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3 years ago

Plane a travels at 900km/h and plane b travels at 250/5.which plane travels faster

Physics

1 answer:

vesna_86 [32]3 years ago

5 0

Explanation:

We have,

Speed of plane a is 900 km/h

Plane b is moving at a rate of $\dfrac{250\ km}{5\ h}=50\ km/h$

It is required to find which plane is faster. To find which plane is faster, we need to compare their speeds.

Speed of a plane a is 900 km/h and that of plane b is 50 km/h. So, we can say that plane a is moving faster.

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The chart below shows the average surface temperature or temperature range for each of the eight planets.

Vaselesa [24]

Answer:

A) Earth and the other inner planets have higher average surface temperatures than the outer planets.

Explanation:

the earth and the other inner planets have higher average surface temperatures than the outer planets.

The reason for this response is due to the distance between the sun and the respective planet, the source of energy generation is the sun and the only way in which the temperature increase of each planet is guaranteed is by radiation, the further away a planet is from its star, its temperature will decrease. Although it is also important to highlight the atmospheric composition of the planet if this planet in its stratosphere has high density clouds that do not allow the entry of solar radiation, the temperature of the planet's surface will not increase, independent of the distance from the sun, but these are more complex cases where specialists in that area enter to perform a study in detail.

4 0

3 years ago

Read 2 more answers

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Lunna [17]

Answer:

7) there is a theory that these animals can see the magnetic field of the Earth due to a compass like mechanism inside their eyes in order to navigate.

8) Earth's magnetic field is a result of the currents found inside the outer core that consist of electricity.

3 0

3 years ago

What speed would a fly with a mass of 0.55g need in order to have a kinetic energy of 7.6 •10^4 j?

masya89 [10]

Answer:

16613 m/s

Explanation:

Given that

mass of the fly, m = 0.55 g = 0.55*10^-3 kg

Kinetic Energy of the fly, E = 7.6*10^4 J

Speed of the fly, v = ? m/s

We know that the Kinetic Energy is that energy that an object, in this case, the fly, possesses due to its motion.

The Kinetic Energy, KE of any object is represented by the formula

KE = 1/2 * m * v²

If we substitute the values in the relation, we have,

7.6*10^4 = 1/2 * 0.55*10^-3 * v²

v² = (15.2*10^4) / 0.55*10^-3

v² = 2.76*10^8

v = √2.76*10^8

v = 16613 m/s

Thus, the fly would need a speed of 16.6 km/s in order to have a Kinetic Energy of 7.6*10^4 J

7 0

3 years ago

In a certain electrolysis experiment involving Al3+ ions, 60.2 g of Al is recovered when a current of 0.352 A is used. How many

yawa3891 [41]

Answer:

30.5 x 10³ min.

Explanation:

Atomic weight of aluminium = 27

For the reaction

Al⁺³ + 3e = Al

3 mole of electron is required by 1 mole of aluminium

3 x 96500 C of charge is required for 27 gram of aluminium

60.2 g aluminium will require

$\frac{3\times96500}{27} \times60.2$

645.47 x 10³ C

So charge required = 645.47 x 10³ C

Now Charge = Current x time

Current ( given ) = 0.352A

Time = charge / current = 645.47 x 10³ / .352

=1833 x 10³ s

30.5 x 10³ minutes

6 0

3 years ago

A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i

AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume $v$ and enthalpy $h$ as,

$h_1 = 95.96Btu/lb$ ( $h_1 = h_f$ at $2psia$ )

$v_1 = 0.016238ft^3/lb$ ( $v_1 = v_f$ at $2psia$ )

$w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb$

$w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb$

$h_3 = 1364.0Btu/lb$

$s_3 = 1.5073Btu/lb.R$

( at $P_3 = 1500psia$ & $T_3 = 800^0F$ )

$P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765$

( $S_f$ & $S_{fg}$ when pressure is 2psia)

$h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb$

$n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb$

Therefore,

$q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb$

To calculate the mass flow rate of steam in the cycle, we use the formula

$W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s$

where $1Kj = 0.947817 Btu$

The power output and the rate of heat addition are calculated thus,

$W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW$

$Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s$

The thermal efficiency of the cycle can be found thus;

$n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343$

= 34.3%

5 0

3 years ago