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3 years ago

a man crossed a road 8.25m wide at a speed of 2.01m/s,how long does it take to get man to cross the road

Physics

1 answer:

Svetradugi [14.3K]3 years ago

5 0

Answer:

t = 4.1 seconds

Explanation:

It is given that,

Width of road which is to be crossed by a man is 8.25 m, it means it is distance to be covered.

Speed of man is 2.01 m/s

We need to find the time taken by the man to cross the road. It is a concept of speed. Speed of a person is given by total distance covered divided by time taken. So,

$v=\dfrac{d}{t}$

t is time taken

$t=\dfrac{d}{v}\\\\t=\dfrac{8.25}{2.01}\\\\t=4.1\ s$

So, the time taken by the man to cross the road is 4.1 seconds.

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The voltage entering a transformer’s primary winding is 120 volts. The primary winding is wrapped around the iron core 10 times.

loris [4]

Answer

48 Volts

Explanation
The question can be solve using the turn rule of a transformer that states;

Np/Ns = Vp/Vs
Where Np ⇒ number of turns in the primary coil.
Ns ⇒number of turns in the seconndary coil
Vp ⇒ primary voltage
Vs ⇒secondary voltage

Np/Ns = Vp/Vs

10/4 = 120/Vp

Vp = (120 × 4)/10

= 480/10
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5 0

3 years ago

Read 2 more answers

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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3 years ago

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iVinArrow [24]

Answer:

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Explanation:

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Answer:

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2 years ago

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2 years ago

a man crossed a road 8.25m wide at a speed of 2.01m/s,how long does it take to get man to cross the road​

a man crossed a road 8.25m wide at a speed of 2.01m/s,how long does it take to get man to cross the road