What are the four kinds of air uplift?

Soft ball,kick ball,wiffle ball??

Anvisha [2.4K]

1. No they aren’t because they all belong to different sports and are used differently

6 0

2 years ago

Read 2 more answers

Define the term DNA

andrew11 [14]

Answer:

It is the carrier of genetic information.

4 0

3 years ago

Read 2 more answers

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

What does itmean to have an acceleration of 8 m/s2

Masteriza [31]

Answer:

The answer is option A.

You speed up 8 m/s every second

Hope this helps you

8 0

2 years ago

Read 2 more answers

A 1. 18 kg gold cube hangs at the end of a 4. 00 m long string. Rhogold = 19. 3 × 103 kg/m3; rhomercury = 13. 6 × 103 kg/m3. Whe

VashaNatasha [74]

When the gold cube is immersed in mercury, the tension in the string in Newtons is 3.142N.

<h3>What is tension?</h3>

Tension is the force acting on the linear object like string, chain or rope due to pulling.

Volume of gold V = mass / density

V = 1.18 /19.3x 10³ =61.1 x 10⁻⁶ m³

Tension in the string after immersing will be

T = [ρ(Gold) -ρ(Hg)] g. V

T =[ 19.3x 10³ - 13.6 x 10³] x 9.81 x 61.1 x 10⁻⁶

T =3.416 N
Thus, the tension in the string is 3.42 N.

Learn more about tension.

brainly.com/question/4087119

#SPJ4

6 0

2 years ago