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3 years ago

12

Ccording to Newton's Second Law, what effect will doubling the mass of an object have?

1 answer:

lord [1]3 years ago

4 0

As the mass of an object is increased, the acceleration of the object is decreased.

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What is law of conservation of mass

timurjin [86]

Explanation:

<h2>law of conservation of mass</h2>

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

5 0

3 years ago

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

Irina-Kira [14]

Answer:

part (a) $\alpha\ =\ -2.5\ rad/s^2$

part (b) N = 79.61 rev

part (c) $\tau\ =\ 23.54\ Nm$

Explanation:

Given,

Initial speed of the wheel = $w_o\ =\ 50.0\ rad/s$
total time taken = t = 20.0 sec

part (a)

Let $\alpha$ be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

$\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2$

part (b)

Let $\theta$ be the total angular displacement of the wheel from initial position till the rest.

$\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad$

We know, 1 revolution = $2\pi$ rad

Let N be the number of revolution covered by the wheel.

$\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev$

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

Mass of the pole = m = 4 kg
Length of the pole = L = 2.5 m
Angle of the pole with the horizontal axis = $\theta\ =\ 60^o$

Now the center of mass of the pole = $d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m$

Weight component of the pole perpendicular to the center of mass = $F\ =\ mgcos\theta$

$\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm$

3 0

4 years ago

A 100 kg box is hanging from two strings. String #1 pulls up and left, making an angle of 80o with the horizontal on the left, a

frutty [35]

Answer:296.76 N

Explanation:

Given

mass of box $m=100 kg$

Let $T_1$ be the Tension in left side and $T_2$ be the Tension in the right side

From diagram

$T_1\cos 80=T_2\cos 65$

$T_1=T_2\cdot \frac{\cos 65}{\cos 80}$

and

$T_1\sin 80+T_2\sin 65=100\cdot g$

$T_2\left [ \tan 80\cdot \cos 65+\sin 65\right ]=100\cdot g$

$T_2=\frac{100\cdot g}{\left [ \tan 80\cdot \cos 65+\sin 65\right ]}$

$T_2=\frac{980}{3.3023}=296.76 N$

3 0

4 years ago

A pulley system has an MA of 4. How much force would be necessary to pull a 200 newton box

Airida [17]

Mechanical advantage is defined as

$MA = \frac{F_{output}}{F_{input}}$

MA = 4

now we will have

$F_{output} = 200 N$

as we need to pull up a box of weight 200 N

now by above formula we can find how much force it requires to pull up the box

$4 = \frac{F_{output}}{F_{input}}$

$4 = \frac{200}{F_{input}}$

$F_{input} = \frac{200}{4} = 50 N$

so it requires 50 N force to pull the weight of 200 N

7 0

3 years ago

Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 4.5 m long, its mass is 0.4

lions [1.4K]

Answer:

density = 0.0933 kg/m

speed = 27.581 m/s

Explanation:

given data

length L = 4.5 m

mass m = 0.42 kg

force F = 71 N

to find out

mass density and speed

solution

we find linear mass density

linear mass density = mass / length

put here all value

density = 0.42 / 4.5

density = 0.0933 kg/m

and

speed of wave

speed = √(F/density)

speed = √(0.42/0.933)

speed = 27.581 m/s

7 0

3 years ago