The famous leaning tower of pisa doesn't topple over because its center of gravity is

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

3 years ago

If a photon has frequency = 2.00 x 1014s-1 and the speed of light = 3.00 x 108ms-1, then what is its wavelength?

butalik [34]

Answer:

The photon has a wavelength of $1.5x10^{-6}m$

Explanation:

The speed of a wave can be defined as:

$v = \nu \cdot \lambda$ (1)

Where v is the speed, $\nu$ is the frequency and $\lambda$ is the wavelength.

Equation 1 can be expressed in the following way for the case of an electromagnetic wave:

$c = \nu \cdot \lambda$ (2)

Where c is the speed of light.

Therefore, $\lamba$ $\lambda$ can be isolated from equation 2 to get the wavelength of the photon.

$\lambda = \frac{c}{\nu}$ (3)

$\lambda = \frac{3.00x10^{8}m/s}{2.00x10^{14}s^{-1}}$

$\lambda = 1.5x10^{-6}m$

Hence, the photon has a wavelength of $1.5x10^{-6}m$

<em>Summary: </em>

Photons are the particles that constitutes light.

3 0

3 years ago

A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is

Rudik [331]

Answer:

$a=16.2m/s^{2}$

Explanation:

From the attached file diagram, the total force acting on the charged box is the downward weight and the repulsive force acting in opposite to the weight force . Hence we can write the total force as

$F=masin\alpha -\frac{kQq}{r^{2}} \\\alpha =35^{0}, m=0.495kg, r=0.61m, Q=2.5*10^{-6}, q=75.0*10^{-6}\\$

When fixed,F=o

Hence

$masin\alpha =\frac{kQq}{r^{2}}\\0.495kg*asin35=\frac{9*10^{9}*2.5*10^{-6}*75.0*10^{-6}}{0.61^{2}} \\0.28a=4.5351\\a=\frac{4.5351}{0.28}\\\\ a=16.2m/s^{2}$

The value of the acceleration is 16.2m/s^2

7 0

3 years ago

Assume that the function x(t) represents the length of tape that has unwound as a function of time. find θ(t), the angle through

bekas [8.4K]

We know that arc length (x(t)) is given with the following formula:
$x(t)=\theta(t) r$
Where r is the radius of the barrel. We must keep in mind that as barrel rolls its radius decreases because less and less tape is left on it.
If we say that the thickness of the tape is D then with every full circle our radius shrinks by d. We can write this down mathematically:
$r(\theta)=r_0-\frac{D\cdot \theta}{2\pi}$
When we plug this back into the first equation we get:
$x(t)=\theta(r_0-\frac{D\theta}{2\pi}})\\ \frac{D\theta^2}{2\pi}-\theta r_0+x(t)=0\\$
We must solve this quadratic equation.
The final solution is:
$\theta=\frac{\pi r_0+\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D},\:\theta=\frac{\pi r_0-\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D}$
It is rather complicated solution. If we asume that the tape has no thickness we get simply:
$x(t)=\theta(r_0-\frac{D\theta}{2\pi}});D=0\\
x(t)=\theta r_0\\
\theta(t)=\frac{x(t)}{r_0}$

8 0

3 years ago

Sports car goes from a velocity of zero to a velocity of 12 m/s East in two seconds. What is the cars acceleration?

bogdanovich [222]

Answer:

6 m/sec

Explanation:

12/2=6

7 0

3 years ago