What's the definition energy?

A piece of plastic has a net charge of +9.9 μC. How many more protons than electrons does this piece of plastic have? (e = 1.60

Leto [7]

Answer:

The number of protons 6.19 more than electron.

Explanation:

Given that,

Charge $Q=+9.9\times10^{-6}\ C$

We know that,

formula of charge

$Q = nc$

Where,

Q = total charge

n = number of protons

e = charge of electron

Put the value into the formula

$n = \dfrac{Q}{e}$

$n=\dfrac{9.9\times10^{-6}}{1.6\times10^{-19}}$

$n=6.1875\times10^{13}\ C$

According to statement of question

Divide the answer by $10^{13}$

$n=\dfrac{6.1875\times10^{13}}{10^{13}}$

$n=6.19\ C$

Hence, The number of protons 6.19 more than electron.

6 0

3 years ago

A basketball has a coefficient of restitution of 0.821 in collisions with the wood floor of a basketball court. The ball is drop

Tanya [424]

Answer: The height of its fourth bounce = 0.43m

Explanation:

The coefficient of restitution denoted by (e), is the ratio that shows the final velocity to initial relative velocity between two objects after collision

IT is given by the formula in terms of height as

Coefficient of Restitution, e = √(2gh))/√(2gH) = √(h/H)

Where

Coefficient of Restitution, e= 0.821

H = 2.07 m

At fourth bounce , we have that

Coefficient of Restitution, e⁴ =√(h₄/H)

Putting the given values and solving , we have,

e⁴ =√(h₄/H)

= 0.821⁴ = √(h₄/2.07)

(0.821⁴ )² =h₄/2.07

0.2064 x 2.07 = 0.427 = 0.43

At fourth bounce, h₄ height = 0.43m

7 0

3 years ago

Two sections, A and B, are 0.5 km apart along a 0.05 m diameter rough concrete pipe. A is 115 m higher than B, the water tempera

KonstantinChe [14]

Answer:

Q = 178.41 m^3 / s

Explanation:

Given:

Length of the pipe L = 0.5 km
Diameter D = 0.05 m
Pressure head @ A (P_a / γ )= 21.7 m
Pressure head @ B (P_b / γ )= 76.1 m
Elevation head Z_a = 115 m
Elevation head Z_b = 0 m
Minor Losses = 0 m
Major Losses = f*L*V^2 / 2*D*g = (500/2*0.05*9.81) *f*V^2 = 509.684*f*V^2
Velocity at cross section A and B: V_a = V_b m/s
Roughness e = 2.5 mm
Dynamic viscosity of water u = 8.9*10^-4 Pa-s
Density of water p = 997 kg/m^3

Find:

Flow Rate Q = pi*V*D^2/4 m^3/s ??

Solution:

We will use the Head Balance as derived from Energy Balance:

(P_a / γ ) - (P_b / γ ) + (V_a^2 - V_b^2) / 2*g + (Z_a - Z_b) = Major Losses

21.7 - 76.1 + 0 + 115-0 = 509.684*f*V^2

f*V^2 = 0.18897199

To find correction factor f which is a function of e / D = 0.05, and Reynold's number which is unknown. In such cases we will guess a value of f and perform iterations as follows:

Guess: f_o = 0.072 (Moody's Chart @ e / D = 0.05 and most turbulent function).

V_o = sqrt(0.18897199 / 0.072) = 1.28504 m/s

Re_o = p*V_o*D / u = 997*1.28504*0.05 / 8.9*10^-4 = 71976.6

1st iteration

f_1 = g (Re_o , e/d) = 0.0718702 (Moody's Chart)

V_1 = sqrt(0.18897199 / 0.0718702) = 1.621528 m/s

Re_1 = p*V_1*D / u = 997*1.621528*0.05 / 8.9*10^-4 = 90823.81698

2nd iteration

f_2 = g (Re_1 , e/d) = 0.0718041 (Moody's Chart)

V_2 = sqrt(0.18897199 / 0.0718041) = 1.662273585 m/s

Re_2 = p*V_2*D / u = 997*1.662273585*0.05 / 8.9*10^-4 = 90865.54854

3rd iteration

f_3 = g (Re_2 , e/d) = 0.0718040 (Moody's Chart)

V_3 = sqrt(0.18897199 / 0.0718040) = 1.622274714 m/s

Re_3 = p*V_3*D / u = 997*1.622274714*0.05 / 8.9*10^-4 = 90865.61182

We can observe the convergence of V to 1.6222 m /s. Hence, the required velocity will be used to calculate the Flow rate Q:

Q = pi*V*D^2/4 = pi*1.6222*0.05^2 / 4

Q = 178.41 m^3 / s

3 0

3 years ago

Un bloque de 2.5kg de masa es empujado 2.2m a lo largo de una mesa horizontal sin fricción por una fuerza constante de 16.0 N di

Tanzania [10]

Answer:

Explanation:

(a) The applied force has two components Fx and Fy. The Fx component is the only one that does work

$W_{x}=F_{x}x=(16N)cos(25)(2.2m)=31.9J$

(b) There in no net force in the vertical component

$F_{N}-F_{y}-F_{g}=0\\F_{N}=F_{y}+F_{g}=(16N)sin(25)+(2.5kg)(9.8\frac{m}{s^{2}})=31.26N$

(c)

$F_{g}=Mg=(2.5kg)(9.8\frac{m}{s^{2}})=24.5N$

(d)

$F_{T}=F_{x}=(16N)cos(25)=14.5N$

I attached an scheme of the force diagram

8 0

3 years ago

A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

Ne4ueva [31]

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4

Tension = 24 N

We need to calculate the maximum friction force

Using formula of friction force

$f=\mu mg$

Put the value into the formula

$f=0.8\times37$

$f=29.6\ N$

So, the tension must exceeds 29.6 N for the block to move

We need to calculate the frictional force acting on the block

Using formula of frictional force

$f = \mu N$

Put the value in to the formula

$f=0.4\times37$

$f=14.8\ N$

Hence, The frictional force acting on the block is 14.8 N.

6 0

3 years ago

What's the definition energy?​

What's the definition energy?