Show that the 1 and 3 laws of motion are collection of 2 law of motion ?

An engine flywheel initially rotates counterclockwise at 6.55 rotations/s. Then, during 20.9 s, its rotation rate changes to 2.1

11Alexandr11 [23.1K]

Answer:

The average angular acceleration is -2.628 rad/s²

Explanation:

Counterclockwise = positive

Clockwise = -negative

Given;

initial rotation of the flywheel, θ₁ = 6.55 rotation/s

final rotation of the flywheel, θ₂ = - 2.19 rotation/s

The average angular acceleration is given by;

$\alpha = \frac{\delta \theta}{\delta t}\\\\ \alpha =\frac{\theta _2 - \theta_ 1}{t}\\\\ \alpha =\frac{-2.19 -6.55}{20.9} \\\\ \alpha =\frac{-8.74}{20.9}\\\\ \alpha = -0.4182 \ rotation / s^2\\\\ \alpha = \frac{-0.4182 \ rotation}{s^2}*\frac{2\pi \ radian}{rotation}\\\\ \alpha = -2.628 \ rad/s^2$

Therefore, the average angular acceleration is -2.628 rad/s²

7 0

3 years ago

The figure (Figure 1) shows the velocity of a solar-powered motorhome (RV) as a function of time. The driver accelerates from a

SpyIntel [72]

Explanation :

It is given that, the driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.

We know that acceleration is defined as the rate of change of velocity.

$a=\dfrac{dv}{dt}$

Where

dv is the change in velocity, dv = 0 - 60 m/s = -60 m/s

dt is the change in time, dt = 40 s - 30 s = 10 s

So, $a=\dfrac{-60\ m/s}{10\ s}$

$a = -6\ m/s^2$

From the graph it is clear that, from 30 s to 40 s the car is decelerating. So, at every second within this time the value of acceleration will be same i.e. $-6\ m/s^2$ .

6 0

3 years ago

Blank is how high or low you think a sound is

vfiekz [6]

<span>The word is "pitch", which is exactly that: How "high" or "low" a sound is.</span>

3 0

3 years ago

Read 2 more answers

Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe

AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

$Mm=44.0 g/mol$ is the molar mass of the carbon dioxide

So, the number of moles of the gas is

$n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol$

Now we can find the pressure of the gas by using the ideal gas equation:

$pV=nRT$

where

p is the pressure

$V=1.00 L = 0.001 m^3$ is the volume

n = 0.023 mol is the number of moles

$R=8.314 J/mol K$ is the gas constant

$T=25.0^{\circ}+273=298 K$ is the temperature of the gas

Solving the equation for p, we find

$p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa$

And since we have

$1 atm = 1.01\cdot 10^5 Pa$

the pressure in atmospheres is

$p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm$

5 0

3 years ago

A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t

julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

$F_k = F_{W,E}$