Which of the following solid fuels has the highest heating value?

A cylindrical block of mass M=50kg and height h=0.2m is hanging on a rope and is in equilibrium. Any difference in atmospheric p

agasfer [191]

Answer:

$\Delta P = 1961.4\,Pa$

Explanation:

The difference of pressure is given by gauge pressure:

$\Delta P = \rho_{w}\cdot g \cdot \Delta h$

$\Delta P = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.2\,m)$

$\Delta P = 1961.4\,Pa$

8 0

4 years ago

Read 2 more answers

What is the appropriate term for the movement of the orange liquid in the lava lamp animation?

Taya2010 [7]

Heat energy is responsible for it.It is because the waxy substance present in the lava lamp get hot at its bottom and cold on its top and this process is mostly like convection

5 0

3 years ago

Suppose a car is traveling at +25.0 m/s, and the driver sees a traffic light turn red. After 0.340 s has elapsed (the reaction t

scoundrel [369]

First, we will get the distance traveled before the driver applied the brakes.
distance = velocity * time
distance = 25*0.34 = 8.5 m

Now, we will calculated the distance that the car traveled after the driver applied the brakes. To do this, we will use the equation of motion:
<span>vf^2 = vi^2 + 2*a*d where:
</span>vf = zero, vi = 25 m/s and a = -7 m/s^2
Note: The negative sign is only to show deceleration
d = <span> 1/2*(625) /(7) = 44.6428 m

The total stopping distance =</span> 8.5 + 44.6428 = 53.1428 m

3 0

3 years ago

A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere

Tanzania [10]

Answer:

The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

Explanation:

Given that,

Radius = 5 cm

Charge density $J= 400\ nC/m^3$

We need to calculate the total charge Q of the sphere

Using formula of charge

$q=\rho V$

Where, $\rho$ = charge density

V = volume

Put the value into the formula

$q=\rho\times(\dfrac{4}{3}\pi r^3)$

Put the value into the formula

$q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3$

$q=2.094\times10^{-10}\ C$

Hence, The total charge Q of the sphere is $2.094\times10^{-10}\ C$ .

6 0

3 years ago

Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc

lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

$E = \frac{V}{L}$

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

$J = \frac{E}{p}$

Substitute (V/L) for E in the above equation of current density.

$J = \frac{V}{pL} ------(1)$

Substitute iR for V in equation (1)

$J = \frac{iR}{pL} ------(2)$

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

$J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2$

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

$p = \frac{RA}{L}$

$A = \frac{pL}{R}$

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A and λ is use for (m/L) in the eqn above

$\lambda = d\frac{p}{\frac{R}{L} } ------(3)$

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m

$\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1$

c) Using the equation (2) current density for aluminum cable is,

$J = \frac{iR}{pL}$

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

$J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2$

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

$\lambda = d\frac{p}{\frac{R}{L} }$

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

$\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m$

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0

4 years ago

Read 2 more answers