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kupik [55]
3 years ago
12

A person travels 1.5 km in 6 minutes, how fast were they traveling in meters per second?

Physics
1 answer:
Evgen [1.6K]3 years ago
7 0

Answer:

4.167m/sec

Explanation:

1km=1000m

1.5km=1500m

1min=60sec

6min=360sec

In 360sec they travel 1500m

In 1 sec they travel=1500m/360

1sec=4.167m

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URGENT!!!!! If a 6 Ω resistor is in a circuit connected to a voltage source, and the current through the circuit is 2 amps, what
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Answer:

Explanation:

according to ohms law we know that

v=IR

given current =2 amps

given resistance =6Ω

so voltage is

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4 years ago
What is the main function of a cell
Andre45 [30]

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5 0
3 years ago
Can you please answers these for me please today is the last day to turn in work and I need this to pass please I’m begging than
Ymorist [56]

Answer:

1.   <u>F = ma</u>  <em>F = 0.2kg * 20m/s² = 4Kg * m/s² =</em> 4N

2.  <u>F = ma</u>  <em>F - 18Kg * 3m/s² = 54Kg * m/s² =</em> 54N

3.  <u>F = ma</u>  <em>F = 0.025Kg * 5m/s² =</em> 0.125N

4.  <u>F = ma</u>  <em>F = 50Kg * 4m/s² =</em> 200N

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6.  <u>F = ma</u>  <em>F = 9Kg * 9.8m/s² =</em> 88.2N

Explanation:

Hope this helps ! ^^

8 0
2 years ago
In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh
horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}

Where,

C_d is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = \frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2

A₂ = Area at the throat

A₂ = \frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}

or

Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0
3 years ago
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