What are wind names according to direction

How far will a free falling object fall in 8.7 secs if it started from rest? Remember acceleration is negative for free fall. Do

sleet_krkn [62]

Answer:

h~=371.26m

Explanation:

when an object falls we use the equations of accelerated motion. There is only one that gives distance.

$x = ut + \frac{1}{2} a {t}^{2}$

Since we have no initial velocity (started from rest) we can get rid of the (ut) term

where a we substitute g (gravitational acceleration, constant for given heights and almost 9.81m/s^2).

$h = \frac{1}{2} g {t}^{2} = \frac{1}{2} \times 9.81 \times {8.7}^{2} = 371.26m$

4 0

3 years ago

Consider two soap bubbles with radius r1 and r2 (r1 <r2) connected via a valve. What happens if we open the valve

Sedaia [141]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The pressure difference of the first bubble is $\Delta P _1 =10 J/m^3$

The pressure difference of the second bubble is $\Delta P _2 =20 J/m^3$

The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger

Explanation:

From the question we are told that

The radius of the first bubble is $r_1 = 10 \ mm=0.01 \ m$

The radius of the second bubble is $r_2 = 5 \ mm = 0.005 \ m$

The surface tension of the soap solution is $s = 25 \ mJ/m^2 = 25*10^{-3} J/m^2$

Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as

$\Delta P = \frac{4 s}{R}$

Now the pressure difference for the first bubble is mathematically evaluated as

$\Delta P _1 = \frac{4 s}{r_1}$

substituting values

$\Delta P _1 = \frac{4 *25 *10^{-3}}{0.01}$

$\Delta P _1 =10 J/m^3$

Now the pressure difference for the second bubble is mathematically evaluated as

$\Delta P _2 = \frac{4 s}{r_1}$

$\Delta P _2 = \frac{4 *25 *10^{-3}}{0.005}$

$\Delta P _2 =20 J/m^3$

3 0

3 years ago

A 0.453 kg pendulum bob passes through the lowest part of its path at a speed of 2.58 m/s. What is the tension in the pendulum c

harkovskaia [24]

Answer with Explanation:

Mass of pendulum bob, m=0.453 kg

Speed, $v_1=$ 2.58 m/s

a.r=75.1 cm= $75.1\times 10^{-2}m$ =0.751 m

$1cm=10^{-2} m$

Tension in the pendulum cable is given by

Tension=Centripetal force+force due to gravity

$T=\frac{mv^2}{r}+mg$

Where $g=9.8 m/s^2$

Substitute the values

$T=\frac{0.453(2.58)^2}{75.1\times 10^{-2}}+0.453\times 9.8$

$T=8.45 N$

b.When the pendulum reaches its highest point,then

Final velocity, $v_2=0$

According to law of conservation of energy

$mgh_1+\frac{1}{2}mv^2_1=mgh_2+\frac{1}{2}mv^2_2$

$gh_1+\frac{1}{2}v^2_1=gh_2+\frac{1}{2}v^2_2$

$h_1=0$

Substitute the values

$9.8\times 0+\frac{1}{2}(2.58)^2=9.8\times h_2+\frac{1}{2}(0)^2$

$3.3282=9.8h_2$

$h_2=\frac{3.3282}{9.8}=0.34 m$

The angle mad by cable with the vertical= $cos\theta=\frac{0.751-0.34}{0.751}=0.55$

$\theta=cos^{-1}(0.55)=56.6^{\circ}$

c.When the pendulum reaches at highest point then

Acceleration, a=0

Therefore, the tension in the pendulum cable

$T=mgcos\theta$

Substitute the values

$T=0.453\times 9.8cos56.6$

$T=2.4 N$

8 0

3 years ago

A power plant uses Uranium<br> to produce energy

antoniya [11.8K]

Answer:

how is that a question?

Explanation:

yeah i dunno the answer cause thats not a question

7 0

3 years ago

A mercedes-benz 300sl (m 5 1700 kg) is parked on a road that rises 158 above the horizontal. what are the magnitudes of (a) the

vaieri [72.5K]

The normal force of a certain thing is the weight in force of the object multiplied by the sine function of the angle of inclination, be ∅ in this case. Given the mass of the car equal to 300 slug or 5.17 kg, then the nrmal force is 5.17*9.81 sin ∅ or 50.72 sin ∅. The frictional force is normal force times ∨k or coefficient of friction

7 0

3 years ago