Ask question

Ask question

All categories

3 years ago

5

A 6 kg bowling ball moves with a speed of 3 m/s. How fast does a 7 kg bowling ball need to move so that it has the same kinetic

energy?

1 answer:

maw [93]3 years ago

4 0

Answer: 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy.

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion.

$K.E=\frac{1}{2}mv^2$

m = mass of object

v= velocity of the object

$K.E=\frac{1}{2}\times 6kg\times (3m/s)^2=27Joules$

b) for a 7 kg bowl to have kinetic energy of 27 Joules:

$27J=\frac{1}{2}\times 7kg\times v^2$

$v^2=7.7$

$v=2.8m/s$

Thus 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy

You might be interested in

Describe an experiment to show that air support burning

noname [10]

Take a small burning candle. ... After few minutes the candle is extinguished. As the supply of air is stopped due to glass jar the burning of candle is also stopped. This experiment proves that air supports burning.

3 0

3 years ago

Atoms with many electron shells will let go of their electrons more easily than those with fewer shells.

Anna007 [38]

True

The more the number of shells will let go of their outer electrons more easily because the effective nuclear charge on the outer (valence) electrons will be lower. This is called 'shielding', the outer electrons will be shielded from the nucleus by the inner electrons.

Hope this Helps

8 0

3 years ago

Read 2 more answers

A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general

valentina_108 [34]

Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength $\lambda_i$

• The wave does not satisfy the boundary conditions $y_i(0;t)=0
$

Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions, $y(0,1)=0
$

and $y(L,t)=0$

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength $\lambda_i$ can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, $y_i(0;t)=0$ and $y_i(L;t)=0$ , that are satisfied only for particular value of $\lambda_i$ .

Given below are the incorrect options about the wave in the string.

• $A_i$ must be chosen so that the wave fits exactly o the string.

• Any one of $A_i$ or $\lambda_i$ or $f_i$ can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies $f_i$ and the wavelength $\lambda_i$ must be such that $y_i(0;t) = y_i(L;t)=0
$

(C)

Expression for the wavelength of the various normal modes for a string is,

$\lambda_n=\frac{2L}{n}$ (1)

When $n=1$ , this is the longest wavelength mode.

Substitute 1 for n in equation (1).

$\lambda_n=\frac{2L}{1}\\\\2L$

When $n=2$ , this is the second longest wavelength mode.

Substitute 2 for n in equation (1).

$\lambda_n=\frac{2L}{2}\\\\L$

When $n=3$ , this is the third longest wavelength mode.

Substitute 3 for n in equation (1).

$\lambda_n=\frac{2L}{3}$

Therefore, the three longest wavelengths are $2L$ , $L$ and $\frac{2L}{3}$ .

(D)

Expression for the frequency of the various normal modes for a string is,

$f_n=\frac{v}{\lambda_n}$

For the case of frequency of the $i^{th}$ normal mode the above equation becomes.

$f_i=\frac{v}{\lambda_i}$

Here, $f_i$ is the frequency of the $i^{th}$ normal mode, v is wave speed, and $\lambda_i$ is the wavelength of $i^{th}$ normal mode.

Therefore, the frequency of $i^{th}$ normal mode is $f_i=\frac{v}{\lambda_i}$

.

6 0

3 years ago

Suppose you have a total charge qtot that you can split in any manner. Once split, the separation distance is fixed. How do you

dybincka [34]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know from the Coulomb's Law that, Coulomb's force is directly proportional to the product of two charges q1 and q2 and inversely proportional to the square of the radius between them.

So,

F = $\frac{Kq1q2}{r^{2} }$

Now, we are asked to get the greatest force. So, in order to do that, product of the charges must be greatest because the force and product of charges are directly proportional.

Let's suppose, q1 = q

So,

if q1 = q

then

q2 = Q-q

Product of Charges = q1 x q2

Now, it is:

Product of Charges = q x (Q-q)

So,

Product of Charges = qQ - $q^{2}$

And the expression qQ - $q^{2}$ is clearly a quadratic expression. And clearly its roots are 0 and Q.

So, the highest value of the quadratic equation will be surely at mid-point between the two roots 0 and Q.

So, the midpoint is:

q = $\frac{Q + 0}{2}$

q = Q/2 and it is the highest value of each charge in order to get the greatest force.

8 0

3 years ago

Which is not a factor that affects the pressure of a gas in a closed container?

maksim [4K]

it is size of of particles because it does not matter about the size in a closed container

7 0

3 years ago