In a collision that is not perfectly elastic, what happens to the mechanical energy of the system?

Which statement about force is incorrect

garik1379 [7]

Answer:

What are the options?

Explanation:

4 0

3 years ago

A string of length L, mass per unit length \mu, and tension T is vibrating at its fundamental frequency. What effect will the fo

viva [34]

The fundamental frequency on a vibrating string is given by:

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension

$\mu$ is the mass per unit length of the string

Keeping this equation in mind, we can now answer the various parts of the question:

(a) The fundamental frequency will halve

In this case, the length of the string is doubled:

L' = 2L

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2(2L)}\sqrt{\frac{T}{\mu}}=\frac{1}{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{2}$

So, the fundamental frequency will halve.

(b) the fundamental frequency will decrease by a factor $\sqrt{2}$

In this case, the mass per unit length is doubled:

$\mu'=2\mu$

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2L}\sqrt{\frac{T}{2 \mu}}=\frac{1}{\sqrt{2}}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{\sqrt{2}}$

So, the fundamental frequency will decrease by a factor $\sqrt{2}$ .

(c) the fundamental frequency will increase by a factor $\sqrt{2}$

In this case, the tension is doubled:

$T'=2T$

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2L}\sqrt{\frac{2T}{\mu}}=\sqrt{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\sqrt{2}f$

So, the fundamental frequency will increase by a factor $\sqrt{2}$ .

8 0

3 years ago

A 5 m3 tank containing 5kg of an unknown ideal gas at 500 kPa is connected through a valve to another tank containing 10 kg of t

Ivan

Answer:

a) $V_{T} = 9\,m^{2}$ , b) $m_{T} = 15\,kg$ , c) $P_{T} = 416.667\,kPa$

Explanation:

a) The equation of state for ideal gas is:

$P \cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

Given the existence of an isothermal process, the following relation is derived:

$\frac{P_{1}\cdot V_{1}}{m_{1}} = \frac{P_{2}\cdot V_{2}}{m_{2}}$

The volume of the other tank is:

$V_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}}\right)\cdot V_{1}$

$V_{2} = \left(\frac{10\,kg}{5\,kg} \right)\cdot \left(\frac{200\,kPa}{500\,kPa}\right)\cdot (5\,m^{3})$

$V_{2} = 4\,m^{3}$

The total volume is:

$V_{T} = V_{1} + V_{2}$

$V_{T} = 5\,m^{3} + 4\,m^{3}$

$V_{T} = 9\,m^{2}$

b) The total mass is:

$m_{T} = m_{1} + m_{2}$

$m_{T} = 5\,kg + 10\,kg$

$m_{T} = 15\,kg$

c) The pressure of the gas in the two tanks is:

$P_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{V_{1}}{V_{2}}\right)\cdot P_{1}$

$P_{T} = \left(\frac{15\,kg}{5\,kg}\right)\cdot \left(\frac{5\,m^{2}}{9\,m^{2}} \right)\cdot (500\,kPa)$

$P_{T} = 416.667\,kPa$

3 0

4 years ago

The mass of an object is 60kg on the surface of the earth what will be its weight on the surface of the moon

iris [78.8K]

Answer:

Wm = 97.2 [N]

Explanation:

We must make it clear that mass and weight are two different terms, the mass is always preserved that is to say this will never vary regardless of the location of the object. While weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

m = mass = 60 [kg]

g = gravity acceleration = 10 [m/s²]

But in order to calculate the weight of the body on the moon, we must know the gravitational acceleration of the moon. Performing a search of this value on the internet, we find that the moon's gravity is.

gm = 1.62 [m/s²]

Wm = 60*1.62

Wm = 97.2 [N]

8 0

3 years ago

A speed skater just finished a race. After she crossed the finish line, she coasted to a complete stop. If her initial speed was

romanna [79]

Answer:

5seconds

Explanation:

because she was going through the same time

6 0

2 years ago