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tatiyna
3 years ago
10

How can I protect my computer from being shot by an MBT

Physics
1 answer:
Pie3 years ago
4 0

Answer: While Technology is Evolving so is Hackers that being said "Norton Vpn" would be beneficial for computer protection.

Explanation:

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Diano4ka-milaya [45]
If it i hot it makes it easy to clean
4 0
3 years ago
Can someone check my answer please? I think that it is A, but I am not sure.
Yuliya22 [10]
Im pretty sure its chemical energy
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4 years ago
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An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
just olya [345]

Answer:

2.78\times 10^{-35}\ \text{kg m/s}

6.178\times 10^{-34}\ \text{m/s}

0.31\times 10^{-4}\ \text{m/s}

Explanation:

\Delta x = Uncertainty in position = 1.9 m

\Delta p = Uncertainty in momentum

h = Planck's constant = 6.626\times 10^{-34}\ \text{Js}

m = Mass of object

From Heisenberg's uncertainty principle we know

\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}

The minimum uncertainty in the momentum of the object is 2.78\times 10^{-35}\ \text{kg m/s}

Golf ball minimum uncertainty in the momentum of the object

m=0.045\ \text{kg}

Uncertainty in velocity is given by

\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}

The minimum uncertainty in the object's velocity is 6.178\times 10^{-34}\ \text{m/s}

Electron

m=9.11\times 10^{-31}\ \text{kg}

\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}

The minimum uncertainty in the object's velocity is 0.31\times 10^{-4}\ \text{m/s}.

6 0
3 years ago
At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until
Rama09 [41]

Answer:

(A) 570 rad

(B) 10 s

(C) 12.5 rad/s²

Explanation:

The equations of motion for circular motions are used.

  • Initial angular velocity,  \omega_0 = 30.0 \text{ rad/s}
  • Angular acceleration, \alpha =35.0 \text{ rad/s}^2

(A)

At <em>t</em> = 2.00 s, the angular displacement, <em>θ</em>, is given by

\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

\omega = \omega_0+\alpha  t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t

Here, \theta_2=440 (the angular displacement during deceleration)

The subscripts, <em>i</em> and <em>f</em>, on <em>ω</em> denote the initial and final angular velocities during deceleration.

\omega_i = 100

\omega_f = 0

t = \dfrac{2\theta_2}{\omega_i} = \dfrac{2\times400}{100} = 8\ \text{s}

This is the time for deceleration. The deceleration began at <em>t</em> = 2 s.

Hence, the wheel stops at <em>t</em> = 2 + 8 = 10 s.

(C)

The deceleration is given by

\alpha_R = \dfrac{\omega_f-\omega_i}{t} = \dfrac{0-100}{8} = -12.5\text{ rad/s}^2

The negative sign appears because it is a deceleration.

4 0
3 years ago
75 pts!
jasenka [17]

again CORRECT ans is 4) light strikes the grooves → different wavelengths of light bend at different angles → diffracted wavelengths reach the eyes → the eyes see different colors.

moderator - plz review the ans as u deleted my right ans n approved the wrong ans :(

6 0
3 years ago
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