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Korolek [52]
2 years ago
5

A ball with a mass of 2.00 kg is dropped from a height of 10.0 m. 2kg 10 m How long will it take to hit the ground?​

Physics
1 answer:
GREYUIT [131]2 years ago
5 0
1.4s.....................
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A bird flew 16 km west in 5 hours , then flew 20 km east in 6 hours . What was the birds velocity?
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During that period of time, the bird's displacement was 4 km east. So its velocity was (4km east)/(11hrs). That's 0.36 km/hour east. (rounded)
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The force that moving, charged particles exert on one another is called __________. . intermolecular force. contact force. gravi
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The correct answer is the last option. The force that moving, charged particles exert on one another is called electromagnetic force. This force involves physical interaction between two electrically charged particles. It is seen as electromagnetic fields such as electric fields, magnetic fields and light.
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United States citizens have many ways to participate in national life. The most common form of participation is?
Travka [436]

Answer:

You would have to give better explanation on subject.

Explanation:

3 0
2 years ago
A car travels in a straight line covering a total distance of 90.0 miles in 60.0 minutes. Which one of the following statements
andrezito [222]

Answer:

E) The average velocity of the car is 90.0 miles per hour in the direction of motion.

Explanation:

Given;

distance covered by the car, d = 90 miles

time taken, t = 60 minutes = 1 hour

The average velocity of the car is given by change in displacement per change in time;

V = Δd / Δt

V = \frac{x_f - x_o}{t_f -t_o} = \frac{90-0}{1-0} \\\\V = 90 \ miles/hour

Therefore, the average velocity of the car is 90.0 miles per hour in the direction of motion.

7 0
3 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}

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5 0
3 years ago
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