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d1i1m1o1n [39]
2 years ago
11

Set local ground level to 700 ft, and record the inches of mercury.

Physics
1 answer:
pashok25 [27]2 years ago
4 0

The altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg

<h3>How to determine the altimeter reading?</h3>

The given parameters are:

  • Ground level = 700 ft i.e. the field elevation
  • Pressure altitude = 1450 ft

The formula to calculate the pressure altitude is:

Altitude = (29.92 – Altimeter reading) * 1,000 + Ground level

Substitute the known values

1450 = (29.92 - Altimeter reading) * 1000 + 700

Evaluate the like terms

750 = (29.92 - Altimeter reading) * 1000

Divide both sides by 1000

0.75 = 29.92 - Altimeter reading

Evaluate the like terms

Altimeter reading = 29.17

Hence, the altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg

Read more about pressure at:

brainly.com/question/26040104

#SPJ1

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U = 3, 9 , v = 4, 2 (a) find the projection of u onto v. (b) find the vector component of u orthogonal to v.
ExtremeBDS [4]

Answer:

(a) At U = 3, 9 , v = 4, 2, the projection of u onto v is w1=<2,8>

(b)At U = 3, 9 , v = 4, 2,  the vector component of u orthogonal to v is w2 = <4,-1>

Explanation:

A

The projection of u onto v is given by:

w1= projvu = (u⋅v||v||2)v

Given that  u= <6,7> and v=<1,4>, we can find the projection of u onto v as shown below:

w1= projvu = (u⋅v||v||2v=(<6,7>⋅<1,4><1,4>⋅<1,)

=(6⋅1+7⋅41⋅1+4⋅4)<1,4>

=3417<1,4>

=<2,8>

Part B

The vector component of u orthogonal to v is given by:

Using the given vectors and the projection found in part (a), we can find the vector component of u orthogonal to v as shown below:

w2=u−projvu

=<6,7≻<2,8>

=<(6−2),(7−8)>

=<4,−1>

To learn more about vector component, click brainly.com/question/17016695

#SPJ4

5 0
2 years ago
If you want to increase the gravitational force between two objects; would you rather double the mass of one object or halve the
Step2247 [10]

To solve this problem we will apply the concepts related to the Force of gravity from Newtonian theory for which it is necessary to

F = \frac{GMm}{R^2}

Where,

G = Gravitational universal constant

M = Mass of Earth

m = mass of Object

x = Distance between center of mass of the objects.

From this equation we can observe that the Force is inversely proportional to the squared distance between the two objects. The greater the distance, the lower the force of gravity and vice versa.

F \propto \frac{1}{R^2}

If you want to increase the force of gravity, you need to reduce the distance of the two. Therefore the correct option is B. Talk to the distance between them.

5 0
3 years ago
If the car has a mass of 1000 kg what is its momentum
Mars2501 [29]
Its momentum is (1000) x (its speed in m/s).
3 0
3 years ago
A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep
salantis [7]
The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force F_f which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,
F-F_f=0
so
F=F_f

The frictional force can be rewritten as
F_f = \mu m g
where m=50 kg, g=9.81 m/s^2. Re-arranging, we can solve this equation to find \mu, the coefficient of dynamic friction:
\mu =  \frac{F}{mg}= \frac{99 N}{(50 kg)(9.81 m/s^2)}  =0.20
4 0
3 years ago
Describe three events that you cannot explain.<br> About energy and matter
ZanzabumX [31]

Answer:

A teacher giving homework on fridays

A teacher giving tests on fridays

A teacher holding you back after class on fridays

Explanation:

because i luv friday

5 0
3 years ago
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