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Anuta_ua [19.1K]

2 years ago

13

A baseball sits motionless near first base on a baseball diamond. What statement best explains why the baseball remains motionle

ss?

1 answer:

oksano4ka [1.4K]2 years ago

3 0

Answer:

Because no one is using that ball, so it is motionless

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Altitude means the what?

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2 years ago

A wheelbarrow can be used to help lift a load, such as a pile of dirt, and then push the load across a distance. A man pushes a

skelet666 [1.2K]

Answer:

B.) a wheel and axle and a lever

Explanation:

P.S - The exact question is -

Given - A wheelbarrow can be used to help lift a load, such as a pile of dirt, and then push the load across a distance. A man pushes a wheelbarrow.

To find - Which simple machines make up a wheelbarrow?

A.) a pulley and an inclined plane

B.) a wheel and axle and a lever

C.) a pulley and a wheel and axle

D.) a lever and a wedge

Proof -

The correct option is - B.) a wheel and axle and a lever

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With this machine, During hauling people can save time.

4 0

2 years ago

Read 2 more answers

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