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3 years ago

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A train travelling at 50km/h approaches another train moving towards the first at 90km/h. If they are 35km apart (on a straight

track) how long does it take before they meet?

1 answer:

Ede4ka [16]3 years ago

8 0

it will take 36 seconds

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The weight of the atmosphere above 1 m- of

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Answer:

1.09 kg.m

Explanation:

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2 years ago

What are stepdown transformers used for

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Answer:

Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.

Explanation:

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2 years ago

A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes

Agata [3.3K]

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

8 0

3 years ago

The average velocity of blood flowing in a certain 4-mm-diameter artery in the human body is 0.28 m/s. The viscosity and density

OLga [1]

Answer:

V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s

Explanation:

The volume flow rate of the blood in the artery can be given by the following formula:

$V = Av$

where,

V = Volume flow rate = ?

A = cross-sectional area of artery = πd²/4 = π(0.004 m)²/4 = 1.26 x 10⁻⁵ m²

v = velcoity = 0.28 m/s

Therefore,

$V = (1.26\ x\ 10^{-5}\ m^2)(0.28\ m/s)$

<u>V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s</u>

4 0

3 years ago

A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas

Nookie1986 [14]

The maximum amount of work performed is

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

Explanation:

The efficiency of a real heat engine is given by the equation:

$\eta = 1-\frac{T_C}{T_H}$ (1)

where

$T_C$ is the temperature of the cold reservoir

$T_H$ is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

$\eta = \frac{W_{max}}{Q_H}$

where

$W_{max}$ is the maximum work done

$Q_H$ is the heat absorbed from the hot reservoir

$Q_H$ can be written as

$Q_H=W_{max}+Q_C$

where

$Q_C$ is the heat released to the cold reservoir

So the previous equation can be also written as

$\eta=\frac{W_{max}}{W_{max}+Q_C}$ (2)

By combining eq.(1) and (2) we get

$1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}$

And re-arranging the equation and solving for $W_{max}$ , we find

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

Learn more about work and heat:

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8 0

3 years ago