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A girl and a boy are riding on a merry go round that is turning at a constant rate. The girl is near the outer edge, and the boy

galina1969 [7]

Answer:

The girl has greater tangential acceleration

Explanation:

The angular acceleration ( $\alpha$ ) of the merry go round is equal to the rate of the change of the angular velocity, $\omega$ :

$\alpha = \frac{d\omega}{dt}$

Since all the points of the merry go round complete 1 circle in the same time, the angular velocity of each point of the merry go round is the same, and so all the points also have the same angular acceleration.

The tangential acceleration instead is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the distance from the centre of the merry go round

Since the girl is near the outer edge and the boy is closer to the centre, the value of r for the girl is larger than for the boy, so the girl has greater tangential acceleration.

5 0

3 years ago

2) What did Rutherford discover in his experiment?<br> a) nucleus<br> b) electrons<br> c) neutrons

OlgaM077 [116]

A) nucleus because some aims reflected off the gold foil proving that there was something bigger inside the atom hence discovering the nucleus

5 0

2 years ago

Review:

Tresset [83]

Answer:

1. 1. A quantity is completely described by magnitude alone. A quantity Is completely described by a magnitude with a direction.

[a]. scalar, vector

b. vector, scalar

2.2. Speed is a velocity is a quantity and quantity.

a. scalar, vector

[b]. vector, scalar

3 0

2 years ago

Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be

avanturin [10]

Answer:

The mass rate of the cooling water required is: $1'072988.5\frac{kg}{h}$

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

${m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}$

Where w refers to the cooling water and s to the steam flow. Reorganizing,

$m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}$

Write the difference of enthalpy for water as Cp (Tout-Tin):

$m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}$

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, ${h_{s}}^{out}=251.40\frac{kJ}{kg}$ and ${h_{s}}^{in}$ can be calculated as:

${h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}$ .

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 $\frac{kJ}{kgK}$ . If the average temperature is actually different, it won't mean a considerable mistake. Also we know that ${T_{w}}^{out}-{T_{w}}^{in}\leq 10$ , so let's work with the limit case, which is ${T_{w}}^{out}-{T_{w}}^{in}=10$ to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

$m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}$

Download pdf

5 0

3 years ago

Emily’s vacuum cleaner has a power rating of 200 watts. If the vacuum cleaner does 360,000 joules of work,

gtnhenbr [62]

Divide 360000 by 200 to get 1800 seconds, or half of hour.

7 0

3 years ago