The amount of matter contained in a given volume for a substance is called

A pendulum on earth swings with angular frequency ω. On an unknown planet, it swings with angular frequency ω/ 4. The accelerati

Afina-wow [57]

Answer:

g / 16

Explanation:

T = 2π $\sqrt{\frac{l}{g} }$

angular frequency ω = 2π /T

= $\sqrt{\frac{g}{l} }$

ω₁ /ω₂ = $\sqrt{\frac{g_1}{g_2} }$

Putting the values

ω₁ = ω , ω₂ = ω / 4

ω₁ /ω₂ = 4

4 = $\sqrt{\frac{g}{g_2} }$

g₂ = g / 16

option d is correct.

6 0

3 years ago

How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A)

Elan Coil [88]

Answer:

Option (D) is correct.

Explanation:

Let the speed is v.

$\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c$

Option (D) is correct.

4 0

3 years ago

When one person shouts at a football game, the sound intensity level at the center of the field is 58.4 dB. When all the people

Tanzania [10]

Answer:

The number of people at game are approximately 22909

Explanation:

Given data

When one person shout $\beta _{1}=58.4dB$

When n number of person shout together $\beta _{n}=102dB$

The sound intensity level during one person shout is given by:

$\beta _{1}=10log(\frac{I_{1}}{I_{o}} )\\58.4=10log(\frac{I_{1}}{I_{o}} )\\5.84=log(\frac{I_{1}}{I_{o}} )\\\frac{I_{1}}{I_{o}} =10^{5.84}\\I_{1}=10^{5.84}*I_{o}$

The sound intensity level during n number of person shout is given by:

$\beta _{n}=10log(\frac{I_{n}}{I_{o}} )\\102=10log(\frac{I_{n}}{I_{o}} )\\10.2=log(\frac{I_{n}}{I_{o}} )\\\frac{I_{n}}{I_{o}}=10^{10.2}\\I_{n}=10^{10.2}*I_{o}$