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4 years ago

When does air become turbulent around a thrown ball?

Physics

1 answer:

kherson [118]4 years ago

8 0

Answer: B (on the rough side)

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The block A and attached rod have a combined mass of 50 kg and are confined to move along the guide under the action of the 796-

Luba_88 [7]

Answer:

The bending moment is 459.16 N.m

Explanation:

From the given information;

Let's assume that the angle is 66°

Then, the free body diagram is draw and attached in the file below.

Now, the calculation of the acceleration from the first part of the free body diagram is:

$\sum F_x = ma_x \\ \\ 796 - 50(9.81) sin 66=50a \\ \\ 796 - 448.094 = 50 a \\ \\ a = \dfrac{347.906}{50} \\ \\ a = 6.96 \ m/s^2$

Bending moment M:

From the second part of the diagram:

$\sum M_B = mad \\ \\ M - (15 \times 9.81) (1.5) = (25 \times 6.96)(1.5 sin 66) \\ \\ M - 220.725 = 238.435 \\ \\ M = 238.435 + 220.725 \\ \\ \mathbf{M = 459.16 \ N.m}$

6 0

3 years ago

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.6 multiply 1010 m (inside t

Doss [256]

We will apply the concepts related to energy conservation to develop this problem. In this way we will consider the distances and the given speed to calculate the final speed on the path from the sun. Assuming that the values exposed when saying 'multiply' is scientific notation we have the following,

$d_1 = 4.6*10^{10}m$

$v_i = 9.3*10^4m/s \rightarrow \text{Initial velocity comet}$

$d_2 = 6*10^{12}m$

The difference of the initial and final energy will be equivalent to the work done in the system, therefore

$E_f = E_i +W$

$K_f +U_f = K_i +U_i + 0$

$\frac{1}{2} mv_f^2+\frac{-GMm}{d_2} = \frac{1}{2} mv_i^2+\frac{-GMm}{d_1}$

Here,

m = Mass

$v_f$ = Final velocity

G = Gravitational Universal Constant

M = Mass of the Sun

m = Mass of the comet

$v_i$ = Initial Velocity

Rearranging to find the final velocity,

$v_f = \sqrt{v_i^2+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

Replacing with our values we have finally,

$v_f = \sqrt{(9.3*10^4)+2(6.7*10^{-11})(1.98*10^{30})(\frac{1}{6*10^{12}}-\frac{1}{4.6*10^{10}})}$

$v_f = 75653.9m/s$

Therefore the speed is 75653m/s

8 0

4 years ago

Fertilization occurs in the fallopian tubes.

Tamiku [17]

What’s the question? Is it true or false?

7 0

3 years ago

Two satellites are orbiting earth at different altitudes. Which satellite orbits at a higher speed v around earth?

aniked [119]

Answer:

Low satellite has high orbital velocity

Explanation:

let v be the orbital speed of the satellite orbiting at a height h is given by

$v=\sqrt{\frac{GM}{R+h}}$

where, M be the mass of planet, r be the radius of planet and h be the height of planet from the surface of planet.

here we observe that more be the height lesser be the orbital velocity.

So, a satellite which is at low height has high orbital velocity.

3 0

4 years ago

Compare an earthquake with a Richter scale magnitude of 5.5 to one with a magnitude of 7.5.

maksim [4K]

The 7.5 is a 100 times stronger than the 5.5 . The 100 comes from the increase in scale from 5.5 to 7.5 which is an increase of two so you multiply the strength of the weaker earthquake 10^7-5=10^2=100 and that gives you the strength of the stronger earthquake!

7 0

3 years ago