Consider a sample containing 1.70 mol of an ideal diatomic gas.

11. A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration ? Is this acceleration higher than that of

Marat540 [252]

$a = \frac{v - u}{t}$

v = final velocity

u = initial velocity

t = time taken

the acceleration of the cyclist is

$\frac{10 - 0}{3} = 3.333333....$

approximately 3.33 m/s^2

the acceleration of the car is

$\frac{40 - 0 }{8} = 5.0$

5.0 m/s^2

$5.0 > 3.33 \\ so \: the \: answer \: is \: no$

6 0

3 years ago

Which of the following is the FINAL step in a forecasting system?

Lelechka [254]

Answer:

gather the data needed to make the forecast

3 0

3 years ago

Learning Goal: To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circ

andreev551 [17]

Answer:

v = 15.8 m/s

Explanation:

Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force

$W_{fr}$ = ΔEm = $Em_{f}$ -Em₀

Let's write the mechanical energy at each point

Initial

Em₀ = Ke = ½ k x²

Final

$Em_{f}$ = K + U = ½ m v² + mg y

Let's use Hooke's law to find compression

F = - k x

x = -F / k

x = 4400/1100

x = - 4 m

Let's write the energy equation

fr d = ½ m v² + mgy - ½ k x²

Let's clear the speed

v² = (fr d + ½ kx² - mg y) 2 / m

v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0

v² = (160 + 8800 - 1470) / 30

v = √ (229.66)

v = 15.8 m/s

5 0

3 years ago

Read 2 more answers

On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the

IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = $\frac{1}{2}\varepsilon _{0}E^{2}$

$=\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119$

= 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = $\frac{B^{2}}{2\mu _{0}}$

$=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}$

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

8 0

3 years ago

Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13

svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

Steady operations exist;
The heat transfer coefficient (h) is uniform over the entire fin surfaces;
Thermal conductivity (k) is constant;
Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

5 0

3 years ago