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Korolek [52]
3 years ago
5

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the plane were to fly in the same 500 m circl

e at a speed of 300 m/s, by what factor would its centripetal acceleration change
Physics
1 answer:
elena-s [515]3 years ago
4 0

Answer:

By Factor of 4

Explanation:

The expression for centripetal acceleration is given as,

a = v²/r........... equation 1

Where a = centripetal acceleration, v = speed, r = radius of the circle.

Case 1

Given: v = 150 m/s, r = 500 m

Substitute into equation 1

a = 150²/500

a = 22500/500

a = 45 m/s².

Case 2

Given: v = 300 m/s, r = 500 m

Also substitute into equation 1

a = 300²/500

a = 180 m/s²

Factor of change of the acceleration = 180/45

Factor of change of the acceleration = 4.

Hence the acceleration will change by a factor of 4

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Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B​
WARRIOR [948]

\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}

Explanation:

Given:

\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}

\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}

The cross product \textbf{A}×\textbf{B} is given by

\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|

=  \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}

= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}

5 0
3 years ago
Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that ap
eduard

Answer:

The object can have zero velocity and, simultaneously, nonzero acceleration.

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

Explanation:

An object in simple harmonic motion has a total mechanical energy (sum of elastic potential energy and kinetic energy) that is constant:

E=U+K=1/2kx^2 + 1/2}mv^2

where,

k is equal to the spring constant

x is equal to the displacement

m is the mass

v is the speed

We can note that the force on the spring is given by Hook's law:

F=-kx

In Newton's law F = ma, this can be also be written as

ma=-kx

a=-k/mx

This implies that the acceleration is proportional to the displacement.

From the first equation, we can now states that:

When the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum

When the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

In all the other intermediate situations, both velocity and acceleration are nonzero.

So the correct answers are

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

The object can have zero velocity and, simultaneously, nonzero acceleration.

4 0
3 years ago
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I hope the answer will help you. 
6 0
3 years ago
) Un círculo de 120 cm de radio gira a 600 rpm. Calcula: a) su velocidad angular
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Responder:

20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz

Explicación:

Dado lo siguiente:

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T = 2π / w = 2π / 20π = 0.1 seg

D.) Frecuencia (f):

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Answer:

B. and D. would be my best guess.

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