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vekshin1
3 years ago
5

A train travels 85 kilometers in 5 hours, and then 63 kilometers in 5 hours what is its average speed?

Physics
1 answer:
Lesechka [4]3 years ago
3 0
We know average speed =total distance/time taken
So avg speed=(85+63)/(5+5)=14.8km/hr
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Two objects with masses m and 3m are connected by a compressed spring so that the energy stored in the spring is 375 joules. If
cluponka [151]

Answer:

281.25 J

Explanation:

We are told that the two objects with masses m and 3m.

Also that energy stored in the spring is 375 joules.

Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3

Thus, from principle of conservation of energy, we have;

½mv² + ½(3m)(v/3)² = 375J

(m + 3m/9)½v² = 375

(4/3)m × ½v² = 375

Multiply both sides by ¾ to get;

½mv² = 375 × ¾

½mv² = 281.25 J

Therefore, energy of lighter body is 281.25 J

7 0
3 years ago
Which object is in static equilibrium
balu736 [363]

Answer:

The body is said to be in static equilibrium if the net force acting on a body at rest is zero.As the net force is zero,the body will not undergo motion.

Explanation:

3 0
3 years ago
Read 2 more answers
Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What p
Temka [501]

Answer:

The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

Explanation:

From General gas equation.

PV = nRT...............................  Equation 1

Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.

n = mass/molar mass .................. Equation 2

substituting equation 2 into equation 1.

PV = (mass/molar mass)RT

⇒ Mass/molar mass = PV/RT..................... Equation 3

But mass = Density × Volume

⇒ M = D × V.................... Equation 4

Where D = density, M = mass

Substituting equation 4 into equation 3

DV/molar mass = PV/RT............ Equation 5

Dividing both side of the equation by Volume (V) in Equation 5

D/molar mass = P/RT .............. Equation 6

Cross multiplying equation 6

D × RT = P × molar mass

∴ Molar mass = (D × RT)/P.................. Equation 7

Where D = 0.518 g/L , R = 0.0821 atm dm³/K.mol,

T = 25°C = 25 + 273 = 298 K,

P =721 mmHg = (721/760) atm= 0.949 atm

Substituting these values into equation 7

Molar mass = (0.518 × 0.0821 × 298)/0.949

Molar mass = 13.35 g/mole

The molar mass of the mixture is =13.35 g/mole

Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.

∴ 13.35 = 4(y) + 32(1-y)

13.35 = 4y + 32 - 32y

Collecting like terms in the equation,

32y - 4y = 32 - 13.35

28y = 18.65

y = 18.65/28

y =0.666

y = 0.666 × 100 = 66.6%

∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

6 0
3 years ago
A 400.0 ohm resistor has a potential difference of 20.0 volts. What is the magnitude of the power dissipated by the resistor
Sergio [31]

P=IV, where P is power, I is resistance, and V is voltage.  Plug in and solve:

P=400(20)

P=8000W

Hope this helps!!

3 0
3 years ago
How are step up transformers used in the transmission of electrical energy
tensa zangetsu [6.8K]

Answer:

Transformers are used to increase or decrease the voltage of AC currents

Explanation:

A transformer is a device consisting of two coils (called primary and secondary coil) wrapped at the two sides of a soft iron core. When an AC current is present in the primary coil, it induces a magnetic field inside the core, and the presence of this changing magnetic field induces a voltage (and a current) into the secondary coil.

The voltages in the primary and the secondary coil are related by the transformer equation:

\frac{V_p}{V_s}=\frac{N_p}{N_s}

where

Vp, Vs are the voltages in the primary and secondary coil

Np, Ns are the number of turns in the primary and secondary coil

There are two types of transformers:

- Step-up transformers: these have N_s > N_p, so that V_s > V_p, which means that they increase the voltage. They are used to increase the voltage of the AC current produced by the power plants, before being sent into the transmission lines.

- Step-down transformers: these have N_s < N_p, so that V_s < V_p, which means that they decrease the voltage. They are used at the end of the transmission lines, before the houses, in order to decrease the voltage and allow the household appliances to work properly (in fact, household appliances need lower voltages to work)

8 0
3 years ago
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