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4 years ago

What is the zone that has just been added to the exchange zone where athletes may now hand off the baton

Engineering

1 answer:

Vanyuwa [196]4 years ago

8 0

Answer:

The zone where athletes may hand off baton in exchange zone is the changeover zone.

Explanation:

A relay race is a track event that involves the exchange of baton among a set of athletes. The baton is majorly exchange within the exchange zone which is 20 meters long. The exchange zone is made up on two other zones: acceleration zone and changeover zone.

Acceleration zone is the region of the track that allows the athlete to take the next leg to start increasing speed before collecting the baton.

Changeover zone is the section where baton exchange between athletes takes place. The baton should only be exchanged within the changeover zone.

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A reversible process and an irreversible process both have the same________ between the same two states. a. Internal energy b. W

Vlad1618 [11]

Answer:

a) Internal energy

Explanation:

As we know that internal energy is a point function so it did not depends on the path ,it depends at the initial and final states of process.All point function property did not depends on the path.Internal energy is a exact function.

Work and heat is a path function so these depend on the path.They have different values for different path between two states.Work and heat are in exact function.

We know that in ir-reversible process entropy will increase so entropy will be different for reversible and ir-reversible processes.

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3 years ago

What is soil texture in civil engineering

Bond [772]

Answer:

It will indicate the percentage of sand, silt, and clay present in the soil.

Explanation:

4 0

3 years ago

You can change lanes during a turn long as there’s no traffic and you driving slowly

Vanyuwa [196]

Your allowed to switch lanes as long as the road is clear and you use signals.

5 0

3 years ago

Air at 80 kPa and 10Â°C enters an adiabatic diffuser steadily with a velocity of 150 m/s and leaves with a low velocity at a pre

il63 [147K]

Answer:

The exit temperature is 293.74 K.

Explanation:

Given that

At inlet condition(1)

P =80 KPa

V=150 m/s

T=10 C

Exit area is 5 times the inlet area

Now

$A_2=5A_1$

If consider that density of air is not changing from inlet to exit then by using continuity equation

$A_1V_1=A_2V_2$

So $A_1\times 150=5A_1V_2$

$V_2=30$ m/s

Now from first law for open system

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

Here Q=0 and w=0

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}$

When air is treating as ideal gas

$h=C_pT$

Noe by putting the values

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}$

$1.005\times 283+\dfrac{150^2}{2000}=1.005\times T_2+\dfrac{30^2}{2000}$

$T_2=293.74K$

So the exit temperature is 293.74 K.

7 0

3 years ago

A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as

katrin2010 [14]

Answer:

Follows are the solution to this question:

Explanation:

Calculating the area under the curve:

A = as

$=\frac{1}{2}(3 +6 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(6+4 \frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(9 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(10\frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(900 \frac{m^2}{s^2})+ \frac{1}{2}(1,000\frac{m^2}{s^2}) \\\\=(450 \frac{m^2}{s^2})+ (500\frac{m^2}{s^2}) \\\\= 950 \ \frac{m^2}{s^2}$