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3 years ago

If a rocket travels 35,200 miles in 2 hours, what is its speed?

Physics

2 answers:

Brrunno [24]3 years ago

8 0

Answer:

Given:

Distance = 35200 miles, time = 2 hr.

So we know = 35200/ 2, = 17600 miles / hr.

muminat3 years ago

4 0

If you mean what is it’s speed per hours (mph)

35,200/2

= 17,600 miles per hour

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Determine the ratio of the flow rate through capillary tubes A and B (that is, Qa/Qb).

Oliga [24]

To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.

It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

The mathematical equation that expresses this concept is

$\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}$

Where

P = Pressure at each point

r = Radius

$\eta =$ Viscosity

l = Length

Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}$

From the problem two terms are given

$R_A = \frac{R_B}{2}$

$L_A = 2L_B$

Replacing we have to

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}$

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}$

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}$

Therefore the ratio of the flow rate through capillary tubes A and B is 1/32

6 0

3 years ago

Which of the following

kirill115 [55]

Answer:

Explanation:

matter and energy van flow and cycle in an ecosystem depending on the temperature

7 0

3 years ago

Physicists and engineers from around the world have come together to build the largest accelerator in the world, the Large Hadro

elena-14-01-66 [18.8K]

Solution :

Energy of photon, E = 6.7 eV

E = $6.7 \times 1.602 \times 10^{-7}$ joule

Kinetic energy, $K.E. =\frac{1}{2} mv^2 = 1.602 \times 6.7 \times 10^{-7}$

$v^2=\frac{2 \times 1.602 \times 6.7 \times 10^{-7}}{1.6726 \times 10^{-27}}$

$=12.834 \times 10^{-20}$

Kinetic energy at high speeds

$(r-1)\times mc^2 = 6.7 \ eV$

$(r-1)=\frac{6.7 \times 1.602 \times 10^{-7}}{1.6726 \times 10^{-27} \times 9 \times 10^{16}}$

r - 1 = 7130

r = 7130 + 1

r = 7131

$\frac{1}{\sqrt{1-\frac{v^2}{C^2}}}=7131$

$1-\frac{v^2}{C^2} = \left(\frac{1}{7131}\right)^2$

$v^2=C^2\left[1-\left(\frac{1}{7131}\right)^2\right]$

$v=0.99999999017C$

Δ = 1 - 0.99999999017

= 0.00000000933

Relative mass, $m_{rel}=r.m$

$=7131 \times 1.6728 \times 10^{-27}$

$=1.1927 \times 10^{-23}$ kg

6 0

3 years ago

A stone is thrown vertically downward with an initial speed of 12.0 m/s from the top of a building. The stone takes 1.54 s to re

mr_godi [17]

(a) The stone travels a vertical distance y of

y = (12.0 m/s) t + 1/2 g t ²

where g = 9.80 m/s² is the acceleration due to gravity. Note that this equation assume the downward direction to be positive, and that y = 0 corresponds to the height from which the stone is thrown.

So if it reaches the ground in t = 1.54 s, then the height of the building y is

y = (12.0 m/s) (1.54 s) + 1/2 (9.80 m/s²) (1.54 s)² ≈ 30.1 m

(b) The stone's (downward) velocity v at time t is

v = 12.0 m/s + g t

so that after t = 1.54 s, its velocity is

v = 12.0 m/s + (9.80 m/s²) (1.54 s) ≈ 27.1 m/s

(and of course, speed is the magnitude of velocity)

8 0

3 years ago

ListenA person on a ledge throws a ball vertically downward, striking the ground below the ledge with 200 joules of kinetic ener

polet [3.4K]

Answer:

A. 200 J

Explanation:

The initial kinetic energy depends on the initial speed, while the gravitational potential energy depends on the height, both balls are thrown with the same initial speed and from the same height. Therefore, due to the law of conservation of energy, the balls must have the same mechanical energy (the sum of both energies) when both impact the ground. Since the potential energy is zero at this point, its final kinetic energy must also be the same.

8 0

3 years ago