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Masteriza [31]
3 years ago
5

What is the maximum number of primary partitions that gpt supports?

Physics
1 answer:
Vikki [24]3 years ago
4 0
It supports 128 primary partitions.
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___________is the tendency of an object to resist change in its state of motion
Mademuasel [1]
Inertia
<span> An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.this tendency to retain its motion is referred to as Inertia </span>
5 0
3 years ago
A block of mass = 4.00 kg is supported by a spring scale with unit of measure of Newtons. This spring scale is attached to top o
faust18 [17]

Answer:

118 N

Explanation:

Given mass of the block, m = 4.00kg.

The acceleration of the elevator, a = 3.0 m/s^2.

As elevotar attaced with spring scale and accelerating upward

(block and elevator), so total force

F_N-mg=ma

Here, mg is the weight of the block downward direction.

or

F_N=ma+mg=m(g+a)

substitute the given value, we get

F_N=4kg(9.8m/s^2+3m/s^2)

     = 117.6 N = 118 N.

Thus, the reading on the spring scale to 3 significant figures is 118 N.

5 0
3 years ago
A man 6 feet tall walks at a rate of 6 feet per second away from a light that is 15 feet above the ground.
Tems11 [23]

Answer:(a)10 ft/s

(b)4 ft/s

Explanation:

Given

height of light =15 feet

height of man=6 feet

\frac{\mathrm{d} x}{\mathrm{d} t}=6 ft/s

From diagram

\frac{15}{y}=\frac{6}{y-x}

5(y-x)=2y

3y=5x

differentiate both sides

3\times \frac{\mathrm{d} y}{\mathrm{d} t}=5\times \frac{\mathrm{d} x}{\mathrm{d} t}

Tip of shadow is moving at the rate of

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{5}{3}\times 6=10 ft/s

(b)rate at which length of his shadow  is changing

Length of shadow is y-x

differentiating w.r.t time

\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} t}-\frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=10-6=4 ft/s

7 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
While standing atop a building 49.6 m tall, you spot a friend standing on a street corner. Using a protractor and a dangling plu
olga nikolaevna [1]

Answer:

75degree don't forget wind and gravity force pulling down

6 0
2 years ago
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