Since the stone was dropped from height, its initial velocity = 0 m/s

Using v² = u² + 2gs.

Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m

v² = u² + 2gs

v² = 0² + 2*10*2.5

v² = 0 + 50

v² = 50

v = √50

v ≈ 7.07 m/s

Hence velocity just before hitting the ground is ≈ 7.07 m/s

6 0

3 years ago

Which term describes the increase in a material's volume due to an increase in temperature?

AlladinOne [14]

Thermal Expansion. best describes the increase in a materials volume

7 0

3 years ago

Read 2 more answers

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

7 0

4 years ago

List small/average stars<br><br>

mario62 [17]

Answer:

Lol, you should do Nate, Bobby, Cindy, Joe, and Beth

Jk, if you want to be series and probably not fail go for these:

If it wants types of small/average stars, then go with

Small star names:

OGLE-TR-122B

Gliese 229 B

TRAPPIST-1

Teegarden's Star

Luyten 726-8 (A and B)

Proxima Centauri

Wolf 359 111400

Ross 248

Barnard's Star

CM Draconis B

Ross 154 167000

CM Draconis A

Kapteyn's Star

7 0

3 years ago