Which is greater, an acceleration from

A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os

Bingel [31]

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

$Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}$

$\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}$

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

$k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\$

4 0

3 years ago

According to the kinetic molecular therory, the pressure of a gas in a container will increase if the

Masja [62]

According to the kinetic molecular theory, the pressure of a gas in a container will increase if the number of collisions with the container wall increases.

<u>Explanation:</u>

Keeping the volume of the vessel constant, if we increase the amount of gas in it; the pressure will increase. This is because when the number of gas particles increases in that limited volume, they hit the walls of the container with more energy and hence, the overall pressure of the gas increases.

If we decrease the amount of gas in the vessel or increase the volume for the same amount of gas, the pressure decreases. As the pressure inside the vessel depends upon the gas supplied in the container.

5 0

3 years ago

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

3 years ago

From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s

Svetradugi [14.3K]

Solution:

Let the slope of the best fit line be represented by ' $m_{best}$ '

and the slope of the worst fit line be represented by ' $m_{worst}$ '

Given that:

$m_{best}$ = 1.35 m/s

$m_{worst}$ = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

$\Delta m = \frac{m_{best}-m_{worst}}{2}$ (1)

Substituting values in eqn (1), we get

$\Delta m = \frac{1.35 - 1.29}{2}$ = 0.03 m/s

8 0

3 years ago

Which person is weightless?

ahrayia [7]

Answer: Option (b) is the correct answer.

Explanation:

The force of gravity acting on an object helps in determining the weight of an object. But a place where there will be no gravity or have zero gravitational pull then it means the person will be weightless.

For example, force of gravity on moon is zero which means any object or person on moon will be weightless.

On the other hand, when a child is in the air as she plays on a trampoline then it means gravitational pull form the earth is acting on it. So, it will definitely has some weight.

Similarly, a scuba diver exploring a deep-sea wreck is under the ground where there will be force of gravity. Hence, it will also have some weight.

Thus, we can conclude that an astronaut on the Moon is the person who is weightless.

7 0

3 years ago