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blagie [28]
4 years ago
10

A stone is dropped from the bridge, it takes 4s to reach the water. what's the height of the bridge?​

Physics
2 answers:
iVinArrow [24]4 years ago
6 0
Is there any other type of information?
blsea [12.9K]4 years ago
5 0

Explanation:

Using Equations of Motion :

s = ut +  \frac{1}{2} g {t}^{2}

Height = 0 * 4 + 4.9 * 16

<u>Height = 78.4 m</u>

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Two resistors of 15 and 30 ω are connected in parallel. if the combination is connected in series with a 9. 0-v battery and a 20
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Resistors Working Together.

Resistors are shown coupled in parallel to a voltage source in Figure 10.3.4. When all of the resistors' ends are connected to one another by a continuous wire of minimal resistance and their other ends are also connected to one another by a continuous wire of minimal resistance, the resistors are said to be in parallel. There is a constant potential drop across all resistors. Ohm's law, I=V/R, can be used to determine the current flowing through each resistor while the voltage is constant across each resistor. For instance, the headlights, radio, and other components of an automobile are linked in parallel so that each subsystem can use the entire voltage of the source and function independently. The wiring in your home or any other structure shares the same

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2 years ago
20.0 m [N] - 15 m [S20degreesE]
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3 years ago
A 16.0kg canoe moving to the left at 12.5m/s makes an elastic head-on collision with a 14.0kg raft moving to the right at 16.0m/
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The canoe is moving at 14.1 m/s to the right after the collision.

Explanation:

According to the law of conservation of momentum, in absence of external forces the total momentum of the system must be conserved before and after the collision. So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 16.0 kg is the mass of the canoe

u_1 = -12.5 m/s is the initial velocity of canoe (we take right as positive direction, and since the canoe is moving to the left, its velocity is negative)

v_1 is the final velocity of the canoe

m_2 = 14.0 kg is the mass of the raft

u_2 = +16.0 m/s is the initial velocity of the raft

v_2 = -14.4 m/s is the final velocity of the raft

Re-arranging the equation and substituting the values, we find: the final velocity of the canoe:

v_1 = \frac{m_1 u_1 + m_2 u_2-m_2 v_2}{m_1}=\frac{(16.0)(-12.5)+(14.0)(16.0)-(14.0)(-14.4)}{16.0}=+14.1 m/s

So, the canoe is moving at 14.1 m/s to the right after the collision.

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5 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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