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2 years ago

1mm is equal to .1 what

Physics

1 answer:

bulgar [2K]2 years ago

8 0

Answer:

When you have to do an English-Metric (SI) length conversion, and you already know the English units of length (miles, yards, feet, inches, etc.), all you need to remember is one simple relationship, and you can readily convert any length in the SI system, to the equivalent length in the other.

1 foot (ft) = 0.3048 meters (m)

BIn this case you need your answer in inches. You (hopefully) know there are 12 inches in a foot, so you just do the following:

1 inch (in) = 1/12 ft = 0.3048/12 m = 0.0254 m

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In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21

Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

$z = 20km$

And the velocity can be written as,

$V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$V = 980.55m/s$

From the properties of standard atmosphere at altitude z = 20km temperature is

$T = 216.66K$

$k = 1.4$

$R = 287 J/kg$

Velocity of sound at this altitude is

$a = \sqrt{kRT}$

$a = \sqrt{(1.4)(287)(216.66)}$

$a = 295.049m/s$

Then the Mach number

$Ma = \frac{V}{a}$

$Ma = \frac{980.55}{296.049}$

$Ma = 3.312$

So front stagnation temperature

$T_0 = T(1+\frac{k-1}{2}Ma^2)$

$T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)$

$T_0 = 689.87K$

Therefore the temperature at its front stagnation point is 689.87K

6 0

3 years ago

4. Johnny exerts a 3.55 N rightward force on a 0.200-kg box to accelerate it across a low-friction track. If the total resistanc

Anon25 [30]

a) $15.2 m/s^2$

b) 1.96 N

c) 1.96 N

Explanation:

To find the box's acceleration, we have to find first the net force acting on the box in the horizontal direction.

We have:

- The forward force of 3.55 N

- The backward, resistive force of 0.52 N

So, the net force forward is

$\sum F=3.55-0.52=3.03 N$

Now we can find the acceleration by using Newton's second law of motion, which states that:

$\sum F=ma$

where

m = 0.200 kg is the mass of the box

a is its acceleration

And solving for a, we find the acceleration:

$a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2$

The gravitational force on an object is the force with which the object is pulled towards the ground by the Earth.

It is given by

$W=mg$

where

m is the mass of the object

g is the gravitational field strength

In this problem we have

m = 0.200 kg is the mass of the box

$g=9.8 m/s^2$ is the gravitational field strength

So, the gravitational force on the box is

$W=(0.200)(9.8)=1.96 N$

The normal force is the reaction force exerted by the floor on the box, in the upward direction.

In order to find the magnitude of this force, we apply Newton's second law of motion along the vertical direction.

We have two forces in this direction:

- The gravitational force, W, downward

- The normal force, N, upward

So the net force is

$\sum F=N-W$

According to Newton's second law,

$\sum F=ma$

However, the box is at rest in the vertical direction, so the vertical acceleration is zero:

$a=0$

This means that the net force is zero:

$\sum F=0$

And so, we can find the normal force:

$N-W=0\\N=W=1.96 N$

4 0

2 years ago

There are two types of addiction, psychological and physical. A. True B. False

Alexus [3.1K]

It is true that there are two types of addiction, psychological and physical. When it comes to physical addiction, it normally refers to drug withdrawal or drug usage, when your body is exhibiting various symptoms of needing drugs. Mental addiction refers to when there is no actual need to use the object of addiction, but rather a mental desire to use it and have it again.

4 0

2 years ago

Which statements accurately describe the shape of Earth’s orbit around the Sun? Check all that apply.

damaskus [11]

<span>The Earth’s orbit is a nearly circular ellipse.
</span><span>The Sun is located at one of the two focal points.</span>

The Earth moves around the Sun in an orbit that is almost but not quite circular. As Kepler proved in the seventeenth century, the orbit is actually an ellipse. A parameter called the eccentricity (e) defines the degree of departure from a circle. A value of e=0 would indicate a circle whereas a value of e=0.9 would indicate a very elongated ellipse. The eccentricity of the Earth's orbit is currently e=0.0167.

4 0

2 years ago

Read 2 more answers

As an aid in working this problem, consult Concept Simulation 11.1. The volume flow rate in an artery supplying the brain is 3.5

klemol [59]

Answer

given,

flow from the artery = 3.5 x 10⁻⁶ m³/s

Radius of artery = 5.80 x 10⁻³m

area = π R²

= π x (5.8 x 10⁻³)²

= 1.06 x 10⁻⁴ m²

$v = \dfrac{Q}{A}$

$v = \dfrac{3.5 \times 10^{-6}}{1.06 \times 10^{-4}}$

v = 0.033 m/s

b) new velocity of flow

Radius = R' = R/4

A V = A' V'

R² V = R'² V'

R^2 V = (\dfrac{R}{4})^2 V'

V' = 16 V

V' = 16 x 0.033

V' =0.528 m/s

5 0

2 years ago