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4 years ago

A 3.00-kg pendulum is 28.84 m long. what is its period on earth?

1 answer:

7 0

$T=2 \pi \sqrt{ \frac{l}{g} }$
This is the period of a pendulum. Notice the mass doesn't matter. Plugging in your values and 9.81m/s^2 for g, gives
T=10.7731s.

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Which of the following is not a tectonic activity?

viktelen [127]

Answer:

Erosion of Mountains

Explanation:

5 0

3 years ago

Read 2 more answers

Consider an ideal monatomic gas of N particles with mass m in thermal equilibrium at a temperature T. The gas is contained in a

harina [27]

Answer:

$K.E.=\dfrac{3}{2}KT$

Explanation:

Given that

Number of particle =N

Equilibrium temperature= T

Side of cube = L

Gravitational acceleration =g

The kinetic energy of an atom given as

$K.E.=\dfrac{3}{2}KT$

Where

Equilibrium temperature= T

Boltzmann constant =K

K =1.380649×10−23 J/K

3 0

3 years ago

visible light is a range of ____ energy EM waves in the electromagnetic spectrum that the human eye can see

Artemon [7]

A range of infrared wavelengths

4 0

3 years ago

The electric field between two parallel plates is uniform, with magnitude 646 N/C. A proton is held stationary at the positive p

Ahat [919]

Answer:

The distance from the positive plate at which the two pass each other is 0.0023 cm.

Explanation:

We need to find the acceleration of each particle first. Let's use the electric force equation.

$F=Eq$

$ma=Eq$

For the proton

$m_{p}a_{p}=Eq_{p}$

$a_{p}=\frac{Eq_{p}}{m_{p}}$

$a_{p}=\frac{646*1.6*10^{-19}}{1.67*10^{−27}}$

$a_{p}=6.19*10^{10}\: m/s^{2}$

For the electron

$m_{e}a_{e}=Eq_{e}$

$a_{e}=\frac{Eq_{e}}{m_{e}}$

$a_{e}=\frac{646*1.6*10^{-19}}{9.1*10^{−31}}$

$a_{e}=1.14*10^{14}\: m/s^{2}$

Now we know that the plate separation is 4.26 cm or 0.0426 m. The travel distance of the proton plus the travel distance of the electron is 0.0426 m.

$x_{p}+x_{e}=0.0426$

Both of them have an initial speed equal to zero. So we have:

$\frac{1}{2}a_{p}t^{2}+\frac{1}{2}a_{e}t^{2}=0.0426$

$t^{2}(a_{p}+a_{e})=2*0.0426$

$t^{2}=\frac{2*0.0426}{a_{p}+a_{e}}$

$t=\sqrt{\frac{2*0.0426}{6.19*10^{10}+1.14*10^{14}}}$

$t=2.73*10^{-8}\: s$

With this time we can find the distance from the positive plate (x(p)).

$x_{p}=\frac{1}{2}a_{p}t^{2}$

$x_{p}=\frac{1}{2}6.19*10^{10}*(2.73*10^{-8})^{2}$

$x_{p}=0.0023\: cm$

Therefore, the distance from the positive plate at which the two pass each other is 0.0023 cm.

I hope it helps you!

7 0

3 years ago

How far does light travel in the time it takes sound to go 1 cm?

aksik [14]

Sound travels at speed v=343 m/s, therefore the time the sound takes to cover a distance of
$S= 1 cm=0.01 m$
is
$t= \frac{S}{v}= \frac{0.01 m}{343 m/s}=2.9 \cdot 10^{-5} s$

Light travels at speed $c=3 \cdot 10^8 m/s$ , therefore the distance it travels during the same time t is
$S=ct=(3 \cdot 10^8 m/s)(2.9 \cdot 10^{-5} s)=8746 m=8.75 km$

6 0

3 years ago