What happens when both sides of the tug-of-war rope are equal in force?What is the net force in this scenario?

Jaiden is writing a report about the structure of the atom. In her report, she says that the atom has three main parts and two s

USPshnik [31]

No because an atom consists of <u>two</u> main parts <em>and</em> <u>three</u> subatomic particles - protons, neutrons, electrons. Each one is smaller than an atom, therefore they are subatomic particles. An atom only requires protons and electrons to be an atom - e.g. Hydrogen has 1 proton and 1 electron. Neutrons do not affect the overall charge of the atom, and only increase the atomic mass.

7 0

2 years ago

Read 2 more answers

A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at

GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0

1 year ago

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init

Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

$h=\frac{gt^2}{2}$

For second stone

$h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$ ---2

Equating 1 &2 we get

$\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$

$\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$26.46t-35.721-52.92t+142.884=0$

$t=4.05 s$

4 0

2 years ago

An infinite charged wire with charge per unit length lambda lies along the central axis of a cylindrical surface of radius r and

serg [7]

<h2>The flux through the infinite charged wire along the central axis of a cylindrical surface of radius r and length l is ∅E = E x 2πrl </h2>

Explanation:

let us consider a thin infinitely long straight wire having a uniform charge density λ Cm⁻¹.To determine the field at a distance r from the line charge , we have cylindrical gaussian surface of radius r, length l,and with its axis along the line charge. it has curved surface S₁ , and flat circular ends S₂ and S₃. Obviously, dS₁//E, dS₂ ⊥E , and dS₃ ⊥ E , so, only the curved surface contributes towards the total flux.

∅E = ∫ E.dS = ∫E.dS₁ +∫E.dS₂ +∫E.dS₃

= ∫EdS₁ cos0⁰ +∫EdS₂ cos 90⁰ +∫Eds₃ cos 90⁰

= E∫ds₁₁ +0+0

= E x area of curved surface

∅E = E x 2πrl

3 0

2 years ago

A Porsche challenges a Honda to a 200-m race.Because the Porsche's acceleration of 3.5 m/s2 is larger than the Honda's 3.0 m/s2,

Blizzard [7]

Answer:

Honda won by 0.14 s

Explanation: