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Ilya [14]
2 years ago
11

A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.8 m/s rel

ative to the ground.
Physics
1 answer:
Svetach [21]2 years ago
4 0

I'm assuming the question is what is the robin's speed relative to to the ground...

Create an equation that describes its relative motion.

rVg = rVa + aVg


Substitute values.

rVg = 12 m/s [N] + 6.8 m/s [E]


Use vector addition.

| rVg | = √ | rVa |² + | aVg |²

| rVg | = √ 144 m²/s² + 46.24 m²/s²

| rVg | = √ 19<u>0</u>.24 m²/s²

| rVg | = 1<u>3</u>.78 m/s


Find direction.

tanФ = aVg / rVa

tanФ = 6.8 m/s / 12 m/s

Ф = 29°


Therefore, the velocity of the robin relative to the ground is 14 m/s [N29°E]

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The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f
tensa zangetsu [6.8K]

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

4 0
1 year ago
What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?
kkurt [141]

Answer:

U = 12,205.5 J

Explanation:

In order to calculate the internal energy of an ideal gas, you take into account the following formula:

U=\frac{3}{2}nRT        (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: temperature of the gas = 100K

You replace the values of the parameters in the equation (1):

U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J

The total internal energy of 10 mol of Oxygen at 100K is 12,205.5 J

6 0
2 years ago
A 13,000-N vehicle is to be lifted by a 25-cm diameter hydraulic piston. What force needs to be applied to a 5.0 cm diameter pis
MaRussiya [10]

Answer:

Explanation:

We shall apply Pascal's Law in fluid mechanics

According to it , pressure is transmitted in liquid from one point to another without any change .

25 cm diameter = 12.5 x 10⁻² m radius

Area = 3.14 x (12.5 x 10⁻²)²

= 490.625 x 10⁻⁴ m²

Pressure by vehicle

Force / area

13000 / 490.625 x 10⁻⁴

= 26.497 x 10⁴ Pa

5 cm diameter = 2.5 x 10⁻² radius

area = 3.14 x (2.5 x 10⁻²)²

= 19.625 x 10⁻⁴ m²

If we assume required force F on this area

Pressure = F / 19.625 x 10⁻⁴ Pa

According to Pascal Law

F / 19.625 x 10⁻⁴  = 26.497 x 10⁴

F = 19.625 x 26.497

= 520 N

4 0
2 years ago
8. The resistance of a bagel toaster is 14 Ω. To prepare a bagel, the toaster is operated for one minute from a 120-V outlet. Ho
IgorC [24]

Answer: The energy delivered to the toaster is 264.490KJ

Explanation:

Here is the complete question:

The resistance of a bagel toaster is 14 ?. To prepare a bagel, the toaster is operated for one minute from a 120-V outlet. How much energy is delivered to the toaster?

Step-by-step explanation:

Please see attachment below

7 0
2 years ago
Read 2 more answers
An observer O is standing on a platform of length L = 90 m on a station. A rocket train passes at a relative (constant) speed of
Natali [406]

Answer:

Explanation:

Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .

Then length of the platform = length of the train rocket .

A )

Time to cross a particular point on the platform

= length of rocket train / .96 x 3 x 10⁸

= 90 /  .96 x 3 x 10⁸

= 31.25 x 10⁻⁸ s

B)  Rest length of the rocket = length of platform = 90 m

C ) length of platform  as viewed by moving observer =

\frac{90}{\sqrt{1-\frac{v^2}{c^2 } } }

= \frac{90}{\sqrt{1-\frac{0.92}{1 } } }

= 321 m

D )  For the observer on platform time taken = 31.25 x 10⁻⁸ s

for the observer in the rocket , time will be dilated so time recorded by observer in motion ,

31.25\times10^{-8} \times \sqrt{1-\frac{.96^2}{1} }

8.75 x 10⁻⁸ s .

8 0
1 year ago
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