What happens to steam as it releases thermal energy inside a radiator?

A 4.0 kg object will have a weight of approximately 14.8 N on Mars. What is the gravitational field strength on M

Pie

Answer:

Gravitational field strength =weight/mass

Explanation:

14.8N/4.0kg

3.7N/kg

3 0

2 years ago

Saturated steam at 125 kpa is compressed adiabatically in a centrifugal compressor to 700 kpa at the rate of 2.5 kg⋅s−1. the com

Tpy6a [65]

M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h = 0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
Work = m(h₂ - h₁) = 2.5(3154.6 - 2685.2
W = 1173.5KW

5 0

3 years ago

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$

$\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}$

Removing $y_o$ and dividing by t (t different of zero)

$\displaystyle 0=v_osin\theta-\frac{gt}{2}$

Then we find the total flight as

$\displaystyle t=\frac{2v_osin\theta}{g}$

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

$\boxed{t=24.5/2=12.25\ seconds}$

4 0

3 years ago

When we apply the energy conversation principle to a cylinder rolling down an incline without sliding, we exclude the work done

NikAS [45]

Answer:

D. the linear velocity of the point of contact (relative to the inclined surface) is zero

Explanation:

The force of friction emerges only when there is relative velocity between two objects . In case of perfect rolling , there is no sliding so relative velocity between the surface and the point of contact is zero . In other words the velocity of point of contact becomes zero , even though , the whole body is in linear motion . It happens due point of contact having two velocities which are equal and opposite . One of the velocity is in forward direction and the other velocity which is due to rotation is in backward direction . So net velocity of point of contact becomes zero . Due to absence of sliding , displacement due to friction becomes zero . Hence work done by friction becomes zero.

5 0

3 years ago

In an electricity experiment, a 1.10 g plastic ball is suspended on a 56.0 cm long string and given an electric charge. A charge

Shalnov [3]

Answer:

Tension, T = 0.0115 N

Explanation:

Given that,

Mass of the plastic ball, m = 1.1 g

Length of the string, l = 56 cm

A charged rod brought near the ball exerts a horizontal electrical force F on it, causing the ball to swing out to a 21.0 degree angle and remain there. According to attached figure :

$T\cos\theta=mg$

T is tension in the string

$T=\dfrac{mg}{\cos\theta}\\\\T=\dfrac{1.1\times 10^{-3}\times 9.8}{\cos(21)}\\\\T=0.0115\ N$

So, the tension in the string is 0.0115 N.

8 0

3 years ago