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Nikolay [14]
4 years ago
10

How long would it take a drag racer to increase her speed from 10m/s to 20 m/s if her car accelerates at a uniform rate of 15 m/

s^2?
Physics
1 answer:
weeeeeb [17]4 years ago
7 0

Answer:

t = 0.67 [s]

Explanation:

To solve this problem we must use the following kinematics equation.

v_{f} =v_{i} +(a*t)\\where:

Vf = final velocity = 20[m/s]

Vi = initial velocity = 10 [m/s]

a = aceleration = 15 [m/s^2]

Now replacing in the equation we have:

20 = 10 + (15*t)

t = (20-10)/15

t = 0.67 [s]

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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:

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Diameter of the pipe = 100mm = 0.1m

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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

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What is the scientific term for rocks formed from lava?
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