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3 years ago

Explain why some objects fall faster than other objects,

1 answer:

8 0

objects that are more dense, or have more mass, fall at a faster rate than less dense objects, due to this air resistance. A feather and brick dropped together. Air resistance causes the feather to fall more slowly.

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A student fills a tank of radius r with water to a height of h1 and pokes a small, 1.0 cm diameter hole at a distance h2 from th

Alik [6]

when a hole is made at the bottom of the container then water will flow out of it

The speed of ejected water can be calculated by help of Bernuolli's equation and Equation of continuity.

By Bernoulli's equation we can write

$Po + \frac{1}{2}\rho v_1^2 + \rho g h = Po + \frac{1}{2}\rho v_2^2 + \rho g *0$

Now by equation of continuity

$A_1v_1 = A_2v_2$

$\pi (0.2)^2 v_1 = \pi (0.01)^2 v_2$

from above equation we can say that speed at the top layer is almost negligible.

$v_1 = 0$

now again by equation of continuity

$\rho g h = \frac{1}{2} \rho v^2$

$v = \sqrt{2 g h}$

here we have

$h = h_1 - h_2$

$h = 0.50 - 0.03 = 0.47m$

now speed is given by

$v = \sqrt{2* 9.8 * 0.47}$

$v = 3.03 m/s$

7 0

3 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

3 years ago

Am I correct?? Will give brainliest

IRISSAK [1]

You are correct earth science is studied to predict planetery changes

4 0

3 years ago

Read 2 more answers

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

8 0

3 years ago

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm . The explorer finds that

larisa86 [58]

Answer:

12.4 m/s²

Explanation:

L = length of the simple pendulum = 53 cm = 0.53 m

n = Number of full swing cycles = 99.0

t = Total time taken = 128 s

T = Time period of the pendulum

g = magnitude of gravitational acceleration on the planet

Time period of the pendulum is given as

$T = \frac{t}{n}$

$T = \frac{128}{99}$

T = 1.3 sec

Time period of the pendulum is also given as

$T = 2\pi \sqrt{\frac{L}{g}}$

$1.3 = 2(3.14) \sqrt{\frac{0.53}{g}}$

g = 12.4 m/s²

4 0

3 years ago