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3 years ago

What's the definition of altitude?

Physics

2 answers:

OleMash [197]3 years ago

6 0

Altitude is the height of something or the point of sea level or ground level.
hope i helped!!

marysya [2.9K]3 years ago

3 0

the height of an object or point in relation to sea level or ground level."flight data including airspeed and altitude"synonyms: height, elevation, distance above the sea/ground; loftiness "clouds are classified according to form and altitude"great height.
or the mechanism can freeze at altitude

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A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele

Fantom [35]

The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, $\Delta t$ :
$I= \frac{Q}{\Delta t}$
First we need to find the net charge flowing at a certain point of the wire in one second, $\Delta t=1.0 s$ . Using I=0.92 A and re-arranging the previous equation, we find
$Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C$

Now we know that each electron carries a charge of $e=1.6 \cdot 10^{-19} C$ , so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
$N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}$

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3 years ago

What is the wave speed of a 6 m wave with a frequency of 20 Hz?

andrew-mc [135]

Answer: Wave speed= frequency x wavelength

=20 x 3

=60 m/s

Explanation:

8 0

3 years ago

An echo is heard from a building 0.4 s after you shout "hello." How many feet away is

UNO [17]

Answer:

Circular motion: find period, find radius, find velocity, find centripetal acceleration 27 V= T a =vºlr=rw

Explanation:

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3 years ago

A potter’s wheel moves from rest to an angular speed of 0.50 rev/s in 28.9 s. Assuming constant angular acceleration, what is it

Fudgin [204]

Answer:

0.108 rad/s².

Explanation:

Given that

Initial angular velocity ,ωi = 0 rad/s

Final angular velocity ωf= 0.5 rev/s

We know that

1 rev/s = 6.28 rad/s

ωf= 3.14 rad/s

t= 28.9 s

We know that (if acceleration is constant)

ωf=ωi + α t

α=Angular acceleration

3.14 = 0 + α x 28.9

$\alpha=\dfrac{3.14}{28.9}\ rad/s^2\\\alpha=0.108\ rad/s^2$

Therefore the acceleration will be 0.108 rad/s².

Therefore the answer will be 0.108 rad/s².

5 0

3 years ago

Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the c

Amanda [17]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is $v_A= 4m/s$

Explanation:

From the question we are told that

The length of the string is $L = 1m$

The initial speed of block A is $u_A$

The final speed of block A is $v_A = \frac{1}{2}u_A$

The initial speed of block B is $u_B = 0$

The mass of block A is $m_A = 7kg$ gh

The mass of block B is $m_B = 2 kg$

According to the principle of conservation of momentum

$m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}$

Since block B at initial is at rest

$m_A u_A = m_Bv_B + m_A \frac{u_A}{2}$

$m_A u_A - m_A \frac{u_A}{2} = m_Bv_B$

$m_A \frac{u_A}{2} = m_Bv_B$

making $v_B$ the subject of the formula

$v_B =m_A \frac{u_A}{2 m_B}$

Substituting values

$v_B =\frac{7 u_A}{4}$

This $v__B$ is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of $v__B'$ at the top of the vertical circle

The angular centripetal acceleration would be mathematically represented

$a= \frac{v^2_{B}'}{L}$

Note that this acceleration would be toward the center of the circle

Now the forces acting at the top of the circle can be represented mathematically as

$T + mg = m \frac{v^2_{B}'}{L}$

Where T is the tension on the string

According to the law of energy conservation

The energy at bottom of the vertical circle = The energy at the top of

the vertical circle

This can be mathematically represented as

$\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L$

From above

$(T + mg) L = m v^2_{B}'$

Substitute this into above equation

$\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = T + 5mgL$

The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero

This is mathematically represented as

$\frac{49 mv_A^2}{16} = 5mgL$

making $v_A$ the subject

$v_A = \sqrt{\frac{80mgL}{49m} }$

substituting values

$v_A = \sqrt{\frac{80* 9.8 *1}{49} }$

$v_A= 4m/s$

6 0

3 years ago