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3 years ago

10

Surface currents and waves are powered by what

1 answer:

Allisa [31]3 years ago

4 0

It is powered by the Earth's rotation and the moon gives a little boost.

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Could someone please help me

BaLLatris [955]

Answer:

the size of the shadow will be smaller due to smaller hands

3 0

3 years ago

Why does immersion oil improve resolution?

Umnica [9.8K]

Answer:

d. It increases numerical aperture and maintains a uniform light speed.

Explanation:

In optical microscopes, various immersion mediums are used to improve or enhance the resolution. Immersion oils like cedar and Leica oils are one of those mediums which are used to improve resolution by increasing the numerical aperture and keeping the speed of light uniform.

7 0

3 years ago

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the

Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

$\theta = 54.6^{\circ}$

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = $\sqrt{\frac{gx}{sin2\theta}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s$

Now,

The angular velocity can be calculated as:

$v = \omega R$

Thus

$\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s$

8 0

3 years ago

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as:

Westkost [7]

Answer:

dorsiflexion

Explanation:

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as: dorsiflexion

7 0

3 years ago

Read 2 more answers

Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un

Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

$K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}$

$\gamma$ = 1, $K_{IC}$ = 24.1

[/tex]\sigma_{y}[/tex] = 570

and, $\sigma_{f} = 570 \times \frac{3}{4}$

= 427.5

Hence, we will calculate the critical crack length as follows.

a = $\frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}$

= $\frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}$

= $10.13 \times 10^{-4}$

Therefore, largest size is as follows.

Largest size = 2a

= $2 \times 10.13 \times 10^{-4}$

= $20.26 \times 10^{-4}$

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is $20.26 \times 10^{-4}$ .

4 0

3 years ago