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Brut [27]
3 years ago
11

What is malware short for?

Physics
1 answer:
yaroslaw [1]3 years ago
8 0
The correct answer is C
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How long should you hold a stretch?<br> 10-30 seconds<br> 90 seconds<br> 5 minutes
galina1969 [7]

I think you should hold a stretch for 10-30 seconds

5 0
3 years ago
Read 2 more answers
If the room radius is 4.5 m, and the rotation frequency is 0.8 revolutions per second when the floor drops out, what is the mini
kondaur [170]
<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration. The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N. a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r. The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec. So a_N = 114 m/sec^2. g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>
3 0
3 years ago
A bus driver heads south with a steady speed of
Nutka1998 [239]

ANS

{ A }

Magnitude

5400.097 \: ✓ \: m

Direction

25.52 \:  ✓ \: ° South  \: of \:  West

{ B }

Average speed in ( m/s )

23.43 \:  ✓  \: m/s

{ C }

Magnitude

14.06 \:  ✓  \: m/s

Direction

25.52 \:  ✓  \: ° South  \: of \:  West

5 0
3 years ago
Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0
3 years ago
. A stationary mass explodes into two parts of mass 4 kg and 40 kg . If the K.E of larger mass is 100 J . The K.E of small mass
alisha [4.7K]

Answer:

10J

Explanation:

KE = (1/2)mv²

100J = (.5)(40kg)v²

v²=(100J)/(20kg)

v²= 5

KE = 5(.5)(4kg)

KE = 10J

3 0
3 years ago
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