20 meters in 2 seconds starting from rest. The acceleration of the ra

givi [52]

Answer:

2.5m/s^2

Explanation:

Step one:

given

distance = 20meters

time = 2 seconds

initial velocity u= 0m/s

let us solve for the final velocity

velocity = distance/time

velocity= 20/2

velocity= 10m/s

$v^2=u^2+2as\\\\10^2=0^2+2*a*20\\\\100=40a$

divide both sides by 40

$a= 100/40\\\\a=10/4\\\\a= 5/2\\\\a= 2.5m/s^2$

5 0

3 years ago

Hooke’s law describes the linear relationship between stress and strain through Young’s modulus. Given two materials under the s

stiks02 [169]

Answer:

The material with higher modulus will stretch less than

The material with lower modulus

Explanation:

A material with a higher modulus is stiffer and has better resistance to deformation. The modulus is defined as the force per unit area required to produce a deformation or in other words the ratio of stress to strain.

E= stress/stain

Hooks law states that provided the elastic limit is not exceeded the extension e of a spring is directly proportional to the load or force attached

F=ke

Where k is the constant which gives the measure of the spring under tension

3 0

3 years ago

Read 2 more answers

Suppose you lived in a pre-industrial society and needed to lift a heavy (20 kg) block a height of 5 m and had two choices for h

igomit [66]

Let's break the question into two parts:

1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.

1. Ramp Scenario:
In an incline, the only component of cart's weight(mg) that is in the direction of motion is $mgsin \alpha$ . Therefore the effort force in this case must be equal or greater than $mgsin \alpha$ .

Now we need to find $\alpha$ . $\alpha$ is the angle between the incline of the ramp and the ground.

Since the height is 5m and the length of the ramp is 8m, $sin \alpha$ would be 5/8 or 0.625. Now that you have $sin \alpha$ , mutiple it with mg.

=> m*g* $sin \alpha$ = 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is 125N.

2. Lever Scenario:
Just apply "moment action" in this case, which is:
$F_{e} d_{e} = F_{r} d_{r}$

$F_{e}$ = ?

$F_{r}$ = mg = 20 * 10 = 200N
$d_{e}$ = 10m
$d_{r}$ = 1m

Plug-in the values in the above equation:
$F_{e}$ = 200/10= 20N

As 20N << 125N, the best choice is to use lever.

4 0

3 years ago

HELP ME please

allochka39001 [22]

Answer:

Ruko zara kuch Time dedo na please

3 0

2 years ago

A heavy rope, 80 ft long and weighing 32 lbs, hangs over the edge of a building 100 ft high. how much work w is done in pulling

allochka39001 [22]

The first thing you should know for this case is that work is defined as the product of force by the distance traveled in the direction of force.
We have then:
W = Fd
The distance varies, so we must integrate:
from 0 to 20:
W = ∫F (x) dx
W = ∫32xdx
W = 32∫xdx
W = 32 (x ^ 2/2) = (16) (20 ^ 2) = 6400 ft * lbs
answer:
6400 ft * lbs is work done pulling the rope up 20 ft

6 0

3 years ago