Is there any possibility to make 100% efficient system

The inner and outer surfaces of a 5m x 6m brick wall of thickness 30 cm and thermal conductivity 0.69 w/m.0 c are maintained at

Volgvan

The working equation to be used here is written below:

Q = kA(T₁ - T₂)/Δx
where
Q is the rate of heat transfer
k is the heat transfer coefficient
A is the cross-sectional area of the wall
T₁ - T₂ is the temperature difference between the sides of the wall
Δx is the thickness of the wall

The solution is as follows:

Q = (0.69 W/m²·°C)(5 m × 6 m)(50°C - 20°C)/(30 cm * 1 m/100 cm)
Q = 2,070 W/m

4 0

3 years ago

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the

Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

$\theta = 54.6^{\circ}$

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = $\sqrt{\frac{gx}{sin2\theta}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s$

Now,

The angular velocity can be calculated as:

$v = \omega R$

Thus

$\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s$

8 0

3 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= $\int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx$

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*0.8(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = $\frac{36}{11}$ kg/m

Mass of rope = weight of rope × change in distance

= $\frac{36}{11}$ kg/m × (11-x)m = 3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*3.27(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0

4 years ago

How long does a ship take to cross the pacific?

lidiya [134]

Between 2 weeks and a month

4 0

3 years ago

A spaceship from a friendly, extragalactic planet flies toward Earth at 0.201 times the speed of light and shines a powerful las

bagirrra123 [75]

Answer:

The wavelength of observed light on earth is 568.5 nm

Explanation:

Given that,

Velocity of spaceship $v= 0.201c$

Wavelength of laser $\lambda= 697\ nm$

We need to calculate the wavelength of observed light on earth

Using formula of wavelength

$\lambda_{0}=\lambda_{e}\times\sqrt{\dfrac{1-\dfrac{v}{c}}{1+\dfrac{v}{c}}}$

$\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-\dfrac{0.201 c}{c}}{1+\dfrac{0.201c}{c}}}$

$\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-0.201}{1+0.201}}$

$\lambda=5.685\times10^{-7}\ m$

$\lambda=568.5\times10^{-9}\ m$

$\lambda=568.5\ nm$

Hence, The wavelength of observed light on earth is 568.5 nm

8 0

4 years ago