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Marizza181 [45]

3 years ago

15

A 2kg bowling ball is 2.5 meter off the ground on a post when it falls. Just before it reaches the ground, it is traveling 7m/s.

Assuming that there is no air resistance, which statement is true

1 answer:

shutvik [7]3 years ago

7 0

<span>The mechanical energy is conserved.

I hope this helps, good luck! :)</span>

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Water flows from a large drainage pipe at a rate of 950 gal/min. What is this volume rate of flow in (a) m3/s , (b) liters/min,

Paladinen [302]

Answer:

$0.05997\ m^3/s$

$3596.1395\ L/min$

$2.11647\ ft^3/s$

Explanation:

$1\ gal/min=\dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=0.05997\ m^3/s$

Volume rate of flow is $0.05997\ m^3/s$

$1\ gal/min=3.78541\ L/min\\\Rightarrow 950\ gal/min=950\times 3.78541=3596.1395\ L/min$

Volume rate of flow is $3596.1395\ L/min$

$1\ gal/min=\dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=2.11647\ ft^3/s$

Volume rate of flow is $2.11647\ ft^3/s$

6 0

3 years ago

Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista

soldi70 [24.7K]

Answer:

$v = 7793150 m/s$

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

$v = \frac{kQ}{r}$

Where $k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }$

and it is satisfy the superposition principle, thus

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}$

$v = 222.73v$

The electrical potential at 10 cm from charge 1 is:

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}$

$v = 395.44 v$

Since the work - energy theorem, we have:

$q\Delta v = \frac{mv^{2} }{2}$

where q is the electron's charge and m is the electron's mass

Therefore:

$v = \sqrt{\frac{2q\Delta v}{m} }$

$v = 7793150 m/s$

6 0

3 years ago

How to calculate ?Help!!!

Alona [7]

How to calculate what?

6 0

3 years ago

A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness

Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

$\sigma_h = \frac{Pd}{2t}$

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

$\sigma_l = \frac{Pd}{4t}$

The letters have the same meaning as before.

Then he hoop stress would be,

$\sigma_h = \frac{Pd}{2t}$

$\sigma_h = \frac{75 \times 8}{2\times 0.25}$

$\sigma_h = 1200psi$

And the longitudinal stress would be

$\sigma_l = \frac{Pd}{4t}$

$\sigma_l = \frac{75\times 8}{4\times 0.25}$

$\sigma_l = 600Psi$

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

$\tau_{max} = \frac{\sigma_h}{2}$

$\tau_{max} = \frac{1200}{2}$

$\tau_{max} = 600Psi$

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0

3 years ago

If you could change one property of a substance ( solid, liquid, and gas) you have to use in your daily life, what would it be a

shepuryov [24]

Answer

Soap into a liquid

Explanation:

4 0

3 years ago

Read 2 more answers