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klasskru [66]
3 years ago
7

Find the magnitude of vector A = i - 2j + 3k O V14 10 O4

Physics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:

|A|=\sqrt{1^2+(-2)^2+(3)^2}=3.74

Explanation:

Given that,

Vector A=i-2j+3k............(1)

We need to find the magnitude of vector A. Let us suppose vector in the form of, R = xi + yj + zk. Its magnitude is given by :

|R|=\sqrt{x^2+y^2+z^2}

In equation (1),

x = 1

y = -2

z = 3

So, the magnitude of vector A is given by :

|A|=\sqrt{1^2+(-2)^2+(3)^2}

|A| = 3.74

So, the magnitude of given vector is 3.74. Hence, this is the required solution.

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A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

\omega= Angular velocity

Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

\Omega = 1.17rad/s

The precession frequency is

\Omega = \frac{2\pi rad}{T}

T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

Therefore the precession period is 5.4s

7 0
3 years ago
Which term describes the quantity of energy transferred from a warmer object to a cooler object?
Mars2501 [29]

Answer: B

Explanation:

6 0
3 years ago
A bicycle rider traveling east at 10 km/hr sees a blue car pass her appearing to travel west at 50 km/hr. What is the blue car's
BigorU [14]

Answer:

v_{bR} = 25 km/h towards west

Explanation:

As we know that the speed of the blue car as appear to the bicycle rider is given as

v_{bc} = 50 km/h towards west

also it is given that bicycle is moving at speed of 10 km/h towards East

so here we have

v_{bc} = v_b - v_c

so we have

v_b = v_{bc} + v_c

v_b = -50 + 10 = 40 km/h towards west

now speed of the red car is given as 15 km/h towards west

so here the relative speed of blue car with respect to red car is given as

v_{bR} = v_b - v_R

v_{bR} = 40 - 15 = 25 km/h towards west

8 0
3 years ago
When is the absorption spectrum of an electron generated
serious [3.7K]
A
Excitation to a higher energy state requires energy which is absorbed from the electromagnetic waves applied.
7 0
3 years ago
Read 2 more answers
A hiker starts at point P and walks 2.0 km due east and then walks at an angle of 30 degrees north of east for 1.4 km.
olchik [2.2K]

Answer:

The magnitude of the hiker’s displacement is 2.96 km

Explanation:

Let the initial displacement of the hiker, = x = 2km

the final displacement of the hiker, = y = 1.4 km

The resultant of the two vectors, According to Pythagorean theorem is the vector sum of the two vectors.

R' = x' + y'

Check the image uploaded for solution;

8 0
3 years ago
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