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4 years ago

A computer technician always touches the metal body of a computer before touching any of its electrical parts. why?

Physics

1 answer:

Nikolay [14]4 years ago

6 0

Answer:

Computer technician always touch the metal body parts of the computer before touch any electrical part of the computer because it will neutralized the body static charge.

The components of the computer are made up of the materials with the positive and negative charge electron and their electric effect basically neutralized the one another.

while touching any electrical hardware make sure that the electricity is off and discharge our body from any static charge. Also avoid wearing wool clothes when touching any electrical hardware parts as they also build static charge.

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A right triangle has sides measuring 5 and 12 inches. If the two vectors have a magnitude of 5 and 12 and are at right angles to

ollegr [7]

Answer:

$x=13$

Explanation:

From the question we are told that

Magnitude 1 $M_1=5$

Magnitude 1 $M_2=12$

Generally the Pythagoras equation for the magnitudes is mathematically given as

$\sqrt{x^2}=\sqrt{M_1^2 +M_2^2}$

$\sqrt{x^2}=\sqrt{5^2 +12^2}$

$x=13$

Therefore resultant magnitude is

$x=13$

8 0

3 years ago

A spring-loaded launcher has a mass of 0.60 kg and is placed on a platform 1.2m above the ground. The force of friction is negl

kobusy [5.1K]

Answer:

C The launcher will fall off the platform and land D/2 to the left of the platform because the launcher is twice the mass of the ball.

Explanation:

The figure is missing: you can find it in attachment.

We can apply the law of conservation of momentum to check that the launcher will leave the platform with a speed which is half the speed of the ball. In fact, the total initial momentum is zero:

$p_i = 0$

while the total final momentum is:

$p_f = m_l v_l + m_b v_b$

where

$m_l = 0.60 kg$ is the mass of the launcher

$m_b = 0.30 kg$ is the mass of the ball

$v_l$ is the velocity of the launcher

$v_b$ is the velocity of the ball

Since the total momentum must be conserved, $p_i=p_f$ , so

$0=m_l v_l + m_b v_b$

Therefore we find

$v_l = - \frac{m_b}{m_l}v_b = -\frac{0.30}{0.60}v_b = -\frac{v_b}{2}$

which means that the launcher leaves the platform with a velocity which is half that of the ball, and in the opposite direction (to the left).

Since the distance covered by both the ball and the launcher only depends on their horizontal velocity, this also means that the launcher will cover half the distance covered by the ball before reaching the ground: therefore, since the ball covers a distance of D, the launcher will cover a distance of D/2.

7 0

4 years ago

Read 2 more answers

Three metal fishing weights, each with a mass of 1.00x102 g and at a temperature of 100.0°C, are placed in 1.00x102 g of water a

worty [1.4K]

Answer:

Approximately $0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ assuming no heat exchange between the mixture and the surroundings.

Explanation:

Consider an object of specific heat capacity $c$ and mass $m$ . Increasing the temperature of this object by $\Delta T$ would require $Q = c\, m \, \Delta T$ .

Look up the specific heat of water: $c(\text{water}) = 4.182\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ .

It is given that the mass of the water in this mixture is $m(\text{water}) = 1.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the water: $\Delta T(\text{water}) = (45 - 35)\; {\rm K} = 10\; {\rm K}$ .

Thus, the water in this mixture would have absorbed :

$\begin{aligned}Q &= c\, m\, \Delta T \\ &= 4.182\; {\rm J \cdot g^{-1}\cdot K^{-1}} \\ &\quad \times 1.00 \times 10^{2}\; {\rm g} \times 10\; {\rm K} \\ &= 4.182 \times 10^{3}\; {\rm J}\end{aligned}$ .

Thus, the energy that water absorbed was: $Q(\text{water}) = 4.182 \times 10^{3}\; {\rm J}$ .

Assuming that there was no heat exchange between the mixture and its surroundings. The energy that the water in this mixture absorbed, $Q(\text{water})$ , would be the opposite of the energy that the metal in this mixture released.

Thus: $Q(\text{metal}) = -Q(\text{water}) = -4.182 \times 10^{3}\; {\rm J}$ (negative because the metal in this mixture released energy rather than absorbing energy.)

Mass of the metal in this mixture: $m(\text{metal}) = 3 \times 1.00 \times 10^{2}\; {\rm g} = 3.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the metal in this mixture: $\Delta T(\text{metal}) = (100 - 45)\; {\rm K} = 55\; {\rm K}$ .

Rearrange the equation $Q = c\, m \, \Delta T$ to obtain an expression for the specific heat capacity: $c = Q / (m\, \Delta T)$ . The (average) specific heat capacity of the metal pieces in this mixture would be:

$\begin{aligned}c &= \frac{Q}{m\, \Delta T} \\ &= \frac{-4.182 \times 10^{3}\; {\rm J}}{3.00 \times 10^{2}\; {\rm g} \times (-55\; {\rm K})} \\ &\approx 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}\end{aligned}$ .

6 0

3 years ago

If a train going 60 m/s hits the brakes, and it takes the train 85 seconds to stop, what is

Anvisha [2.4K]

Question :-

If a Train going with 60 m/s Speed hits the Brakes, and takes 85 sec to stop. What is the Acceleration of the Train ?

Answer :-

Acceleration of Train is 0.70 m/s² .

Explanation :-

As per the provided information in the given question, we have been given that the Speed of the Train is 60 m/s . Time taken to stop the Train is 85 sec . And, we have been asked to calculate the Acceleration of the Train .