A car in front of the school goes from rest to 27 m/s in 3.0 seconds. What is its acceleration (assuming

When you ask you friend for hoemwork answers and he answers like a year later hahahah

Hitman42 [59]

G that’s a nice friend

5 0

3 years ago

An aircraft performs a maneuver called an "aileron roll." During this maneuver, the plane turns like a screw as it maintains a s

Dmitriy789 [7]

Answer:

0.17724 m/s²

Explanation:

D = Diameter of roll = Length of wing = 11 m

T = Time it takes to complete the circle = 35 s

Velocity

$v=\frac{2\pi R}{T}\\\Rightarrow v=\frac{\pi D}{T}\\\Rightarrow v=\frac{\pi\times 11}{35}\\\Rightarrow v=0.98735\ m/s$

Acceleration

$a=\frac{v^2}{R}\\\Rightarrow a=\frac{0.98735^2}{\frac{11}{2}}\\\Rightarrow a=0.17724\ m/s^2$

Acceleration of the tip of the plane is 0.17724 m/s²

3 0

3 years ago

If a 15g irregular shaped object is submerged in a graduated cylinder of water, the level rises from 10 ml to 25 ml, what is the

SashulF [63]

Answer:

one im so sry i have no idea. I have been researshing for about 30min and i cant find anything im so sry:/

Explanation:

6 0

4 years ago

in a closed system three objects have the following momentum: 11 kg* m/s, -65 kg*m/s and -100 kg m/s. the objects collide and mo

guajiro [1.7K]

Explanation:

The momentum of the three objects are as follow :

11 kg-m/s, -65 kg-m/s and -100 kg-m/s

Before collision, the momentum of the system is :

$P_i=11+(-65)+(-100)\\\\P_i=-154\ kg-m/s$

After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.

It would mean that, after collision, momentum of the system is equal to the initial momentum.

Hence, final momentum = -154 kg-m/s.

4 0

4 years ago

A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of 384,000 km. Sup

Tema [17]

Answer:

a) v = 19,149.6 m/s

b) f = 95%

c) t = 346.5min

Explanation:

First put all values in metric units:

$15.8 min*\frac{60s}{1min}=948s$

The equation of motion you need is:

$v_f = a*t+v_0$

where $v_f$ is the final velocity, a is acceleration and t is time in hours.

Since the spaceship starts from 0 velocity:

$v_f = a*t = 20.2*948 = 19,149.6 m/s$

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

$x=\frac{v_0+v_f}{2}t$